Abstract A graph is said to be ‐free if it does not contain any subdivision of as an induced subgraph. Lévêque, Maffray and Trotignon conjectured that every ‐free graph is 4‐colorable. In this paper, we show that this conjecture is true for the class of {, diamond, bowtie}‐free graphs, where a diamond is the graph obtained from by removing one edge and a bowtie is the graph consisting of two triangles with one vertex identified.
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Three coloring via triangle counting
In the first partial result toward Steinberg’s now-disproved three coloring conjecture, Abbott and Zhou used a counting argument to show that every planar graph without cycles of lengths 4 through 11 is 3-colorable. Implicit in their proof is a fact about plane graphs: in any plane graph of minimum degree 3, if no two triangles share an edge, then triangles make up strictly fewer than 2/3 of the faces. We show how this result, combined with Kostochka and Yancey’s resolution of Ore’s conjecture for k = 4, implies that every planar graph without cycles of lengths 4 through 8 is 3-colorable.
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- Award ID(s):
- 2054423
- PAR ID:
- 10511020
- Editor(s):
- Albert, Michael; Billington, Elizabeth J
- Publisher / Repository:
- University of Queensland
- Date Published:
- Journal Name:
- The Australasian journal of combinatorics
- Volume:
- 87
- Issue:
- 2
- ISSN:
- 2202-3518
- Page Range / eLocation ID:
- 352-356
- Format(s):
- Medium: X
- Sponsoring Org:
- National Science Foundation
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