2017
DOI: 10.1007/s00493-017-3553-8
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Three-Coloring and List Three-Coloring of Graphs Without Induced Paths on Seven Vertices

Abstract: We present an algorithm to 3-colour a graph G without triangles or induced paths on seven vertices in O(|V (G)| 7 ) time. In fact, our algorithm solves the list 3-colouring problem, where each vertex is assigned a subset of {1, 2, 3} as its admissible colours.

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Cited by 81 publications
(143 citation statements)
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“…Their proof can be used to show that List 3 ‐Colouring is polynomial‐time solvable on (P2+P4)‐free graphs. Recently, Bonomo, Chudnovsky, Maceli, Schaudt, Stein and Zhong gave an O(|Vfalse|23|) time algorithm for List 3‐Colouring on P 7 ‐free graphs (thereby solving Problem 17 in and Problem 56 in ). The same authors showed that 3‐Colouring can be solved in O(|Vfalse|7|) time on (C3,P7)‐free graphs. …”
Section: Results and Open Problems For H‐free Graphsmentioning
confidence: 99%
“…Their proof can be used to show that List 3 ‐Colouring is polynomial‐time solvable on (P2+P4)‐free graphs. Recently, Bonomo, Chudnovsky, Maceli, Schaudt, Stein and Zhong gave an O(|Vfalse|23|) time algorithm for List 3‐Colouring on P 7 ‐free graphs (thereby solving Problem 17 in and Problem 56 in ). The same authors showed that 3‐Colouring can be solved in O(|Vfalse|7|) time on (C3,P7)‐free graphs. …”
Section: Results and Open Problems For H‐free Graphsmentioning
confidence: 99%
“…This proves (5.2). 2 is an induced P 6 in G, a contradiction. This shows that y 1 ∈ Z k and k ∈ {i, j}, or that y 2 ∈ Z and ∈ {i, j}.…”
Section: Contradiction This Proves (422) This Proves (4)mentioning
confidence: 77%
“…Thus we may assume that x is nonadjacent to v 2 , and therefore we obtain outcome (2.1) for i = 0. This proves (2 …”
Section: Lemmamentioning
confidence: 81%
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“…As coloring is polynomial‐time solvable for (P5,trueP5¯)‐free graphs , it would be interesting to solve the following two open problems (see also Table ): is there an integer k such that k‐ coloring for (P6,trueP6¯)‐free graphs is sans-serif𝖭𝖯‐complete? is there an integer k such that k‐ coloring for (P7,trueP7¯)‐free graphs is sans-serif𝖭𝖯‐complete? As 4‐ coloring is polynomial‐time solvable for P6‐free graphs , k must be at least 5 in an affirmative answer for the first question. As 3‐ coloring is polynomial‐time solvable for P7‐free graphs , k must be at least 4 in an affirmative answer for the second question. We refer to the end of Section for some further remarks on these two open problems.…”
Section: Introductionmentioning
confidence: 99%