{"title":"无反完备环图中的诱导路径","authors":"Tung Nguyen , Alex Scott , Paul Seymour","doi":"10.1016/j.jctb.2023.10.003","DOIUrl":null,"url":null,"abstract":"<div><p>Let us say a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span><em>-free</em>, where <span><math><mi>s</mi><mo>≥</mo><mn>1</mn></math></span> is an integer, if there do not exist <em>s</em> cycles of the graph that are pairwise vertex-disjoint and have no edges joining them. The structure of such graphs, even when <span><math><mi>s</mi><mo>=</mo><mn>2</mn></math></span>, is not well understood. For instance, until now we did not know how to test whether a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mn>2</mn></mrow></msub></math></span>-free in polynomial time; and there was an open conjecture, due to Ngoc Khang Le, that <span><math><msub><mrow><mi>O</mi></mrow><mrow><mn>2</mn></mrow></msub></math></span>-free graphs have only a polynomial number of induced paths.</p><p>In this paper we prove Le's conjecture; indeed, we will show that for all <span><math><mi>s</mi><mo>≥</mo><mn>1</mn></math></span>, there exists <span><math><mi>c</mi><mo>></mo><mn>0</mn></math></span> such that every <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free graph <em>G</em> has at most <span><math><mo>|</mo><mi>G</mi><msup><mrow><mo>|</mo></mrow><mrow><mi>c</mi></mrow></msup></math></span> induced paths, where <span><math><mo>|</mo><mi>G</mi><mo>|</mo></math></span> is the number of vertices. This provides a poly-time algorithm to test if a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free, for all fixed <em>s</em>.</p><p>The proof has three parts. First, there is a short and beautiful proof, due to Le, that reduces the question to proving the same thing for graphs with no cycles of length four. Second, there is a recent result of Bonamy, Bonnet, Déprés, Esperet, Geniet, Hilaire, Thomassé and Wesolek, that in every <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free graph <em>G</em> with no cycle of length four, there is a set of vertices that intersects every cycle, with size logarithmic in <span><math><mo>|</mo><mi>G</mi><mo>|</mo></math></span>. And third, there is an argument that uses the result of Bonamy et al. to deduce the theorem. The last is the main content of this paper.</p></div>","PeriodicalId":1,"journal":{"name":"Accounts of Chemical Research","volume":null,"pages":null},"PeriodicalIF":16.4000,"publicationDate":"2023-10-20","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":"{\"title\":\"Induced paths in graphs without anticomplete cycles\",\"authors\":\"Tung Nguyen , Alex Scott , Paul Seymour\",\"doi\":\"10.1016/j.jctb.2023.10.003\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"<div><p>Let us say a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span><em>-free</em>, where <span><math><mi>s</mi><mo>≥</mo><mn>1</mn></math></span> is an integer, if there do not exist <em>s</em> cycles of the graph that are pairwise vertex-disjoint and have no edges joining them. The structure of such graphs, even when <span><math><mi>s</mi><mo>=</mo><mn>2</mn></math></span>, is not well understood. For instance, until now we did not know how to test whether a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mn>2</mn></mrow></msub></math></span>-free in polynomial time; and there was an open conjecture, due to Ngoc Khang Le, that <span><math><msub><mrow><mi>O</mi></mrow><mrow><mn>2</mn></mrow></msub></math></span>-free graphs have only a polynomial number of induced paths.</p><p>In this paper we prove Le's conjecture; indeed, we will show that for all <span><math><mi>s</mi><mo>≥</mo><mn>1</mn></math></span>, there exists <span><math><mi>c</mi><mo>></mo><mn>0</mn></math></span> such that every <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free graph <em>G</em> has at most <span><math><mo>|</mo><mi>G</mi><msup><mrow><mo>|</mo></mrow><mrow><mi>c</mi></mrow></msup></math></span> induced paths, where <span><math><mo>|</mo><mi>G</mi><mo>|</mo></math></span> is the number of vertices. This provides a poly-time algorithm to test if a graph is <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free, for all fixed <em>s</em>.</p><p>The proof has three parts. First, there is a short and beautiful proof, due to Le, that reduces the question to proving the same thing for graphs with no cycles of length four. Second, there is a recent result of Bonamy, Bonnet, Déprés, Esperet, Geniet, Hilaire, Thomassé and Wesolek, that in every <span><math><msub><mrow><mi>O</mi></mrow><mrow><mi>s</mi></mrow></msub></math></span>-free graph <em>G</em> with no cycle of length four, there is a set of vertices that intersects every cycle, with size logarithmic in <span><math><mo>|</mo><mi>G</mi><mo>|</mo></math></span>. And third, there is an argument that uses the result of Bonamy et al. to deduce the theorem. The last is the main content of this paper.</p></div>\",\"PeriodicalId\":1,\"journal\":{\"name\":\"Accounts of Chemical Research\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":16.4000,\"publicationDate\":\"2023-10-20\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"1\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"Accounts of Chemical Research\",\"FirstCategoryId\":\"100\",\"ListUrlMain\":\"https://www.sciencedirect.com/science/article/pii/S0095895623000850\",\"RegionNum\":1,\"RegionCategory\":\"化学\",\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"Q1\",\"JCRName\":\"CHEMISTRY, MULTIDISCIPLINARY\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"Accounts of Chemical Research","FirstCategoryId":"100","ListUrlMain":"https://www.sciencedirect.com/science/article/pii/S0095895623000850","RegionNum":1,"RegionCategory":"化学","ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q1","JCRName":"CHEMISTRY, MULTIDISCIPLINARY","Score":null,"Total":0}
Induced paths in graphs without anticomplete cycles
Let us say a graph is -free, where is an integer, if there do not exist s cycles of the graph that are pairwise vertex-disjoint and have no edges joining them. The structure of such graphs, even when , is not well understood. For instance, until now we did not know how to test whether a graph is -free in polynomial time; and there was an open conjecture, due to Ngoc Khang Le, that -free graphs have only a polynomial number of induced paths.
In this paper we prove Le's conjecture; indeed, we will show that for all , there exists such that every -free graph G has at most induced paths, where is the number of vertices. This provides a poly-time algorithm to test if a graph is -free, for all fixed s.
The proof has three parts. First, there is a short and beautiful proof, due to Le, that reduces the question to proving the same thing for graphs with no cycles of length four. Second, there is a recent result of Bonamy, Bonnet, Déprés, Esperet, Geniet, Hilaire, Thomassé and Wesolek, that in every -free graph G with no cycle of length four, there is a set of vertices that intersects every cycle, with size logarithmic in . And third, there is an argument that uses the result of Bonamy et al. to deduce the theorem. The last is the main content of this paper.
期刊介绍:
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