Robert Šámal, A. Montejano, Sebastián González Hermosillo de la Maza, Matt DeVos, R. Aharoni
{"title":"A rainbow version of Mantel's Theorem","authors":"Robert Šámal, A. Montejano, Sebastián González Hermosillo de la Maza, Matt DeVos, R. Aharoni","doi":"10.19086/aic.12043","DOIUrl":null,"url":null,"abstract":"Mantel's Theorem from 1907 is one of the oldest results in graph theory: every simple $n$-vertex graph with more than $\\frac{1}{4}n^2$ edges contains a triangle. The theorem has been generalized in many different ways, including other subgraphs, minimum degree conditions, etc. This article deals with a generalization to edge-colored multigraphs, which can be viewed as a union of simple graphs, each corresponding to an edge-color class.\n\nThe case of two colors is the same as the original setting: Diwan and Mubayi proved that any two graphs $G_1$ and $G_2$ on the same set of $n$ vertices, each containing more than $\\frac{1}{4}n^2$ edges, give rise to a triangle with one edge from $G_1$ and two edges from $G_2$. The situation is however different for three colors. Fix $\\tau=\\frac{4-\\sqrt{7}}{9}$ and split the $n$ vertices into three sets $A$, $B$ and $C$, such that $|B|=|C|=\\lfloor\\tau n\\rfloor$ and $|A|=n-|B|-|C|$. The graph $G_1$ contains all edges inside $A$ and inside $B$, the graph $G_2$ contains all edges inside $A$ and inside $C$, and the graph $G_3$ contains all edges between $A$ and $B\\cup C$ and inside $B\\cup C$. It is easy to check that there is no triangle with one edge from $G_1$, one from $G_2$ and one from $G_3$; each of the graphs has about $\\frac{1+\\tau^2}{4}n^2=\\frac{26-2\\sqrt{7}}{81}n^2\\approx 0.25566n^2$ edges. The main result of the article is that this construction is optimal: any three graphs $G_1$, $G_2$ and $G_3$ on the same set of $n$ vertices, each containing more than $\\frac{1+\\tau^2}{4}n^2$ edges, give rise to a triangle with one edge from each of the graphs $G_1$, $G_2$ and $G_3$. A computer-assisted proof of the same bound has been found by Culver, Lidický, Pfender and Volec.","PeriodicalId":36338,"journal":{"name":"Advances in Combinatorics","volume":" ","pages":""},"PeriodicalIF":0.0000,"publicationDate":"2018-12-31","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"37","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Advances in Combinatorics","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.19086/aic.12043","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q2","JCRName":"Mathematics","Score":null,"Total":0}
引用次数: 37
Abstract
Mantel's Theorem from 1907 is one of the oldest results in graph theory: every simple $n$-vertex graph with more than $\frac{1}{4}n^2$ edges contains a triangle. The theorem has been generalized in many different ways, including other subgraphs, minimum degree conditions, etc. This article deals with a generalization to edge-colored multigraphs, which can be viewed as a union of simple graphs, each corresponding to an edge-color class.
The case of two colors is the same as the original setting: Diwan and Mubayi proved that any two graphs $G_1$ and $G_2$ on the same set of $n$ vertices, each containing more than $\frac{1}{4}n^2$ edges, give rise to a triangle with one edge from $G_1$ and two edges from $G_2$. The situation is however different for three colors. Fix $\tau=\frac{4-\sqrt{7}}{9}$ and split the $n$ vertices into three sets $A$, $B$ and $C$, such that $|B|=|C|=\lfloor\tau n\rfloor$ and $|A|=n-|B|-|C|$. The graph $G_1$ contains all edges inside $A$ and inside $B$, the graph $G_2$ contains all edges inside $A$ and inside $C$, and the graph $G_3$ contains all edges between $A$ and $B\cup C$ and inside $B\cup C$. It is easy to check that there is no triangle with one edge from $G_1$, one from $G_2$ and one from $G_3$; each of the graphs has about $\frac{1+\tau^2}{4}n^2=\frac{26-2\sqrt{7}}{81}n^2\approx 0.25566n^2$ edges. The main result of the article is that this construction is optimal: any three graphs $G_1$, $G_2$ and $G_3$ on the same set of $n$ vertices, each containing more than $\frac{1+\tau^2}{4}n^2$ edges, give rise to a triangle with one edge from each of the graphs $G_1$, $G_2$ and $G_3$. A computer-assisted proof of the same bound has been found by Culver, Lidický, Pfender and Volec.