Mantel定理的彩虹版本

Q2 Mathematics
Robert Šámal, A. Montejano, Sebastián González Hermosillo de la Maza, Matt DeVos, R. Aharoni
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引用次数: 37

摘要

1907年的曼特尔定理是图论中最古老的结果之一:每一个简单的$n$-顶点图都有$\frac{1}以上{4}n^2$边包含一个三角形。该定理已经以许多不同的方式得到了推广,包括其他子图、最小度条件等。本文讨论了边色多重图的推广,它可以被视为简单图的并集,每个图都对应于一个边色类。两种颜色的情况与原始设置相同:Diwan和Mubayi证明了在同一组$n$顶点上的任何两个图$G_1$和$G_2$,每个顶点都包含超过$\frac{1}{4}n^2$边,产生一个三角形,其中一条边从$G_1$开始,两条边从$G_2$开始。然而,三种颜色的情况不同。修复$\tau=\frac{4-\sqrt{7}}{9}$并将$n$顶点拆分为三个集合$A$、$B$和$C$,使得$|B|=|C|=\lfloor\taun\rfloor$和$|A|=n-|B|-|C|$。图$G_1$包含$A$内部和$B$内部的所有边,图$G_2$包含$A+内部和$C$内部的全部边,而图$G_3$包含在$A$和$B\cup C$之间和$B\cup C$内的全部边。很容易检查是否不存在一个边从$G_1$、一个从$G_2$和一个从$G_3$的三角形;每个图都有大约$\frac{1+\tau^2}{4}n^2=\frac{26-2\sqrt{7}}{81}n^2\约0.25566n^2$边缘。本文的主要结果是这种构造是最优的:在同一组$n$顶点上的任意三个图$G_1$、$G_2$和$G_3$,每个图都包含超过$\frac{1+\tau^2}{4}n^2$边,从图$G_1$、$G_2$和$G_3$中的每一个产生具有一条边的三角形。卡尔弗(Culver)、利迪克(Lidický)、普芬德(Pfender)和沃尔克(Volec)发现了计算机辅助的同一界的证明。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
A rainbow version of Mantel's Theorem
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.
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来源期刊
Advances in Combinatorics
Advances in Combinatorics Mathematics-Discrete Mathematics and Combinatorics
CiteScore
3.10
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7
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