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arXiv 2401.08271math.CO
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边-围长正则图的穷举生成

Exhaustive generation of edge-girth-regular graphs

Jan Goedgebeur, Jorik Jooken

AI总结:

本文研究边-围长正则图的最小阶数$n(k,g,λ)$,提出线性时间计数算法和穷举生成算法,通过大规模计算得到多个新的极值图并改进上下界,同时否证了相关猜想。

摘要:

边-围长正则图(简称$egr$图)是一类高度正则的图。具体而言,对于整数$v$、$k$、$g$和$λ$,一个$egr(v,k,g,λ)$图是一个具有$v$个顶点、围长$g$的$k$-正则图,使得每条边恰好包含在$λ$个长度为$g$的圈中。本文的核心问题是确定$n(k,g,λ)$,即存在$egr(v,k,g,λ)$图的最小整数$v$(若不存在这样的图则为$\infty$),以及确定相应的极值图。我们提出了一种线性时间算法,用于在给定围长$g$的图中计算一条边被包含在长度为$g$的圈中的次数。我们以此作为构建块之一,提出了另一种算法,能够针对固定参数$v, k, g$和$λ$穷举生成所有$egr(v,k,g,λ)$图。我们实现了该算法,并将其用于大规模计算,获得了若干新的极值图,并改进了文献中关于$n(k,g,λ)$的下界和上界。其中,我们证明了$n(3,6,2)=24$、$n(3,8,8)=40$、$n(3,9,6)=60$、$n(3,9,8)=60$、$n(4,5,1)=30$、$n(4,6,9)=35$、$n(6,5,20)=42$,并否证了Araujo-Pardo和Leemans在[Discrete Math. 345(10):112991 (2022)]中针对三次围长8和围长12情形提出的猜想。基于我们的计算,我们猜想$n(3,7,6)=n(3,8,10)=n(3,8,12)=n(3,8,14)=\infty$。

英文摘要:

Edge-girth-regular graphs (abbreviated as $egr$ graphs) are a class of highly regular graphs. More specifically, for integers $v$, $k$, $g$ and $λ$ an $egr(v,k,g,λ)$ graph is a $k$-regular graph with girth $g$ on $v$ vertices such that every edge is contained in exactly $λ$ cycles of length $g$. The central problem in this paper is determining $n(k,g,λ)$, which is defined as the smallest integer $v$ such that an $egr(v,k,g,λ)$ graph exists (or $\infty$ if no such graph exists) as well as determining the corresponding extremal graphs. We propose a linear time algorithm for computing how often an edge is contained in a cycle of length $g$, given a graph with girth $g$. We use this as one of the building blocks to propose another algorithm that can exhaustively generate all $egr(v,k,g,λ)$ graphs for fixed parameters $v, k, g$ and $λ$. We implement this algorithm and use it in a large-scale computation to obtain several new extremal graphs and improvements for lower and upper bounds from the literature for $n(k,g,λ)$. Among others, we show that $n(3,6,2)=24, n(3,8,8)=40, n(3,9,6)=60, n(3,9,8)=60, n(4,5,1)=30, n(4,6,9)=35, n(6,5,20)=42$ and we disprove a conjecture made by Araujo-Pardo and Leemans [Discrete Math. 345(10):112991 (2022)] for the cubic girth 8 and girth 12 cases. Based on our computations, we conjecture that $n(3,7,6)=n(3,8,10)=n(3,8,12)=n(3,8,14)=\infty.$

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