Exhaustive generation of edge-girth-regular graphs

TL;DR

This paper introduces algorithms for exhaustively generating edge-girth-regular graphs, enabling the determination of minimal vertex counts and providing new bounds and counterexamples for specific parameters.

Contribution

The authors develop a linear time algorithm for cycle containment counting and an exhaustive generation method for edge-girth-regular graphs, advancing the understanding of their minimal sizes.

Findings

Determined exact values of n(k,g,λ) for several parameters.

Disproved a conjecture related to cubic girth 8 and 12 cases.

Provided new bounds and extremal graphs for edge-girth-regular graphs.

Abstract

Edge-girth-regular graphs (abbreviated as $egr$ graphs) are a class of highly regular graphs. More specifically, for integers $v$, $k$, $g$ and $\lambda$ an $egr(v,k,g,\lambda)$ graph is a $k$-regular graph with girth $g$ on $v$ vertices such that every edge is contained in exactly $\lambda$ cycles of length $g$. The central problem in this paper is determining $n(k,g,\lambda)$, which is defined as the smallest integer $v$ such that an $egr(v,k,g,\lambda)$ 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 $egr(v,k,g,\lambda)$ graphs for fixed parameters $v, k, g$ and $\lambda$. 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 $n(k,g,\lambda)$. 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 $n(3,7,6)=n(3,8,10)=n(3,8,12)=n(3,8,14)=\infty.$