Twin-free $K_r$-saturated Graphs and Maximally Independent Sets in $K_3$-free Graphs

TL;DR

This paper investigates twin-free $K_r$-saturated graphs, establishing bounds on their minimum edges, and explores related problems on maximally independent sets in $K_3$-free graphs, revealing deep connections between these topics.

Contribution

It introduces bounds for twin-free $K_r$-saturated graphs and links these to problems on maximally independent sets in $K_3$-free graphs, providing new insights and bounds.

Findings

Bounds for $tsat(n,K_3)$: $(5 +2/3)n + o(n) ext{ to } 6n+o(n)$

Bounds for $tsat(n,K_r)$: $(r+2)n + o(n) ext{ to } (r+3)n+o(n)$ for $r \\geq 4$

Connections between saturated graphs and maximally independent sets in $K_3$-free graphs.

Abstract

We say that two vertices are twins if they have the same neighbourhood and that a graph is $K_r$-saturated if it does not contain $K_r$ but adding any new edge to it creates a $K_r$. In 1964, Erd\H{o}s, Hajnal and Moon showed that $sat(n,K_r)=(r-2)n+o(n)$ for $r \geq 3$, where $sat(n,K_r)$ is the minimum number of edges in a $K_r$-saturated graph on $n$ vertices, and determined the unique extremal graph. This graph has many twins, leading us to define $tsat(n,K_r)$ to be the minimum number of edges in a twin-free $K_r$-saturated graph on $n$ vertices. We show that $(5 +2/3)n + o(n) \leq tsat(n,K_3) \leq 6n+o(n)$ and that $\left(r+2\right)n + o(n) \leq tsat(n,K_r) \leq (r+3)n+o(n)$ for $r \geq 4$. We also consider a variant of this problem where we additionally require the graphs to have large minimum degree. Both of these problems turn out be intimately related to two other problems regarding maximally independent sets of a given size in $K_3$-free graphs and generalisations of these problems to $K_r$ with $r \geq 4$. The first problem is to maximise the number of maximally independent sets given the number of vertices and the second problem is to minimise the number of edges given the number of maximally independent sets. They are interesting in their own right.