Order plus size of $\tau$-critical graphs

TL;DR

This paper establishes a sharp combined bound on the order and size of $ au$-critical graphs, extending previous bounds and characterizing all extremal graphs with equality.

Contribution

It introduces a new combined bound on the number of vertices and edges in $ au$-critical graphs and characterizes all graphs that attain this bound.

Findings

Proved the bound |E|+|V| ≤ (t+2 choose 2).

Characterized all graphs with equality in the bound.

Extended classical bounds by Erdős and Gallai, Hajnal, and Moon.

Abstract

Let $G=(V,E)$ be a $\tau$-critical graph with $\tau(G)=t$. Erd\H{o}s and Gallai proved that $|V|\leq 2t$ and the bound $|E|\leq {t+1\choose 2}$ was obtained by Erd\H{o}s, Hajnal and Moon. We give here the sharp combined bound $|E|+|V|\leq {t+2\choose 2}$ and find all graphs with equality.

Equations

$$|V|\le 2tand|E|\le \left(\genfrac{}{}{0ex}{}{t+1}{2}\right).$$

$$|V|+|E|\le \left(\genfrac{}{}{0ex}{}{t+2}{2}\right).$$

$$|V|+|E|\le n+\frac{n(2t-n+1)}{2},$$

Full text

A set of vertices meeting every edge of a graph $G$ is called a transversal set of $G$. The transversal number of $G$, $\tau(G)$, is defined to be the the minimum cardinality of a transversal set of $G$. A simple graph $G=(V,E)$ with no isolated vertex is called $\tau$-critical if $\tau(G-e)=\tau(G)-1$, for every $e\in E$ (where $G-e=(V,E\setminus\{e\})$). The primary sources for the properties of $\tau$-critical graphs are Lovász and Plummer [5, Chapter 12.1], and Lovász [4, Chapter 8, Exercises 10–25].

The tight bounds for the number of edges and the number of vertices in a $\tau$-critical graph $G=(V,E)$ with $\tau(G)=t$ are:

[TABLE]

The vertex bound is due to Erdős and Gallai [1], and the edge bound was obtained by Erdős, Hajnal, and Moon [2]. Here we derive the combined bound $|E|+|V|\leq{t+2\choose 2}$ and determine all extremal graphs (Theorem 1). Note that the combined bound immediately gives the edge bound in (1) since $|V|\geq t+1$. The proof of the combined bound comes easily from the next degree bound.

Theorem A. [Hajnal [3]]

*Let $G=(V,E)$ be a $\tau$-critical graph of order $n$ with $\tau(G)=t$. Then $deg_{G}(x)\leq 2t-n+1$ for every $x\in V$. * $\Box$

[TABLE]

Furthermore, the bound is tight if and only if $G\cong K_{n}$, $n\geq 2$, or $G\cong 2K_{2}$ or $G\cong C_{5}$.

[TABLE]

with equality if $G$ is a $(2t-n+1)$-regular graph. To prove (2), we show that the right hand side of (3) is at most ${t+2\choose 2}$. This is equivalent to $0\leq(n-t-1)(n-t-2)$ which is clearly true, since $n\geq t+1$, with equality only for $n=t+1$ or $n=t+2$. Thus equality in (2) is possible only for $(n-1)$-regular and for $(n-3)$-regular graphs.

In the first case $G=K_{n}$. In the second case the candidates are the graphs whose complements are $2$-regular (and have at least four vertices). Since these graphs are $\tau$-critical with $t=n-2$, the deletion of any edge creates a set of three vertices inducing no edges; equivalently, including an edge in their complements produces a triangle. This implies that the complement of such a graph $G$ must be a single cycle $C$, since otherwise, deletion of an edge between two cycles creates no triangle. In addition, $C$ has at most five vertices because deletion of a long diagonal would not create a triangle. Thus $C$ is a four-cycle (and then $G=2K_{2}$), or $C$ (and its complement $G$) is a five cycle. $\Box$