The average distance of spanning trees in terms of independence number

TL;DR

This paper improves bounds on the minimum average distance of spanning trees in connected graphs based on the independence number, providing tighter bounds and asymptotic optimality.

Contribution

The authors establish new, tighter upper bounds for the average distance of spanning trees in terms of the graph's independence number, improving previous results.

Findings

Improved the upper bound to μ(T) < α + 1 for α ≥ 1.

Further refined bounds depending on the size of α, especially for larger α.

Demonstrated that the new bounds are asymptotically optimal.

Abstract

Let $G$ be a connected graph with vertex set $V(G)$, and denote by $d_G(u,v)$ the distance from $u$ to $v$ in $G$, for any $u,v \in V(G)$. The average distance of an $n$-vertex connected graph $G$, denoted by $\mu(G)$, is defined to be the average of all distances between all pairs of vertices in $G$, i.e., $\mu (G) = \binom{n}{2}^{-1} \sum_{\{u,v\} \subset V(G)}d_G(u,v)$. The problem of finding a spanning tree of minimum average distance is known to be NP-hard, so establishing an upper bound for the minimum average distance among all spanning trees is of particular interest. Mukwembi (J. Graph Theory, 2014) showed that if $G$ is a connected graph of order $n$ with independence number $\alpha$, where $n > 2 \alpha - 1$, then $G$ has a spanning tree $T$ such that $\mu(T) \le \alpha + 2$. In this paper, we first improve the upper bound to $\mu(T) < \alpha + 1$ for $\alpha \ge 1$, and then we find the bound could be further improved when $\alpha$ becomes larger, so a better upper bound \[ \mu(T) < \left\{ \begin{array}{ll} \alpha+1 & \hbox{if } 1\le\alpha\le 6,\\ \alpha+\frac12+\frac{4(\alpha-1)}{\alpha^2} & \hbox{if } \alpha\ge 7, \end{array} \right. \] is established later. In the end, we give a remark to indicate our new upper bound is best possible in the sense of asymptotics (when $n$ and $\alpha$ are large enough).