Buneman's theorem for trees with exatcly n vertices

TL;DR

This paper investigates the conditions under which a family of distances can be realized by a positive-weighted tree with exactly n vertices, showing that the known four-point condition is necessary but not sufficient, and introducing two additional conditions.

Contribution

The paper demonstrates that the four-point condition alone is insufficient for trees with exactly n vertices and proposes two new conditions to fully characterize such families.

Findings

Four-point condition is necessary but not sufficient.

Two additional conditions are introduced for exact n-vertex trees.

Complete characterization of 2-dissimilarity families for n-vertex trees.

Abstract

Let ${\cal T}=(T,w)$ be a positive-weighted tree with at least $n$ vertices. For any $i,j \in \{1,...,n\}$, let $D_{i,j} ({\cal T})$ be the weight of the unique path in $T$ connecting $i$ and $j$. The $D_{i,j} ({\cal T})$ are called $2$-weights of ${\cal T}$ and, if we put in order the $2$-weights, the vector which has the $D_{i,j} ({\cal T})$ as components is called \emph{$2$-dissimilarity vector} of $ {\cal T}$. Given a family of positive real numbers $\{D_{i,j}\}_{i,j \in \{1,...,n\}}$, we say that a positive-weighted tree ${\cal T}=(T,w)$ realizes the family if $\{1,...,n\} \subset V(T)$ and $D_{i,j}({\cal T})=D_{i,j}$ for any $ i,j \in \{1,...,n\}$. A characterization of $2$-dissimilarity families of positive weighted trees is already known (see \cite{B}, \cite{SimP} or \cite{St}): the families must satisfy the well-known \emph{four-point condition}. However we can wonder when there exists a positive-weighted tree with \emph{exactly} $n$ vertices, $1,...,n,$ and realizing the family $\{D_{i,j}\}$. In this paper we will show that the four-point condition is necessary but no more sufficient, and so we will introduce two additional conditions (see Theorem \ref{thm:ThmAgne}).