The diameter of random spanning trees interpolating between the UST and the MST of the complete graph

TL;DR

This paper introduces a model of weighted spanning trees on complete graphs that interpolates between uniform and minimum spanning trees, analyzing phase transitions and the global geometry, especially the diameter growth, as the model parameter varies.

Contribution

It defines a new interpolating model of weighted spanning trees on complete graphs and characterizes phase transitions and diameter growth behavior across different regimes of the model parameter.

Findings

Diameter grows like Θ(n^{1/3}) for large β_n, similar to MST.

Diameter grows like Θ(n^{1/2}) for small β_n, similar to UST.

Identifies phase transition thresholds for algorithm agreement and edge similarity.

Abstract

We introduce $\mathsf{WST}^{\beta_n}(K_n)$ as the weighted spanning tree of the complete graph $K_n$ w.r.t. the random electric network of conductances $\{\exp(-\beta_nU_{e})\}_{e\in E(K_n)}$ with $\mathrm{Unif}[0,1]$ i.i.d. $U_e$'s. Moving from $\beta_n\equiv 0$ to faster and faster growing $\beta_n$'s, the model interpolates between the \emph{uniform} and the \emph{minimum} spanning trees: $\mathsf{WST}^0(K_n)=\mathsf{UST}(K_n)$, and there are phase transitions for $\mathsf{WST}^{\beta_n}(K_n)$ behaving more and more like $\mathsf{MST}(K_n)$: - around $\beta_n=n^{3+o(1)}$ regarding the agreement of the two standard algorithms generating these models : Aldous-Broder and Prim's invasion algorithms, - around $\beta_n=n^{2+o(1)}$ regarding the models consisting of exactly the same edges, and - around $\beta_n=n^{1+o(1)}$ regarding the expected total length $\mathbb{E}\left[\sum_{e\in \mathsf{WST}^{\beta_n}(K_n)}U_e\right]$. But most importantly, we study the global geometry of the model: we prove that the typical diameter of $\mathsf{WST}^{\beta_n}(K_n)$ grows like $\Theta(n^{1/3})$ for $\beta_n\ge n^{4/3+o(1)}$ likewise the $\mathsf{MST}(K_n)$ case, and it grows like $\Theta(n^{1/2})$ for $\beta_n\le n^{1+o(1)}$ similarly to the $\mathsf{UST}(K_n)$ case. For $\beta_n=n^{\alpha}$ with $1<\alpha<4/3$, the behavior of the typical diameter is a more delicate open question, but we conjecture that its exponent strictly between 1/2 and 1/3.