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

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.