Subquadratic-time algorithm for the diameter and all eccentricities on median graphs

TL;DR

This paper introduces the first combinatorial subquadratic algorithm to compute all eccentricities in median graphs, leveraging their structural properties to surpass previous quadratic time barriers.

Contribution

It presents a novel subquadratic algorithm for exact eccentricity computation in median graphs, extending previous results limited to bounded dimension cases.

Findings

Algorithm runs in O(n^{1.6408} log^{O(1)} n) time.

Enumerates all eccentricities in simplex graphs with quasilinear time.

Provides an efficient method to compute reach centralities in median graphs.

Abstract

On sparse graphs, Roditty and Williams [2013] proved that no $O(n^{2-\varepsilon})$-time algorithm achieves an approximation factor smaller than $\frac{3}{2}$ for the diameter problem unless SETH fails. In this article, we solve an open question formulated in the literature: can we use the structural properties of median graphs to break this global quadratic barrier? We propose the first combinatiorial algorithm computing exactly all eccentricities of a median graph in truly subquadratic time. Median graphs constitute the family of graphs which is the most studied in metric graph theory because their structure represents many other discrete and geometric concepts, such as CAT(0) cube complexes. Our result generalizes a recent one, stating that there is a linear-time algorithm for all eccentricities in median graphs with bounded dimension $d$, i.e. the dimension of the largest induced hypercube. This prerequisite on $d$ is not necessarily anymore to determine all eccentricities in subquadratic time. The execution time of our algorithm is $O(n^{1.6408}\log^{O(1)} n)$. We provide also some satellite outcomes related to this general result. In particular, restricted to simplex graphs, this algorithm enumerates all eccentricities with a quasilinear running time. Moreover, an algorithm is proposed to compute exactly all reach centralities in time $O(2^{3d}n\log^{O(1)}n)$.