Un algorithme de test pour la connexit\'e temporelle des graphes dynamiques de faible densit\'e

TL;DR

This paper presents an efficient algorithm for testing temporal connectivity in low-density dynamic graphs, improving over existing methods especially when the graph's instant density remains low over time.

Contribution

It introduces a new algorithm with better time complexity for testing temporal connectivity in dynamic graphs, applicable to both strict and non-strict journeys, especially in low-density scenarios.

Findings

The algorithm runs in O(kμn) time, where k is time steps, μ is maximum instant density, and n is number of vertices.

It outperforms adapted existing algorithms in low-density cases for strict journeys.

A similar approach is effective for non-strict journeys, with no prior known algorithms.

Abstract

We address the problem of testing whether a dynamic graph is temporally connected, i.e. a temporal path ({\em journey}) exists between all pairs of vertices. We consider a discrete version of the problem, where the topology is given as an evolving graph $\G=\{G_1,G_2,...,G_{k}\}$ in which only the set of (directed) edges varies. Two cases are studied, depending on whether a single edge or an unlimited number of edges can be crossed in a same $G_i$ (strict journeys {\it vs} non-strict journeys). For strict journeys, two existing algorithms designed for other problems can be adapted. However, we show that a dedicated approach achieves a better time complexity than one of these two algorithms in all cases, and than the other one for those graphs whose density is low at any time (though arbitrary over time). The time complexity of our algorithm is $O(k\mu n)$, where $k=|\G|$ is the number of time steps and $\mu=max(|E_i|)$ is the maximum {\em instant} density, to be contrasted with $m=|\cup E_i|$, the {\em cumulated} density. Indeed, it is not uncommon for a mobility scenario to satisfy, for instance, both $\mu=o(n)$ and $m=\Theta(n^2)$. We characterize the key values of $k, \mu$ and $m$ for which our algorithm should be used. For non-strict journeys, for which no algorithm is known, we show that a similar strategy can be used to answer the question, still in $O(k\mu n)$ time.