Network exploration by silent and oblivious robots

  • Authors:
  • Jérémie Chalopin;Paola Flocchini;Bernard Mans;Nicola Santoro

  • Affiliations:
  • Laboratoire d'Informatique Fondamentale de Marseille, CNRS & Aix-Marseille Université, Marseille, France;SITE, University of Ottawa, Ottawa, Canada;Macquarie University, Sydney, Australia;School of Computer Science, Carleton University, Ottawa, Canada

  • Venue:
  • WG'10 Proceedings of the 36th international conference on Graph-theoretic concepts in computer science
  • Year:
  • 2010

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Abstract

In this paper we investigate the basic problem of Exploration of a graph by a group of identical mobile computational entities, called robots, operating autonomously and asynchronously. In particular we are concerned with what graphs can be explored, and how, if the robots do not remember the past and have no explicit means of communication. This model of robots is used when the spatial universe in which the robots operate is continuous (e.g., a curve, a polygonal region, a plane, etc.). The case when the spatial universe is discrete (i.e., a graph) has been also studied but only for the classes of acyclic graphs and of simple cycles. In this paper we consider networks of arbitrary topology modeled as connected graphs with local orientation (locally distinct edge labels). We concentrate on class Hk of asymmetric configurations with k robots. Our results indicate that the explorability of graphs in this class depends on the number k of robots participating in the exploration. In particular, exploration is impossible for k k = 3 robots, only a subset of H3 can be explored; we provide a complete characterization of the networks that can be explored. When there are k = 4 robots, we prove that all networks in H4 can be explored. Finally, we prove that for any odd k 4 all networks in Hk can be explored by presenting a general algorithm. The determination of which networks can be explored when k 4 is even, is still open but can be reduced to the existence of a gathering algorithm for Hk.