How to meet when you forget: log-space rendezvous in arbitrary graphs

  • Authors:

  • Jurek Czyzowicz;Adrian Kosowski;Andrzej Pelc

  • Affiliations:

  • Universite du Québec en Outaouais, Gatineau, PQ, Canada;Gdańsk University of Technology, Gdańsk, Poland;Universite du Québec en Outaouais, Gatineau, PQ, Canada

  • Venue:
  • Proceedings of the 29th ACM SIGACT-SIGOPS symposium on Principles of distributed computing
  • Year:
  • 2010

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Abstract

Two identical (anonymous) mobile agents start from arbitrary nodes in an a priori unknown graph and move synchronously from node to node with the goal of meeting. This rendezvous problem has been thoroughly studied, both for anonymous and for labeled agents, along with another basic task, that of exploring graphs by mobile agents. Intuitively, the rendezvous problem is more difficult than exploration, as it reduces to the latter, if one of the agents is inert. A well-known recent result on exploration, due to Reingold, states that deterministic exploration of arbitrary graphs can be performed in log-space, i.e., using an agent equipped with O(log n) bits of memory, where n is the size of the graph. In this paper we study the size of memory of mobile agents that permits us to solve the rendezvous problem deterministically. Our main result establishes the minimum size of the memory of anonymous agents that guarantees deterministic rendezvous when it is feasible. We show that this minimum size is Θ(log n), where n is the size of the graph, regardless of the delay between the starting times of the agents. More precisely, we construct identical agents equipped with Θ(log n) memory bits that solve the rendezvous problem in all graphs with at most n nodes, if they start with any delay τ, and we prove a matching lower bound Ω(log n) on the number of memory bits needed to accomplish rendezvous, even for simultaneous start. In fact, this lower bound is achieved already on the class of rings. This shows a significant contrast between rendezvous and exploration: e.g., while exploration of rings (without stopping) can be done using constant memory, rendezvous, even with simultaneous start, requires logarithmic memory.