Two mobile agents, starting from different nodes of an @math -node network at possibly different times, have to meet at the same node. This problem is known as rendezvous. Agents move in synchronous rounds using a deterministic algorithm. In each round, an agent decides to either remain idle or to move to one of the adjacent nodes. Each agent has a distinct integer label from the set @math , which it can use in the execution of the algorithm, but it does not know the label of the other agent. If @math is the distance between the initial positions of the agents, then @math is an obvious lower bound on the time of rendezvous. However, if each agent has no initial knowledge other than its label, time @math is usually impossible to achieve. We study the minimum amount of information that has to be available a priori to the agents to achieve rendezvous in optimal time @math . This information is provided to the agents at the start by an oracle knowing the entire instance of the problem, i.e., the network, the starting positions of the agents, their wake-up rounds, and both of their labels. The oracle helps the agents by providing them with the same binary string called advice, which can be used by the agents during their navigation. The length of this string is called the size of advice. Our goal is to find the smallest size of advice which enables the agents to meet in time @math . We show that this optimal size of advice is @math . The upper bound is proved by constructing an advice string of this size, and providing a natural rendezvous algorithm using this advice that works in time @math for all networks. The matching lower bound, which is the main contribution of this paper, is proved by exhibiting classes of networks for which it is impossible to achieve rendezvous in time @math with smaller advice.
Memory required by the agents to achieve deterministic rendezvous was studied in @cite_14 for trees and in @cite_0 for general graphs. Memory needed for randomized rendezvous in the ring is discussed, e.g., in @cite_28 .
@cite_0 @cite_28 @cite_14
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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. The rendezvous problem is known to be not easier than graph exploration. 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. As a by-product of our techniques introduced to obtain log-space rendezvous we get the first algorithm to find a quotient graph of a given unlabeled graph in polynomial time, by means of a mobile agent moving around the graph. We present a tradeoff between the expected time for two identical agents to rendez-vous on a synchronous, anonymous, oriented ring and the memory requirements of the agents. In particular, we show that there exists a 2t state agent, which can achieve rendez-vous on an n node ring in expected time O(n2 2t + 2t) and that any t 2 state agent requires expected time Ω(n2 2t). As a corollary we observe that Θ(log log n) bits of memory are necessary and sufficient to achieve rendezvous in linear time. The aim of rendezvous in a graph is meeting of two mobile agents at some node of an unknown anonymous connected graph. In this article, we focus on rendezvous in trees, and, analogously to the efforts that have been made for solving the exploration problem with compact automata, we study the size of memory of mobile agents that permits to solve the rendezvous problem deterministically. We assume that the agents are identical, and move in synchronous rounds. We first show that if the delay between the starting times of the agents is arbitrary, then the lower bound on memory required for rendezvous is Ω(log n) bits, even for the line of length n. This lower bound meets a previously known upper bound of O(log n) bits for rendezvous in arbitrary graphs of size at most n. Our main result is a proof that the amount of memory needed for rendezvous with simultaneous start depends essentially on the number e of leaves of the tree, and is exponentially less impacted by the number n of nodes. Indeed, we present two identical agents with O(log e + log log n) bits of memory that solve the rendezvous problem in all trees with at most n nodes and at most e leaves. Hence, for the class of trees with polylogarithmically many leaves, there is an exponential gap in minimum memory size needed for rendezvous between the scenario with arbitrary delay and the scenario with delay zero. Moreover, we show that our upper bound is optimal by proving that Ω(log e + log log n) bits of memory are required for rendezvous, even in the class of trees with degrees bounded by 3.