We consider the task of graph exploration. An @math -node graph has unlabeled nodes, and all ports at any node of degree @math are arbitrarily numbered @math . A mobile agent has to visit all nodes and stop. The exploration time is the number of edge traversals. We consider the problem of how much knowledge the agent has to have a priori, in order to explore the graph in a given time, using a deterministic algorithm. This a priori information (advice) is provided to the agent by an oracle, in the form of a binary string, whose length is called the size of advice. We consider two types of oracles. The instance oracle knows the entire instance of the exploration problem, i.e., the port-numbered map of the graph and the starting node of the agent in this map. The map oracle knows the port-numbered map of the graph but does not know the starting node of the agent. We first consider exploration in polynomial time, and determine the exact minimum size of advice to achieve it. This size is @math , for both types of oracles. When advice is large, there are two natural time thresholds: @math for a map oracle, and @math for an instance oracle, that can be achieved with sufficiently large advice. We show that, with a map oracle, time @math cannot be improved in general, regardless of the size of advice. We also show that the smallest size of advice to achieve this time is larger than @math , for any @math . For an instance oracle, advice of size @math is enough to achieve time @math . We show that, with any advice of size @math , the time of exploration must be at least @math , for any @math , and with any advice of size @math , the time must be @math . We also investigate minimum advice sufficient for fast exploration of hamiltonian graphs.
Another direction of research concerns exploration of anonymous graphs. In this case it is impossible to explore arbitrary graphs and stop after exploration, if no marking of nodes is allowed, and if nothing is known about the graph. Hence some authors @cite_19 @cite_24 allow pebbles which the agent can drop on nodes to recognize already visited ones, and then remove them and drop them in other places. A more restrictive scenario assumes a stationary token that is fixed at the starting node of the agent @cite_16 @cite_34 . Exploring anonymous graphs without the possibility of marking nodes (and thus possibly without stopping) is investigated, e.g., in @cite_22 @cite_0 . The authors concentrate attention not on the cost of exploration but on the minimum amount of memory sufficient to carry out this task. In the absence of marking nodes, in order to guarantee stopping after exploration, some knowledge about the graph is required, e.g., an upper bound on its size @cite_16 @cite_32 .
@cite_22 @cite_32 @cite_34 @cite_24 @cite_19 @cite_0 @cite_16
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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 study the problem of the amount of information required to draw a complete or a partial map of a graph with unlabeled nodes and arbitrarily labeled ports. A mobile agent, starting at any node of an unknown connected graph and walking in it, has to accomplish one of the following tasks: draw a complete map of the graph, i.e., find an isomorphic copy of it including port numbering, or draw a partial map, i.e., a spanning tree, again with port numbering. The agent executes a deterministic algorithm and cannot mark visited nodes in any way. None of these map drawing tasks is feasible without any additional information, unless the graph is a tree. Hence we investigate the minimum number of bits of information (minimum size of advice) that has to be given to the agent to complete these tasks. It turns out that this minimum size of advice depends on the number n of nodes or the number m of edges of the graph, and on a crucial parameter @m, called the multiplicity of the graph, which measures the number of nodes that have an identical view of the graph. We give bounds on the minimum size of advice for both above tasks. For @m=1 our bounds are asymptotically tight for both tasks and show that the minimum size of advice is very small. For @m>1 the minimum size of advice increases abruptly. In this case our bounds are asymptotically tight for topology recognition and asymptotically almost tight for spanning tree construction. 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. We study the size of memory of mobile agents that permits to solve deterministically the rendezvous problem, i.e., the task of meeting at some node, for two identical agents moving from node to node along the edges of an unknown anonymous connected graph. The rendezvous problem is unsolvable in the class of arbitrary connected graphs, as witnessed by the example of the cycle. Hence we restrict attention to rendezvous in trees, where rendezvous is feasible if and only if the initial positions of the agents are not symmetric. We prove that the minimum memory size guaranteeing rendezvous in all trees of size at most nis i¾?(logn) bits. The upper bound is provided by an algorithm for abstract state machines accomplishing rendezvous in all trees, and using O(logn) bits of memory in trees of size at most n. The lower bound is a consequence of the need to distinguish between up to ni¾? 1 links incident to a node. Thus, in the second part of the paper, we focus on the potential existence of pairs of finiteagents (i.e., finite automata) capable of accomplishing rendezvous in all bounded degreetrees. We show that, as opposed to what has been proved for the graph exploration problem, there are no finite agents capable of accomplishing rendezvous in all bounded degree trees.