![]() This size turns out to depend on the initial distance D and on the ratio g = e/ C, which is the relative cost gain due to advice. Our goal is to find the smallest size of advice which enables the agents to solve rendezvous and treasure hunt at a given cost C in a network with e edges. The length of the string given to the agent in treasure hunt and the sum of the lengths of strings given to the agents in rendezvous is called the size of advice. In the case of rendezvous, the advice given to each agent can be different. The oracle assists the agents by providing them with a binary string called advice, which can be used by each agent during the algorithm execution. Following the paradigm of algorithms with advice, this information is provided to the agents at the start of their navigation by an oracle knowing the network, the starting positions of the agents, and, in the case of treasure hunt, the node where the treasure is hidden. We study tradeoffs between the amount of information available a priori to the agents and the cost of rendezvous and treasure hunt. ![]() If the agents have no information about the network, the cost of both rendezvous and treasure hunt can be as large as Θ( e) for networks with e edges. The cost of a treasure hunt algorithm is the worst-case number of edge traversals performed by the agent until the treasure is found. The cost of a rendezvous algorithm is the worst-case total number of edge traversals performed by both agents until meeting. ![]() The network is modeled as an undirected connected graph whose nodes have distinct identities. In treasure hunt, a single agent has to find a stationary target (treasure) situated at an unknown node. In rendezvous, two agents, initially located at distinct nodes of the network, traverse edges in synchronous rounds and have to meet at some node. Rendezvous and treasure hunt are two basic tasks performed by mobile agents in networks. ![]()
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