The paper considers a team of robots which has to explore a graph G, where some nodes can be harmful. Robots are initially located at the so-called home base node. The dangerous nodes are the so-called black hole nodes, and once a robot enters in one of them, it is destroyed. The goal is to find a strategy in order to explore Gin such a way that minimum number of robots is wasted. The exploration ends if there is at least one surviving robot which knows all the edges leading to the black holes. As many variations of the problem have been considered so far, the solution and its measure heavily depend on the initial knowledge and the capabilities of the robots. In this paper, we assume that G is a directed graph, the robots are associated with unique identifiers, they know the number of nodes n of G (or at least an upper bound on n), and they know the number of edges Delta leading to the black holes. Each node is associated with a whiteboard where robots can read and write information in a mutual exclusive way. A recently posed question [J. Czyzowicz, S. Dobrev, R. Kralovic, S. Miklik, D. Pardubska, Black hole search in directed graphs, in: Proc. of 16th International Colloquium on Structural Information and Communication Complexity, SIROCCO, LNCS, vol. 5869, 2009, pp. 182-194.] is whether some number of robots, expressed as a function of parameter Delta only, is sufficient to detect black holes in directed graphs of arbitrarily large order n. We give a positive answer to this question for the synchronous case, i.e., when the robots share a common clock, showing that O(Delta . 2(Delta)) robots are sufficient to solve the problem. This bound is nearly tight, since it is known that at least 2(Delta) robots are required for some instances. Quite surprisingly, we also show that unlike in the case of undirected graphs, for the directed version of the problem, synchronization can sometimes make a difference: for Delta = 2, in the synchronous case 4 robots are always sufficient, whereas in the asynchronous case at least 5 robots are sometimes required.
Synchronous black hole search in directed graphs
NAVARRA, Alfredo;PINOTTI, Maria Cristina
2011
Abstract
The paper considers a team of robots which has to explore a graph G, where some nodes can be harmful. Robots are initially located at the so-called home base node. The dangerous nodes are the so-called black hole nodes, and once a robot enters in one of them, it is destroyed. The goal is to find a strategy in order to explore Gin such a way that minimum number of robots is wasted. The exploration ends if there is at least one surviving robot which knows all the edges leading to the black holes. As many variations of the problem have been considered so far, the solution and its measure heavily depend on the initial knowledge and the capabilities of the robots. In this paper, we assume that G is a directed graph, the robots are associated with unique identifiers, they know the number of nodes n of G (or at least an upper bound on n), and they know the number of edges Delta leading to the black holes. Each node is associated with a whiteboard where robots can read and write information in a mutual exclusive way. A recently posed question [J. Czyzowicz, S. Dobrev, R. Kralovic, S. Miklik, D. Pardubska, Black hole search in directed graphs, in: Proc. of 16th International Colloquium on Structural Information and Communication Complexity, SIROCCO, LNCS, vol. 5869, 2009, pp. 182-194.] is whether some number of robots, expressed as a function of parameter Delta only, is sufficient to detect black holes in directed graphs of arbitrarily large order n. We give a positive answer to this question for the synchronous case, i.e., when the robots share a common clock, showing that O(Delta . 2(Delta)) robots are sufficient to solve the problem. This bound is nearly tight, since it is known that at least 2(Delta) robots are required for some instances. Quite surprisingly, we also show that unlike in the case of undirected graphs, for the directed version of the problem, synchronization can sometimes make a difference: for Delta = 2, in the synchronous case 4 robots are always sufficient, whereas in the asynchronous case at least 5 robots are sometimes required.I documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.
