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Searching for a Non-adversarial, Uncooperative Agent on a Cycle

机译:在周期中搜索非对抗,不合作的特工

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Assume k robots are placed on a cycle-the perimeter of a unit (radius) disk-at a position of our choosing and can move on the cycle with maximum speed 1. A non-adversarial, uncooperative agent, called bus, is moving with constant speed s along the perimeter of the cycle. The robots are searching for the moving bus but do not know its exact location; during the search they can move anywhere on the perimeter of the cycle. We give algorithms which minimize the worst-case search time required for at least one of the robots to find the bus. The following results are obtained for one robot. (1) If the robot knows the speed s of the bus but does not know its direction of movement then the optimal search time is shown to be exactly (la) 2π, if s > 1, (lb) 4π/(s+1), if 1/3≤ s≤ 1, and (1c) 2π/(1-s), if s ≤ 1/3. (2) If the robot does not know neither the speed nor the direction of movement of the bus then the optimal search time is shown to be 27π(l + 1/(s+1)). Moreover, for all e > 0 there exists a speed s such that any algorithm knowing neither the bus speed nor its direction will need time at least 4π - c to meet the bus. These results are also generalized to k > 2 robots and analogous tight upper and lower bounds are proved depending on the knowledge the robots have about the speed and direction of movement of the bus.
机译:假设将k个机器人放置在我们选择的位置上的一个循环(单位(半径)磁盘的周长)上,并且可以以最大速度1在循环上移动。沿循环周长的恒定速度s。机器人正在搜索行驶中的公交车,但不知道其确切位置;在搜索过程中,它们可以移动到循环周长的任何位置。我们提供的算法可最大程度地减少至少一个机器人找到总线所需的最坏情况搜索时间。对于一个机器人,可获得以下结果。 (1)如果机器人知道总线的速度s但不知道其运动方向,则最佳搜索时间将精确显示为(la)2π,如果s> 1,(lb)4π/(s + 1 ),如果1 /3≤s≤1,则(1c)2π/(1-s),如果s≤1/3。 (2)如果机器人既不知道总线的速度也不知道其移动方向,则最佳搜索时间将显示为27π(l + 1 /(s + 1))。而且,对于所有e> 0,存在一个速度s,使得任何不知道总线速度或其方向的算法都需要至少4π-c的时间来满足总线。这些结果也可以推广到k> 2个机器人,并且根据机器人对公交车的速度和运动方向的了解,证明了类似的严格上下限。

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