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Simple Strategies Versus Optimal Schedules in Multi-agent Patrolling

机译:多代理巡逻中的简单策略与最佳计划

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Suppose that we want to patrol a fence (line segment) using k mobile agents with given speeds v_1,..., v_k so that every point on the fence is visited by an agent at least once in every unit time period. A simple strategy where the ith agent moves back and forth in a segment of length v_i/2 patrols the length (v_1 +…+ v_k)/2, but it has been shown recently that this is not always optimal. Thus a natural question is to determine the smallest c such that a fence of length c(v_1 +…+ v_k)/2 cannot be patrolled. We give an example showing c ≥ 4/3 (and conjecture that this is the best possible). We also consider a variant of this problem where we want to patrol a circle and the agents can move only clockwise. We can patrol a circle of perimeter rv_r by a simple strategy where the r fastest agents move at the same speed, but it has been shown recently that this is not always optimal. We conjecture that this is not even a constant-approximation strategy. To tackle this conjecture, we relate it to what we call constant gap families. Using this relation, we give another example where the simple strategy is not optimal. We propose another variant where we want to patrol a single point under the constraint that for each i = 1,..., k, the time between two consecutive visits of agent i should be a_i or longer. This problem can be reduced to the discretized version where the a_i are integers and the goal is to visit the point at every integer time. It is easy to see that this discretized patrolling is impossible if 1/a_1 +…+ 1/a_k < 1, and that there is a simple strategy if 1/a_1 +…+ 1/a_k ≥ 2. Thus we are interested in the smallest c such that patrolling is always possible if 1/a_1 +…+ 1/a_k ≥ c. We prove that α ≤ c < 1.546, where a = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.
机译:假设我们要使用k个移动代理以给定的速度v_1,...,v_k来对栅栏(线段)进行巡逻,以使代理在每个单位时间段至少访问一次栅栏上的每个点。第i个代理在一段长度为v_i / 2的段中来回移动的简单策略巡逻了长度(v_1 +…+ v_k)/ 2,但是最近已证明这并不总是最佳的。因此,自然的问题是确定最小的c,以使巡逻长度为c(v_1 +…+ v_k)/ 2的栅栏。我们给出一个显示c≥4/3的示例(并推测这是最好的)。我们还考虑了此问题的一种变体,其中我们想巡逻一个圆,而特工只能顺时针移动。我们可以通过一种简单的策略巡逻一个圆周rv_r的圆周,其中r个最快的特工以相同的速度移动,但是最近已证明这并不总是最佳的。我们推测这甚至不是恒定近似策略。为了解决这个猜想,我们将其与恒定间隙族联系起来。使用这种关系,我们给出了另一个示例,其中简单策略并非最优。我们提出了另一种变体,其中我们希望在每个i = 1,...,k的约束下巡逻单个点,两次连续两次访问代理i的时间应为a_i或更长时间。该问题可以简化为a_i为整数且目标是在每个整数时间访问该点的离散版本。很容易看出,如果1 / a_1 +…+ 1 / a_k <1,那么这种离散巡逻是不可能的;如果1 / a_1 +…+ 1 / a_k≥2,则有一个简单的策略。因此,我们对最小c,因此如果1 / a_1 +…+ 1 / a_k≥c,则总是可以巡逻。我们证明α≤c <1.546,其中a = 1.264 ...(我们推测c =α)。我们还将讨论相关问题的计算复杂性。

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