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Simple Strategies Versus Optimal Schedules in Multi-agent Patrolling

机译:Multi-Agent巡逻的简单策略与最佳时间表

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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 α = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.
机译:假设我们希望使用给定速度v_1,...,v_k的k移动代理巡逻围栏(线段),以便在每个单位时间段内至少一次访问栅栏上的每个点。 iTh代理在长度v_i / 2段中来回移动的简单策略巡逻长度(v_1 + ... + v_k)/ 2,但最近已经显示,这并不总是最佳的。因此,自然问题是确定最小的C,使得长度C(V_1 + ... + V_K)/ 2不能巡逻。我们给出了一个例子,显示了c≥4/3(并猜测这是最好的)。我们还要考虑这个问题的变体,我们希望巡逻一个圆圈,并且代理可以仅顺时针移动。我们可以通过一个简单的策略巡逻一个周边RV_R的圈子,其中r最快的代理以相同的速度移动,但最近已经显示出这并不总是最佳的。我们猜想这甚至不是常数逼近的策略。为了解决这个猜想,我们将它与我们称之为恒定的差距家庭。使用这一关系,我们给了一个简单的策略不是最佳的举例。我们提出了另一个变体,在那里我们希望在每个I = 1,...,k的约束下巡逻一个点,...,k,代理的两个连续访问之间的时间应该是a_i或更长的时间。此问题可以减少到可离散的版本,其中A_I是整数,目标是访问每个整数时间的点。很容易看出,如果1 / a_1 + ... + 1 / a_k <1,则不可能是不可能的,如果1 / a_1 + ... + 1 /a_k≥2,则存在简单的策略。因此,我们感兴趣在最小的C中,如果1 / a_1 + ... + 1 /a_k≥c,始终可以巡逻距离。我们证明α≤C<1.546,其中α= 1.264 ......(我们猜测C =α)。我们还讨论了相关问题的计算复杂性。

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