确定性寻宝在平面与角度提示
Deterministic Treasure Hunt in the Plane with Angular Hints
摘要:
一个配备了指南针和长度测量的移动代理必须在欧几里得平面上找到一个惰性的宝藏。代理和宝藏都被建模为点。开始时,智能体与宝藏的距离最多为$D>0$,但既不知道距离,也不知道宝藏的边界。找到宝藏意味着与宝藏的距离最多为1。智能体进行一系列移动。它们中的每一个都包括在选定的方向上以选定的距离直线移动。在开始和每次移动之后,智能体都会得到一个提示,该提示由一个小于$2π$的正角组成,其顶点位于智能体的当前位置,并且宝藏包含在其中。我们研究了这些提示如何允许智能体使用确定性算法降低寻找宝藏的成本的问题,其中成本是智能体轨迹的最坏情况总长度。众所周知,在没有任何提示的情况下,最优(最坏情况)成本是$Θ(D^2)$。我们表明,如果所有的角度作为提示是最$π$,那么成本可以降低到$O(D)$,这是最优的。如果所有角度都至多为$β$,其中$β<2π$是智能体未知的常数,则对于某些$ε>0$,成本至多为$O(D^{2-ε})$。对于这两个积极的结果,我们提出了确定性算法实现上述成本。最后,如果作为提示的角度可以是任意的,小于$2π$,那么我们表明,成本$Θ(D^2)$不能被击败。
英文摘要:
A mobile agent equipped with a compass and a measure of length has to find an inert treasure in the Euclidean plane. Both the agent and the treasure are modeled as points. In the beginning, the agent is at a distance at most $D>0$ from the treasure, but knows neither the distance nor any bound on it. Finding the treasure means getting at distance at most 1 from it. The agent makes a series of moves. Each of them consists in moving straight in a chosen direction at a chosen distance. In the beginning and after each move the agent gets a hint consisting of a positive angle smaller than $2π$ whose vertex is at the current position of the agent and within which the treasure is contained. We investigate the problem of how these hints permit the agent to lower the cost of finding the treasure, using a deterministic algorithm, where the cost is the worst-case total length of the agent's trajectory. It is well known that without any hint the optimal (worst case) cost is $Θ(D^2)$. We show that if all angles given as hints are at most $π$, then the cost can be lowered to $O(D)$, which is optimal. If all angles are at most $β$, where $β<2π$ is a constant unknown to the agent, then the cost is at most $O(D^{2-ε})$, for some $ε>0$. For both these positive results we present deterministic algorithms achieving the above costs. Finally, if angles given as hints can be arbitrary, smaller than $2π$, then we show that cost $Θ(D^2)$ cannot be beaten.
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