鹅卵石引导寻宝在飞机上

Adri Bhattacharya, B. Gorain, P. Mandal
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引用次数: 0

摘要

研究了一个移动智能体在欧氏平面上的寻宝问题。代理人的初始位置和宝藏的位置被建模为欧几里得平面上的特殊点。宝藏位于离代理人初始位置最多$D>0$的距离。代理人有一个完美的指南针,但对手控制代理人的速度。因此,智能体无法测量在给定时间内移动了多少距离。代理商只有在到达宝藏的确切位置时才能找到宝藏。寻宝的成本定义为agent在找到宝藏之前走过的总距离。代理人事先不知道宝藏的位置或$D$的价值。一个知道宝藏的位置和代理的初始位置的神谕,放置一些鹅卵石来引导代理走向宝藏。一旦决定沿着某个指定的角度方向移动,代理只能在遇到鹅卵石或特殊点时决定改变其方向。我们在本文中提出以下中心问题:对于给定$k \ge 0$,如果Oracle最多放置$k$个鹅卵石,那么最便宜的寻宝算法是什么?“我们表明,对于$k=1$,不存在任何以有限成本找到宝藏的寻宝算法。我们证明了对$k=2$的代价为$O(D)$的算法的存在性。对于$k>8$,我们设计了一个算法,使用$k$许多鹅卵石找到宝藏,成本为$O(k^{2}) + D(\sin\theta' + \cos\theta')$,其中$\theta'=\frac{\pi}{2^{k-8}}$。第二个结果表明,对于足够大的$D$值,存在一个代价任意接近$D$的算法。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Pebble guided Treasure Hunt in Plane
We study the problem of treasure hunt in a Euclidean plane by a mobile agent with the guidance of pebbles. The initial position of the agent and position of the treasure are modeled as special points in the Euclidean plane. The treasure is situated at a distance at most $D>0$ from the initial position of the agent. The agent has a perfect compass, but an adversary controls the speed of the agent. Hence, the agent can not measure how much distance it traveled for a given time. The agent can find the treasure only when it reaches the exact position of the treasure. The cost of the treasure hunt is defined as the total distance traveled by the agent before it finds the treasure. The agent has no prior knowledge of the position of the treasure or the value of $D$. An Oracle, which knows the treasure's position and the agent's initial location, places some pebbles to guide the agent towards the treasure. Once decided to move along some specified angular direction, the agent can decide to change its direction only when it encounters a pebble or a special point. We ask the following central question in this paper: ``For given $k \ge 0$, What is cheapest treasure hunt algorithm if at most $k$ pebbles are placed by the Oracle?"We show that for $k=1$, there does not exist any treasure hunt algorithm that finds the treasure with finite cost. We show the existence of an algorithm with cost $O(D)$ for $k=2$. For $k>8$ we have designed an algorithm that uses $k$ many pebbles to find the treasure with cost $O(k^{2}) + D(\sin\theta' + \cos\theta')$, where $\theta'=\frac{\pi}{2^{k-8}}$. The second result shows the existence of an algorithm with cost arbitrarily close to $D$ for sufficiently large values of $D$.
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