A robot is located at a point in the plane. A treasure and an exit, both stationary, are located at unknown (to the robot) positions both at distance one from the robot. Starting from its initial position, the robot aims to fetch the treasure to the exit. At any time the robot can move anywhere on the disk with constant speed. The robot detects an interesting point (treasure or exit) only if it passes over the exact location of that point. Given that an adversary controls the locations of both the treasure and the exit on the perimeter, we are interested in designing algorithms that minimize the treasure-evacuation time, i.e. the time it takes for the treasure to be discovered and brought to the exit by the robot. This seemingly simple treasure evacuation problem turns out to be surprisingly challenging.

In this paper we differentiate how the robot's knowledge of the distance between the two interesting points affects the overall evacuation time. We demonstrate the difference between knowing only a lower bound of the distance versus knowing its the exact value, and provide search algorithms for both cases. In the former case we provide an algorithm which is shown to be optimal. In the latter case we give an algorithm which is off from the optimal algorithm (that does not know the locations of the treasure and the exit) by no more than $\frac{4\sqrt{2}+3\pi +2}{6\sqrt{2}+2\pi +2}\le 1.019$ multiplicatively, or $\frac{\pi }{2}-\sqrt{2}\le 0.157$ additively.

机器人位于飞机上的某个点。宝物和出口（均固定）位于（相对于机器人而言）未知的位置，且距离机器人都只有一个距离。从其初始位置开始，机器人的目标是将财宝拿到出口。机器人可以随时以恒定速度移动到磁盘上的任何位置。仅当机器人经过某个兴趣点的确切位置时，它才会检测到该点。假设对手控制着宝藏和出口在外围的位置，我们对设计算法感兴趣，可以最大程度地减少宝藏撤离时间，即由宝藏发现和转移宝藏所需的时间。机器人。这个看似简单的宝藏疏散问题却极具挑战性。

在本文中，我们区分了机器人对两个有趣点之间距离的了解如何影响总体疏散时间。我们证明了仅知道距离的下限与知道其确切值之间的区别，并针对这两种情况提供了搜索算法。在前一种情况下，我们提供了一种最佳算法。在后一种情况下，我们给出的算法与最佳算法（不知道宝藏和出口的位置）相差不超过$\frac{4\sqrt{2}+3\pi +2}{6\sqrt{2}+2\pi +2}\le 1.019$ 相乘地，或 $\frac{\pi }{2}-\sqrt{2}\le 0.157$ 加性地。