嗅探有助于相遇：网格中匿名智能体的确定性 rendezvous
Sniffing Helps to Meet: Deterministic Rendezvous of Anonymous Agents in the Grid
AI总结：
本文研究网格中匿名智能体的确定性 rendezvous，证明在已知距离或同时启动时可 O(D) 完成，在无先验知识且任意延迟时最优复杂度为 O(D^2)，并给出下界。
摘要：
两个相同的匿名移动智能体必须在一个节点上相遇，该节点位于无限有向网格中，网格节点未标记。这个问题被称为 rendezvous。智能体执行相同的确定性算法。时间被划分为轮次，在每一轮中，每个智能体可以停留在当前节点或移动到相邻节点。对手将智能体放置在网格中距离至多 $D$ 的两个节点上，并可能在不同的轮次唤醒它们。每个智能体在其唤醒轮次开始执行算法。如果智能体不能在访问过的节点上留下任何标记，那么它们永远无法相遇，即使它们同时从相邻节点开始并且知道这一点。因此，我们假设每个智能体标记它访问的任何未标记节点，并且智能体可以区分它访问的节点是否已被标记。Rendezvous 算法的时间是较晚唤醒的智能体唤醒与相遇之间的轮数。我们提出问题：标记节点的能力是否能使智能体相遇，如果能，最快的 rendezvous 算法是什么。我们在三种场景下考虑这个问题。首先，智能体知道 $D$ 但可能以任意延迟开始。其次，它们同时开始但没有任何先验知识。第三，最困难的场景，我们不做上述任何简化假设。智能体以任意延迟开始，并且没有任何先验知识。我们证明在前两种场景中，rendezvous 可以在时间 $O(D)$ 内完成。这显然是最优的。对于第三种场景，我们证明不存在任何在时间 $o(D^{\sqrt{2}})$ 内工作的 rendezvous 算法，并且我们展示了一个在时间 $O(D^2)$ 内工作的算法。上述否定结果显示了两个较简单场景中的最优复杂度与最困难场景中的最优复杂度之间的分离。
英文摘要：
Two identical anonymous mobile agents have to meet at a node of the infinite oriented grid whose nodes are unlabeled. This problem is known as rendezvous. The agents execute the same deterministic algorithm. Time is divided into rounds, and in each round each agent can either stay idle at the current node or move to an adjacent node. An adversary places the agents at two nodes of the grid at a distance at most $D$, and wakes them up in possibly different rounds. Each agent starts executing the algorithm in its wakeup round. If agents cannot leave any marks on visited nodes then they can never meet, even if they start simultaneously at adjacent nodes and know it. Hence, we assume that each agent marks any unmarked node it visits, and that an agent can distinguish if a node it visits has been previously marked or not. The time of a rendezvous algorithm is the number of rounds between the wakeup of the later agent and rendezvous. We ask the question whether the capability of marking nodes enables the agents to meet, and if so, what is the fastest rendezvous algorithm. We consider this problem under three scenarios. First, agents know $D$ but may start with arbitrary delay. Second, they start simultaneously but do not have any {\em a priori} knowledge. Third, most difficult scenario, we do not make any of the above facilitating assumptions. Agents start with arbitrary delay and they do not have any a priori knowledge. We prove that in the first two scenarios rendezvous can be accomplished in time $O(D)$. This is clearly optimal. For the third scenario, we prove that there does not exist any rendezvous algorithm working in time $o(D^{\sqrt{2}})$, and we show an algorithm working in time $O(D^2)$. The above negative result shows a separation between the optimal complexity in the two easier scenarios and the optimal complexity in the most difficult scenario.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
