The Freeze-Tag Problem, introduced in Arkin et al. (SODA'02) consists of waking up a swarm of $n$ robots, starting from a single active robot. In the basic geometric version, every robot is given coordinates in the plane. As soon as a robot is awakened, it can move towards inactive robots to wake them up. The goal is to minimize the wake-up time of the last robot, the makespan. Despite significant progress on the computational complexity of this problem and on approximation algorithms, the characterization of exact bounds on the makespan remains one of the main open questions. In this paper, we settle this question for the $\ell_1$-norm, showing that a makespan of at most $5r$ can always be achieved, where $r$ is the maximum distance between the initial active robot and any sleeping robot. Moreover, a schedule achieving a makespan of at most $5r$ can be computed in optimal time $O(n)$. Both bounds, the time and the makespan are optimal. This implies a new upper bound of $5\sqrt{2}r \approx 7.07r$ on the makespan in the $\ell_2$-norm, improving the best known bound so far $(5+2\sqrt{2}+\sqrt{5})r \approx 10.06r$.

翻译：冻结标记问题（Freeze-Tag Problem）最早由Arkin等人在SODA'02中提出，该问题要求在平面上唤醒由n个机器人组成的群体，初始时仅有一个活跃机器人。在基本几何版本中，每个机器人具有给定的平面坐标。一旦机器人被唤醒，即可移动至非活跃机器人处将其唤醒。目标是最小化最后一个机器人被唤醒的时间（即制造期）。尽管该问题的计算复杂度与近似算法研究已取得显著进展，但对制造期精确上界的刻画仍是最主要的开放问题之一。本文在ℓ₁范数下解决了该问题，证明总能在至多5r的时间内完成唤醒，其中r表示初始活跃机器人与任意休眠机器人的最大距离。进一步地，可在最优时间复杂度O(n)内计算生成一个制造期不超过5r的调度方案。时间复杂度和制造期上界均为最优。该结果导出ℓ₂范数下制造期的新上界为5√2r≈7.07r，将现有最佳上界(5+2√2+√5)r≈10.06r予以改进。