← 研踪 PaperTrailarXiv:2611.09747 · v1
Contention Resolution Against an Adversarial Scheduler
首次提交 2026-11-07 01:17
版本更新 2026-11-07 01:17
雷达观测 2026-11-09 13:22
摘要
In the classical contention-resolution problem, a set of players share a broadcast channel, and a player succeeds in a given time step if it is the only player to broadcast during that step. Implicit in the classical problem is the assumption that every player is free to act on every step: once a player arrives, it participates until it succeeds. We study what happens when that assumption is removed. In our model, an adversarial \emph{scheduler} decides, on each step, which subset of the players gets scheduled to participate in the step. An unscheduled player simply sleeps through the step. We measure a protocol by its \emph{efficiency}: the number of players divided by the expected number of steps in which at least one player was active. Our first result says that, so long as the players are given access to a global clock and to shared randomness, they can achieve efficiency $Ω(1/\log n)$. The players are able to achieve this bound without requiring any channel feedback, and even if the adversarial scheduler is adaptive. We also give a matching lower bound of $O(1 / \log n)$, which holds even with channel feedback, and even for an oblivious scheduler. Our upper bounds all require the players to have access to a shared global clock and to shared randomness. Our final result says that this type of implicit shared state is necessary. If the players have neither a global clock nor shared randomness, then no protocol can achieve efficiency better than $O(\log n / n)$; and this holds even if the other aspects of the problem are designed to be as favorable for the players as possible (the channel gives ternary feedback and the adversary is oblivious).
作者备注：15 pages
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v1 · 2026-11-07 01:17 · 最新观测版本
Contention Resolution Against an Adversarial Scheduler
In the classical contention-resolution problem, a set of players share a broadcast channel, and a player succeeds in a given time step if it is the only player to broadcast during that step. Implicit in the classical problem is the assumption that every player is free to act on every step: once a player arrives, it participates until it succeeds. We study what happens when that assumption is removed. In our model, an adversarial \emph{scheduler} decides, on each step, which subset of the players gets scheduled to participate in the step. An unscheduled player simply sleeps through the step. We measure a protocol by its \emph{efficiency}: the number of players divided by the expected number of steps in which at least one player was active. Our first result says that, so long as the players are given access to a global clock and to shared randomness, they can achieve efficiency $Ω(1/\log n)$. The players are able to achieve this bound without requiring any channel feedback, and even if the adversarial scheduler is adaptive. We also give a matching lower bound of $O(1 / \log n)$, which holds even with channel feedback, and even for an oblivious scheduler. Our upper bounds all require the players to have access to a shared global clock and to shared randomness. Our final result says that this type of implicit shared state is necessary. If the players have neither a global clock nor shared randomness, then no protocol can achieve efficiency better than $O(\log n / n)$; and this holds even if the other aspects of the problem are designed to be as favorable for the players as possible (the channel gives ternary feedback and the adversary is oblivious).
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