The wakeup problem addresses the fundamental challenge of symmetry breaking. Initially, n devices share a time-slotted multiple access channel, which models wireless communication. A transmission succeeds if exactly one device sends in a slot; if two or more transmit, a collision occurs and none succeed. The goal is to achieve a single successful transmission efficiently. Prior work on wakeup primarily analyzes latency -- the number of slots until the first success. However, in many modern systems, each collision incurs a nontrivial delay, C, which prior analyses neglect. Consequently, although existing algorithms achieve polylogarithmic-in-n latency, they still suffer a delay of \Omega(C) due to collisions. Here, we design and analyze a randomized wakeup algorithm, Aim-High. When C is sufficiently large with respect to n, Aim-High has expected latency and expected total cost of collisions that are nearly O(\sqrt{C}); otherwise, both quantities are O(poly{\log n}). Finally, for a well-studied class of algorithms, we establish a trade-off between latency and expected total cost of collisions.
翻译:唤醒问题涉及对称性破缺的基本挑战。初始状态下,n个设备共享一个时分多址信道,该信道用于建模无线通信。当且仅当单个设备在时隙中发送信号时传输成功;若两个及以上设备同时传输,则发生碰撞且均告失败。目标在于高效实现单次成功传输。现有关于唤醒的研究主要分析延迟——即首次成功传输前的时隙数量。然而,在许多现代系统中,每次碰撞都会产生不可忽略的延迟C,这一因素被既往分析所忽视。因此,尽管现有算法实现了n的多对数级延迟,它们仍会因碰撞而承受Ω(C)的延迟。本文设计并分析了随机化唤醒算法Aim-High。当C相对于n足够大时,Aim-High的期望延迟与期望碰撞总成本均接近O(√C);否则,这两个量值均为O(poly{logn})。最后,针对一类经过深入研究的算法,我们建立了延迟与期望碰撞总成本之间的权衡关系。