修正模拟:一种证明空间下界的新方法

Faith Ellen, Rati Gelashvili, Leqi Zhu
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引用次数: 11

摘要

确定解决x≤k的无x阻碍(或随机无等待)k集协议所需的寄存器数是一个开放的问题,它突出了我们对同步的空间复杂性的理解中的重要差距。在无x阻塞协议中,要求进程在执行时返回至多x个进程执行的步骤。在n b> k个进程中,解决这个问题所需的寄存器数的上界是n-k+x个寄存器。没有比2更好的一般下界。我们证明了在n b> k个进程中解决k集协议的任何无x阻碍协议必须使用n-x/k+1-x \rfloor +1个或更多的寄存器。我们的主要工具是一个模拟,它可以减少确定性无等待k集协议的不可能性。特别是,我们表明,如果协议使用较少的寄存器,那么k+1进程可以模拟协议并以无等待的方式确定性地解决k集协议,这是不可能的。模拟的一个重要方面是模拟过程修正模拟过程过去的能力。我们引入了一个增强快照对象,它有助于实现这一点。我们还证明了无阻塞协议所使用的寄存器数的任何下界都适用于满足不确定性单独终止的协议。因此,对于无阻塞情况(即x = 1),我们的n-1/k + 1的下界也适用于随机无等待协议。特别是,我们得到了一个精确的n个寄存器的紧下界,用于解决无阻碍和随机无等待的共识。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Revisionist Simulations: A New Approach to Proving Space Lower Bounds
Determining the number of registers required for solving x-obstruction-free (or randomized wait-free) k-set agreement for x ≤ k is an open problem that highlights important gaps in our understanding of the space complexity of synchronization. In x-obstruction-free protocols, processes are required to return in executions where at most x processes take steps. The best known upper bound on the number of registers needed to solve this problem among n>k processes is n-k+x registers. No general lower bound better than 2 was known. We prove that any x-obstruction-free protocol solving k-set agreement among n > k processes must use n-x/k+1-x \rfloor + 1 or more registers. Our main tool is a simulation that serves as a reduction from the impossibility of deterministic wait-free k-set agreement. In particular, we show that, if a protocol uses fewer registers, then it is possible for k+1 processes to simulate the protocol and deterministically solve k-set agreement in a wait-free manner, which is impossible. An important aspect of the simulation is the ability of simulating processes to revise the past of simulated processes. We introduce an augmented snapshot object, which facilitates this. We also prove that any lower bound on the number of registers used by obstruction-free protocols applies to protocols that satisfy nondeterministic solo termination. Hence, our lower bound of n-1/k + 1 for the obstruction-free case (i.e., x = 1) also holds for randomized wait-free protocols. In particular, we get a tight lower bound of exactly n registers for solving obstruction-free and randomized wait-free consensus.
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