从Bezout恒等式到匿名存储系统的空间最优选择

Emmanuel Godard, Damien Imbs, M. Raynal, G. Taubenfeld
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引用次数: 3

摘要

匿名共享内存REG可以看作是一个原子寄存器数组,这样进程之间在寄存器的名称上就没有先验的一致性。例如,一个非常相同的物理寄存器可以被进程p称为REG[x],被另一个进程q称为REG[y](其中y≠x)。此外,进程p称为REG[a]的寄存器和进程q称为REG[b]的寄存器可以是同一个物理寄存器。假设每个进程都有一个唯一的标识符,只能比较是否相等。本文讨论的是如何解决d-election问题,即在这样一个匿名共享内存系统中,需要选出至少一个最多d个领导者。我们注意到1-选举问题是熟悉的领导人选举问题。设n为进程的数目,m为匿名内存的大小(原子寄存器的数目)。本文证明了条件gcd(m, n)≤d对于解决通过读/写或读+修改+写寄存器进行通信的d-选举问题是充分必要的。用来证明充分条件的算法依赖于Bezout恒等式——一个根据最大公约数将数联系起来的丢芬图方程。进一步,在证明充分条件的过程中,证明了1-leader选举可以只使用一个读/写寄存器来解决(这驳斥了1989年关于需要三个非匿名寄存器的猜想),并且当且仅当gcd(m, n)除d时,可以解决精确的d-election问题,其中必须选举恰好d个leader。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
From Bezout's Identity to Space-Optimal Election in Anonymous Memory Systems
An anonymous shared memory REG can be seen as an array of atomic registers such that there is no a priori agreement among the processes on the names of the registers. As an example a very same physical register can be known as REG[x] by a process p and as REG[y] (where y ≠ x) by another process q. Moreover, the register known as REG[a] by a process p and the register known as REG[b] by a process q can be the same physical register. It is assumed that each process has a unique identifier that can only be compared for equality. This article is on solving the d-election problem, in which it is required to elect at least one and at most d leaders, in such an anonymous shared memory system. We notice that the 1-election problem is the familiar leader election problem. Let n be the number of processes and m the size of the anonymous memory (number of atomic registers). The article shows that the condition gcd(m, n) ≤ d is necessary and sufficient for solving the d-election problem, where communication is through read/write or read+modify+write registers. The algorithm used to prove the sufficient condition relies on Bezout's Identity - a Diophantine equation relating numbers according to their Greatest Common Divisor. Furthermore, in the process of proving the sufficient condition, it is shown that 1-leader election can be solved using only a single read/write register (which refutes a 1989 conjecture stating that three non-anonymous registers are necessary), and that the exact d-election problem, where exactly d leaders must be elected, can be solved if and only if gcd(m, n) divides d.
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