In balanced allocations, the goal is to place $m$ balls into $n$ bins, so as to minimize the gap (difference of max to average load). The One-Choice process places each ball to a bin sampled independently and uniformly at random. The Two-Choice process places balls in the least loaded of two sampled bins. Finally, the $(1+\beta)$-process mixes these processes, meaning each ball is allocated using Two-Choice with probability $\beta\in(0,1)$, and using One-Choice otherwise. Despite Two-Choice being optimal in the sequential setting, it has been observed in practice that it does not perform well in a parallel environment, where load information may be outdated. Following [BCEFN12], we study such a parallel setting where balls are allocated in batches of size $b$, and balls within the same batch are allocated with the same strategy and based on the same load information. For small batch sizes $b\in[n,n\log n]$, it was shown in [LS22a] that Two-Choice achieves an asymptotically optimal gap among all processes with a constant number of samples. In this work, we focus on larger batch sizes $b\in[n\log n,n^3]$. It was proved in [LS22c] that Two-Choice leads to a gap of $\Theta(b/n)$. As our main result, we prove that the gap reduces to $O(\sqrt{(b/n)\cdot\log n})$, if one runs the $(1+\beta)$-process with an appropriately chosen $\beta$ (in fact this result holds for a larger class of processes). This not only proves the phenomenon that Two-Choice is not the best (leading to the formation of "towers" over previously light bins), but also that mixing two processes (One-Choice and Two-Choice) leads to a process which achieves a gap that is asymptotically smaller than both. We also derive a matching lower bound of $\Omega(\sqrt{(b/n)\cdot\log n})$ for any allocation process, which demonstrates that the above $(1+\beta)$-process is asymptotically optimal.
翻译:在平衡分配中,目标是将$m$个球分配到$n$个盒子中,以尽量减小差距(最大负载与平均负载之差)。单一选择过程将每个球分配到一个随机且独立的盒子中。双重选择过程将球放入两个已选择最小负载的盒子中的一个。最后,“$(1+\beta)$”过程混合这两种过程,也就是每个球按照概率$\beta\in(0,1)$使用双重选择,否则使用单一选择。尽管在顺序设置中双重选择是最佳的,但在并行环境中,它不会表现得很好,因为负载信息可能已过时。根据[BCEFN12]的方法,我们研究了这样一种并行设置,即球按批次大小$b$分配,同一批次内的球基于相同的策略和相同的负载信息进行分配。在小批次大小$b\in[n,n\log n]$的情况下,[LS22a]表明,双重选择在使用少量样本时可以实现渐近最优的差距。在这项工作中,我们关注更大的批次大小$b\in[n\log n,n^3]$。[LS22c]证明双重选择会导致$\Theta(b/n)$的差距。作为我们的主要结果,我们证明,如果使用适当选择的$\beta$运行“$(1+\beta)$”过程(事实上,这个结果适用于更大的过程类别),差距将缩小到$O(\sqrt{(b/n)\cdot\log n})$。这不仅证明了双重选择不是最佳选择(导致了先前轻负载的盒子形成“塔”),而且证明了混合两个过程(单一选择和双重选择)会形成一个过程,其差距渐近较小。我们还为任何分配过程导出了匹配的下界$\Omega(\sqrt{(b/n)\cdot\log n})$,这证明了上述“$(1+\beta)$”过程是渐近最优的。