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it has been observed in practice that it does not perform well in a parallel environment, where load information may be outdated.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：在批量负载均衡问题中，目标是将$m$个球放入$n$个箱子，以最小化间隙（最大负载与平均负载之差）。单选择过程将每个球独立均匀随机地放入一个箱子。双选择过程则将球放入两个采样箱子中负载较轻的一个。最后，$(1+\beta)$过程混合了这些策略：每个球以概率$\beta\in(0,1)$采用双选择，否则采用单选择。尽管双选择在顺序场景中是最优的，但实践中发现它在并行环境中表现不佳，因为负载信息可能过时。依据[BCEFN12]，我们研究这样的并行设置：球以大小为$b$的批次分配，同一批次内的球采用相同策略并基于相同的负载信息。对于小批次大小$b\in[n,n\log n]$，[LS22a]已证明双选择在具有常数个采样的所有过程中能达到渐近最优间隙。本文关注更大批次大小$b\in[n\log n,n^3]$。[LS22c]证明双选择导致间隙为$\Theta(b/n)$。作为主要结果，我们证明若采用适当选择$\beta$的$(1+\beta)$过程，间隙可降至$O(\sqrt{(b/n)\cdot\log n})$（实际上该结论对更大类别的过程成立）。这不仅证实了双选择并非最优的现象（导致"塔"结构形成于先前轻载的箱子上），还表明混合两种过程（单选择和双选择）能产生间隙渐近小于两者的过程。我们还为任何分配过程推导出匹配的下界$\Omega(\sqrt{(b/n)\cdot\log n})$，证明上述$(1+\beta)$过程是渐近最优的。