异构容器的平衡分配：内存的力量
Balanced Allocations with Heterogeneous Bins: The Power of Memory
摘要：
我们考虑将$m$球（作业）分配到$n$箱（服务器）中。在标准的二选一过程中，在每一步$t=1,2,\ldots,m$中，我们首先随机均匀地采样两个箱，并将一个球放置在负载最小的箱中。众所周知，对于任何$m \geq n$，这会导致$\log_2 \log n + Θ(1)$的间隙（最大负载和平均负载之间的差异）（概率很高）。在这项工作中，我们考虑的内存过程[Mitzenmacher，Prabhakar和沙阿2002年]，而不是两个选择，我们只采样一个bin每一步，但我们可以访问一个缓存，可以存储一个bin的位置。Mitzenmacher、Prabhakar和Shah表明，在轻负载情况下（$m = n$），记忆过程实现了$\mathcal{O}(\log \log n)$的间隙。以两种方式扩展Mitzenmacher等人的设置，我们首先允许球数$m$是任意的，这包括具有挑战性的重负载情况，其中$m \geq n$。其次，我们遵循Wieder [Wieder 2007]的异质箱模型，其中箱的采样分布可以被偏置到某个任意的乘法常数。有些令人惊讶的是，我们证明，即使在这种情况下，记忆过程仍然达到了$\mathcal{O}(\log \log n)$的间隙界限。这与双选择（或任何$d$ -选择与$d=\mathcal{O}(1)$）过程形成鲜明对比，已知间隙发散为$m \rightarrow \infty$ [Wieder 2007]。此外，我们表明，对于任何采样分布独立于$m$（但可能依赖于$n$）的内存过程有一个间隙，可以绑定独立于$m$。最后，我们证明了一个紧间隙界的$\mathcal{O}(\log n)$的内存在另一个宽松的设置异构（加权）球和缓存，只能维持两个步骤。
英文摘要：
We consider the allocation of $m$ balls (jobs) into $n$ bins (servers). In the standard Two-Choice process, at each step $t=1,2,\ldots,m$ we first sample two bins uniformly at random and place a ball in the least loaded bin. It is well-known that for any $m \geq n$, this results in a gap (difference between the maximum and average load) of $\log_2 \log n + Θ(1)$ (with high probability). In this work, we consider the Memory process [Mitzenmacher, Prabhakar and Shah 2002] where instead of two choices, we only sample one bin per step but we have access to a cache which can store the location of one bin. Mitzenmacher, Prabhakar and Shah showed that in the lightly loaded case ($m = n$), the Memory process achieves a gap of $\mathcal{O}(\log \log n)$. Extending the setting of Mitzenmacher et al. in two ways, we first allow the number of balls $m$ to be arbitrary, which includes the challenging heavily loaded case where $m \geq n$. Secondly, we follow the heterogeneous bins model of Wieder [Wieder 2007], where the sampling distribution of bins can be biased up to some arbitrary multiplicative constant. Somewhat surprisingly, we prove that even in this setting, the Memory process still achieves an $\mathcal{O}(\log \log n)$ gap bound. This is in stark contrast with the Two-Choice (or any $d$-Choice with $d=\mathcal{O}(1)$) process, where it is known that the gap diverges as $m \rightarrow \infty$ [Wieder 2007]. Further, we show that for any sampling distribution independent of $m$ (but possibly dependent on $n$) the Memory process has a gap that can be bounded independently of $m$. Finally, we prove a tight gap bound of $\mathcal{O}(\log n)$ for Memory in another relaxed setting with heterogeneous (weighted) balls and a cache which can only be maintained for two steps.
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