Balanced Allocations with Incomplete Information: The Power of Two Queries

We consider the problem of allocating $m$ balls into $n$ bins with incomplete information. In the classical two-choice process, a ball first queries the load of $\textit{two}$ randomly chosen bins and is then placed in the least loaded bin. In our setting, each ball also samples two random bins but can only estimate a bin's load by sending $\textit{binary queries}$ of the form "Is the load at least the median?" or "Is the load at least $100$?". For the lightly loaded case $m=O(n)$, one can achieve an $O(\sqrt{\log n/\log \log n})$ maximum load with one query per chosen bin using an oblivious strategy, as shown by Feldheim and Gurel-Gurevich (2018). For the case $m=\Omega(n)$, the authors conjectured that the same strategy achieves a maximum load of $m/n+O(\sqrt{\log n/\log \log n})$. In this work, we disprove this conjecture by showing a lower bound of $m/n+\Omega( \sqrt{\log n})$ for a fixed $m=\Theta(n \sqrt{\log n})$, and a lower bound of $m/n+\Omega(\log n/\log\log n)$ for some $m$ depending on the used strategy. Surprisingly, these lower bounds hold even for any $\textit{adaptive strategy}$ with one query, i.e., queries may depend on the full history of the process. We complement this negative result by proving a positive result for multiple queries. In particular, we show that with only two binary queries per chosen bin, there is an oblivious strategy which ensures a maximum load of $m/n+O(\sqrt{\log n})$ whp for any $m \geq 1$. For any $k=O(\log \log n)$ binary queries, the upper bound on the maximum load improves to $m/n+O(k(\log n)^{1/k})$ whp for any $m \geq 1$. Hence for $k=\Theta(\log\log n)$, we recover the two-choice result up to a constant multiplicative factor, including the heavily loaded case where $m=\Omega(n)$. One novel aspect of our proof techniques is the use of multiple super-exponential potential functions, which might be of use in future work.