Nearly Tight Bounds for Exploration in Streaming Multi-armed Bandits with Known Optimality Gap

TL;DR

This paper establishes nearly tight bounds on the number of passes needed for exploration in streaming multi-armed bandits with known optimality gap, showing that $ heta( ext{log} n)$ passes are necessary and sufficient.

Contribution

It provides the first tight bounds on pass complexity for streaming multi-armed bandits with known gaps, including a lower bound and a nearly matching algorithm.

Findings

Any algorithm with sublinear memory requires at least $rac{ ext{log} n}{ ext{log} ext{log} n}$ passes.

A nearly matching algorithm achieves the optimal sample complexity with a single arm memory.

The results clarify the trade-offs between passes, memory, and sample complexity in streaming bandit exploration.

Abstract

We investigate the sample-memory-pass trade-offs for pure exploration in multi-pass streaming multi-armed bandits (MABs) with the *a priori* knowledge of the optimality gap $\Delta_{[2]}$. Here, and throughout, the optimality gap $\Delta_{[i]}$ is defined as the mean reward gap between the best and the $i$-th best arms. A recent line of results by Jin, Huang, Tang, and Xiao [ICML'21] and Assadi and Wang [COLT'24] have shown that if there is no known $\Delta_{[2]}$, a pass complexity of $\Theta(\log(1/\Delta_{[2]}))$ (up to $\log\log(1/\Delta_{[2]})$ terms) is necessary and sufficient to obtain the *worst-case optimal* sample complexity of $O(n/\Delta^{2}_{[2]})$ with a single-arm memory. However, our understanding of multi-pass algorithms with known $\Delta_{[2]}$ is still limited. Here, the key open problem is how many passes are required to achieve the complexity, i.e., $O( \sum_{i=2}^{n}1/\Delta^2_{[i]})$ arm pulls, with a sublinear memory size. In this work, we show that the ``right answer'' for the question is $\Theta(\log{n})$ passes (up to $\log\log{n}$ terms). We first present a lower bound, showing that any algorithm that finds the best arm with slightly sublinear memory -- a memory of $o({n}/{\text{polylog}({n})})$ arms -- and $O(\sum_{i=2}^{n}{1}/{\Delta^{2}_{[i]}}\cdot \log{(n)})$ arm pulls has to make $\Omega(\frac{\log{n}}{\log\log{n}})$ passes over the stream. We then show a nearly-matching algorithm that assuming the knowledge of $\Delta_{[2]}$, finds the best arm with $O( \sum_{i=2}^{n}1/\Delta^2_{[i]} \cdot \log{n})$ arm pulls and a *single arm* memory.