Simple Combinatorial Algorithms for Combinatorial Bandits: Corruptions and Approximations

TL;DR

This paper introduces a simple, combinatorial algorithm for stochastic combinatorial semi-bandits with adversarial corruptions, achieving near-optimal regret bounds with lower complexity and weaker assumptions compared to prior methods.

Contribution

The paper presents a new combinatorial algorithm that improves regret bounds for corrupted semi-bandit problems, simplifying implementation and reducing computational complexity.

Findings

Achieves regret of  C + d^2K/elta_{min} with adversarial corruptions.

Outperforms previous combinatorial algorithms in regret bounds.

Requires weaker assumptions and has lower oracle complexity than existing methods.

Abstract

We consider the stochastic combinatorial semi-bandit problem with adversarial corruptions. We provide a simple combinatorial algorithm that can achieve a regret of $\tilde{O}\left(C+d^2K/\Delta_{min}\right)$ where $C$ is the total amount of corruptions, $d$ is the maximal number of arms one can play in each round, $K$ is the number of arms. If one selects only one arm in each round, we achieves a regret of $\tilde{O}\left(C+\sum_{\Delta_i>0}(1/\Delta_i)\right)$. Our algorithm is combinatorial and improves on the previous combinatorial algorithm by [Gupta et al., COLT2019] (their bound is $\tilde{O}\left(KC+\sum_{\Delta_i>0}(1/\Delta_i)\right)$), and almost matches the best known bounds obtained by [Zimmert et al., ICML2019] and [Zimmert and Seldin, AISTATS2019] (up to logarithmic factor). Note that the algorithms in [Zimmert et al., ICML2019] and [Zimmert and Seldin, AISTATS2019] require one to solve complex convex programs while our algorithm is combinatorial, very easy to implement, requires weaker assumptions and has very low oracle complexity and running time. We also study the setting where we only get access to an approximation oracle for the stochastic combinatorial semi-bandit problem. Our algorithm achieves an (approximation) regret bound of $\tilde{O}\left(d\sqrt{KT}\right)$. Our algorithm is very simple, only worse than the best known regret bound by $\sqrt{d}$, and has much lower oracle complexity than previous work.