Adversarial Combinatorial Bandits with Switching Costs
- URL: http://arxiv.org/abs/2404.01883v1
- Date: Tue, 2 Apr 2024 12:15:37 GMT
- Title: Adversarial Combinatorial Bandits with Switching Costs
- Authors: Yanyan Dong, Vincent Y. F. Tan,
- Abstract summary: We study the problem of adversarial bandit with a switching cost $lambda$ for a switch of each selected arm in each round.
We derive lower bounds for the minimax regret and design algorithms to approach them.
- Score: 55.2480439325792
- License: http://creativecommons.org/licenses/by/4.0/
- Abstract: We study the problem of adversarial combinatorial bandit with a switching cost $\lambda$ for a switch of each selected arm in each round, considering both the bandit feedback and semi-bandit feedback settings. In the oblivious adversarial case with $K$ base arms and time horizon $T$, we derive lower bounds for the minimax regret and design algorithms to approach them. To prove these lower bounds, we design stochastic loss sequences for both feedback settings, building on an idea from previous work in Dekel et al. (2014). The lower bound for bandit feedback is $ \tilde{\Omega}\big( (\lambda K)^{\frac{1}{3}} (TI)^{\frac{2}{3}}\big)$ while that for semi-bandit feedback is $ \tilde{\Omega}\big( (\lambda K I)^{\frac{1}{3}} T^{\frac{2}{3}}\big)$ where $I$ is the number of base arms in the combinatorial arm played in each round. To approach these lower bounds, we design algorithms that operate in batches by dividing the time horizon into batches to restrict the number of switches between actions. For the bandit feedback setting, where only the total loss of the combinatorial arm is observed, we introduce the Batched-Exp2 algorithm which achieves a regret upper bound of $\tilde{O}\big((\lambda K)^{\frac{1}{3}}T^{\frac{2}{3}}I^{\frac{4}{3}}\big)$ as $T$ tends to infinity. In the semi-bandit feedback setting, where all losses for the combinatorial arm are observed, we propose the Batched-BROAD algorithm which achieves a regret upper bound of $\tilde{O}\big( (\lambda K)^{\frac{1}{3}} (TI)^{\frac{2}{3}}\big)$.
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