  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.
