In the storied Colonel Blotto game, two colonels allocate $a$ and $b$ troops, respectively, to $k$ distinct battlefields. A colonel wins a battle if they assign more troops to that particular battle, and each colonel seeks to maximize their total number of victories. Despite the problem's formulation in 1921, the first polynomial-time algorithm to compute Nash equilibrium (NE) strategies for this game was discovered only quite recently. In 2016, \citep{ahmadinejad_dehghani_hajiaghayi_lucier_mahini_seddighin_2019} formulated a breakthrough algorithm to compute NE strategies for the Colonel Blotto game\footnote{To the best of our knowledge, the algorithm from \citep{ahmadinejad_dehghani_hajiaghayi_lucier_mahini_seddighin_2019} has computational complexity $O(k^{14}\max\{a,b\}^{13})$}, receiving substantial media coverage (e.g. \citep{Insider}, \citep{NSF}, \citep{ScienceDaily}). In this work, we present the first known $\epsilon$-approximation algorithm to compute NE strategies in the two-player Colonel Blotto game in runtime $\widetilde{O}(\epsilon^{-4} k^8 \max\{a,b\}^2)$ for arbitrary settings of these parameters. Moreover, this algorithm computes approximate coarse correlated equilibrium strategies in the multiplayer (continuous and discrete) Colonel Blotto game (when there are $\ell > 2$ colonels) with runtime $\widetilde{O}(\ell \epsilon^{-4} k^8 n^2 + \ell^2 \epsilon^{-2} k^3 n (n+k))$, where $n$ is the maximum troop count. Before this work, no polynomial-time algorithm was known to compute exact or approximate equilibrium (in any sense) strategies for multiplayer Colonel Blotto with arbitrary parameters. Our algorithm computes these approximate equilibria by a novel (to the author's knowledge) sampling technique with which we implicitly perform multiplicative weights update over the exponentially many strategies available to each player.
翻译:在存储的布洛托上校的游戏中, 两位上校将美元和美元分别分配到不同的战场上。 一位上校如果派更多的部队参加这场特定战斗, 将赢得一场战斗, 每位上校将最大限度地实现他们的胜利总数。 尽管这个问题在1921年的提法, 但这个游戏的计算纳什平衡( NE) 的第一次多边时间算法直到最近才被发现。 在2016年, 将2美元计算到2美元, 2美元, 2美元, 2美元, 3美元, 4美元, 4美元, 4美元, 4美元, 5美元, 5美元, 5美元, 5美元, 3美元, 3美元, 4美元, 3美元, 3美元, 3美元, 3美元, 3美元, 3美元, 3美元, 3美元, 3美元, 4美元, 3美元, 3美元, 3美元, 3美元, 3美元, 4美元, 4美元, com, com, i, i, i, i, i, i, i, i, com, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, i, si, si, si, si, si, si, si, si, i, i, si, si, si, si, si, si, si, si, si, si, si, si, si, si, si, si, si, li, li, li, li, li, li, li, si, si