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Settling Some Open Problems on 2-Player Symmetric Nash Equilibria

机译:在2人对称纳什均衡解决一些打开问题

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Over the years, researchers have studied the complexity of several decision versions of Nash equilibrium in (symmetric) two-player games (bimatrix games). To the best of our knowledge, the last remaining open problem of this sort is the following; it was stated by Papadimitriou in 2007: find a non-symmetric Nash equilibrium (NE) in a symmetric game. We show that this problem is NP-complete and the problem of counting the number of non-symmetric NE in a symmetric game is #P-complete. In 2005, Kannan and Theobald defined the rank of a bimatrix game represented by matrices (A, B) to be rank(A + B) and asked whether a NE can be computed in rank 1 games in polynomial time. Observe that the rank 0 case is precisely the zero sum case, for which a polynomial time algorithm follows from von Neumann's reduction of such games to linear programming. In 2011, Adsul et al. obtained an algorithm for rank 1 games; however, it does not guarantee symmetric NE in symmetric rank 1 game. We resolve this problem.
机译:多年来,研究人员已经研究了纳什均衡的多个决定版本(对称)双人游戏(Bimatrix游戏)的复杂性。据我们所知,这一排序的最后一个剩下的突出问题是以下; 2007年Papadimitriou表示:在对称游戏中找到一个非对称的NASH均衡(NE)。我们展示了这个问题是NP完整的,并且计算对称游戏中的非对称网元数的问题是#p-temply。 2005年,kannan和theobald定义了由矩阵(a,b)表示的Bimatrix游戏的等级,以等级(a + b),并询问NE可以在多项式时间中排名1游戏。观察到秩0案例正是零和案例,其中von neumann减少此类游戏的多项式时间算法到线性编程。 2011年,Adsul等人。获得了秩1游戏的算法;但是,它不保证对称级别1游戏中的对称NE。我们解决这个问题。

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