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arXiv 1412.0969cs.GTcs.CC
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关于2-局人对称Nash均衡的几个公开问题的解决

Settling Some Open Problems on 2-Player Symmetric Nash Equilibria

Ruta Mehta, Vijay V. Vazirani, Sadra Yazdanbod

摘要:

多年来,研究人员已经研究了(对称)两人博弈(双矩阵博弈)中纳什均衡的几个决策版本的复杂性。据我们所知,这类最后一个悬而未决的问题是:在对称博弈中找到一个非对称的纳什均衡(NE)。我们证明了这个问题是NP-完全的,而在对称博弈中计算非对称NE的个数的问题是P-完全的。在2005年,Kannan和Theobald定义了由矩阵(A,B)表示的“双矩阵博弈的秩”为秩(A+B),并询问了在多项式时间内是否可以计算秩1博弈的NE。观察到秩为0的情况正好是零和的情况,对于这种情况,从冯·诺依曼将这种博弈简化为线性规划,可以得到多项式时间算法。2011年,Adsul et.等人获得了秩1游戏的算法;然而,它没有解决对称秩1游戏的情况。我们解决这个问题。

英文摘要:

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 solve the case of symmetric rank 1 games. We resolve this problem.

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