Constant rank bimatrix games are PPAD-hard

R. Mehta
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引用次数: 37

Abstract

The rank of a bimatrix game (A, B) is defined as rank(A + B). Computing a Nash equilibrium (NE) of a rank-0, i.e., zero-sum game is equivalent to linear programming (von Neumann'28, Dantzig'51). In 2005, Kannan and Theobald gave an FPTAS for constant rank games, and asked if there exists a polynomial time algorithm to compute an exact NE. Adsul et. al. (2011) answered this question affirmatively for rank-1 games, leaving rank-2 and beyond unresolved. In this paper we show that NE computation in games with rank ≥ 3, is PPAD-hard, settling a decade long open problem. Interestingly, this is the first instance that a problem with an FPTAS turns out to be PPAD-hard. Our reduction bypasses graphical games and game gadgets, and provides a simpler proof of PPAD-hardness for NE computation in bimatrix games. In addition, we get: • An equivalence between 2D-Linear-FIXP and PPAD, improving a result by Etessami and Yannakakis (2007) on equivalence between Linear-FIXP and PPAD. • NE computation in a bimatrix game with convex set of Nash equilibria is as hard as solving a simple stochastic game [12]. • Computing a symmetric NE of a symmetric bimatrix game with rank ≥ 6 is PPAD-hard. • Computing a 1/poly(n)-approximate fixed-point of a (Linear-FIXP) piecewise-linear function is PPAD-hard. The status of rank-2 games remains unresolved.
固定等级双矩阵游戏是ppad难度的
双矩阵博弈(a, B)的秩定义为秩(a + B)。计算秩为0的纳什均衡(NE),即零和博弈等价于线性规划(von Neumann'28, Dantzig'51)。2005年,Kannan和Theobald给出了常秩博弈的FPTAS,并询问是否存在多项式时间算法来计算精确的NE。Adsul等人(2011)对排名第一的游戏肯定地回答了这个问题,而对排名第二或更高的游戏则没有答案。在本文中,我们证明了秩≥3的博弈中的NE计算是PPAD-hard的,解决了一个长达十年的开放问题。有趣的是,这是FPTAS问题第一次被证明是ppad难题。我们的简化绕过了图形游戏和游戏小工具,并为双矩阵游戏中的NE计算提供了更简单的ppad -硬度证明。•2D-Linear-FIXP与PPAD之间的等价性,改进了Etessami和Yannakakis(2007)关于Linear-FIXP与PPAD之间等价性的结果。•在具有纳什均衡凸集的双矩阵博弈中计算NE与求解简单随机博弈一样困难[12]。•计算秩≥6的对称双矩阵博弈的对称网元是PPAD-hard。•计算(Linear-FIXP)分段线性函数的1/poly(n)-近似不动点是PPAD-hard。排名第二的游戏的地位仍然悬而未决。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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