Finding a simple Nash Equilibrium

Non-cooperative game theory has been tightly combined with Computational Methods, in which the Nash Equilibrium (NE) is undoubtedly the most important solution concept. While players face several Nash Equilibriums, their most pragmatic strategy may not be the most optimal one, but one they prefer for other reasons. Players may choose a sub-optimal strategy instead of a more optimal but more complex one which might be difficult to learn or to implement. In one word, players prefer strategies as simple as possible. The problem is whether there exists a simple NE in a 2player game, and if so, how to find it. To solve this problem, this paper proposes a search method. Comparing with finding a NE, it is much easier to compute whether a NE exists in a special support for each player. Based on this, our algorithm repeats checking the existence of NE in a limited and ordered support space until a NE is found. There are 3 important steps in our algorithm: first limiting the search space; second ordering the search space; lastly, checking in turn (according to the order) whether there exists a NE with this support for each player. We are not the first ones who search support space to find a NE. Dickhaut and Kaplan proposed an algorithm to find all Nash Equlibria. Porter, Nudelman and Shoham designed an algorithm to find a sample NE. However, the innovation in this paper is on pruning the search space. Our pruning rules are (1) no conditionally dominant strategies in any NE, (2) there exists such a NE whose support is less than k+1 (k is the rank of corresponding payoff matrix). These rules ensure that our search space is smaller than that of [3], but the disadvantage is that ranks of payoff matrixes must be computed. The rest of this paper is structured as follows: first we formulate the problem and give the relevant definitions, and then we describe the algorithm. After that, we compare the execution of our algorithm and the LemkeHowson algorithm. In the final section, we conclude our work.