A Note on Strategy Elimination in Bimatrix Games

A two person (or bimatrix) game is a pair of m x n matrices A. B, with integer entries. This game is played between two players A and B. Player A chooses a row t. player B simultaneously chooses a column j. As a result. A receives a,, (dollars, say), and B receives b,,. An easy way to simplify a bimatrix game is to eliminate from both A and B any strategy (row or column) that is dominated by another. We say that strategy i of player A dominates strategy i' of the same player if a,1 > a,-, for j = 1...., n. Similarly, strategy j of player B dominates strategy j' of the same player if b,j, > b,, for i = 1,.... m. Obviously, both player A and B can avoid playing dominated strategies, with no deterioration of the outcome. Thus, all dominated strategies can be eliminated. Furthermore, deletion of dominated strategies may result in new domination, and so on, until we are left with a reduced game, in which no further elimination -is possible. Strategy domination, and its obvious generalization to many players. has long been a well-known notion of simplification in games. More recently, a new concept of rationality was proposed, based on strategy domination [1,41. In this note we look at the algorithmic aspects of the process of producIng a reduced game. The obvious algorithm for doing this is to repeatedly test each pair of strategies for domination, and to delete the strategies that were found to be dominated. Since there are O(m2 + n2 ) pairs to be examined, each test entails checking n (or m) inequalities, and the whole process must be repeated after each row or column deletion, the most straightforward algorithm takes O(mn(m + n)2) = O(m + n) 4 time overall. But there is a faster (O(m + n) 3 ) algorithm, based on a rather well-known technique that can be called 'obstruction counts'. For each pair of strategies, we can maintain and update a count of the opponent strategies that prevent the first from dominating the second. We do not know how to improve this algorithm asymptotically. In fact, it is not at all obvious how to determine in O((m + n )) time whether there is any domination (in other words, whether or not