Exploitability Lite

Problem

Compute the exploitability of a strategy profile in a two-player zero-sum game.

A is player 1's payoff matrix. br1 is the best pure response to q, and br2 is player 2's best response to p (minimizing player 1's payoff). Return (br1 - br2) / 2, which is zero at a Nash equilibrium.

Hint

Exploitability halves the gap between the two best-response values.

Test cases

case 1

in: [[[1,-1],[-1,1]],[0.5,0.5],[0.5,0.5]]

expected: 0

case 2

in: [[[3,0],[0,1]],[1,0],[1,0]]

expected: 1.5

case 3

in: [[[0,2],[1,0]],[0.5,0.5],[0.5,0.5]]

expected: 0.25

case 4

in: [[[1,0],[0,1]],[1,0],[0,1]]

expected: 0.5

Starter codePython

def exploitability_lite(A, p, q):
    # Your code here
    pass

4 test cases · run your code against them in the editor