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Currently, fictitious play stores count vectors and recomputes each time the payoff of each move by matrix multiplication before selecting the best move.
Currently, fictitious play stores count vectors and recomputes each time the payoff of each move by matrix multiplication before selecting the best move.
move = payoff.argmax()
costscount[move] += 1
costspayoff = matrix @ count
costsInstead, one could just update the payoff vector with$\mathcal{O}(n)$ .
payoff += matrix[:, move]
inEven better, the matrix can be stored transposed to be able to write
matrix[move]
which is more cache efficient.The text was updated successfully, but these errors were encountered: