In the Iterated Prisoner’s Dilemma, two strategies play against each other over multiple rounds. Every round, the strategies can decide to either Coorperate (C) or Defect (D). If both cooperate, they both get 3 points. If both defect, they both get 1 point. If one cooperates and one defects, the one that defects gets 6 points, and the one that cooperates gets nothing. The goal for each strategy is to score as many points as possible.

Below a simple version of the Iterated Prisoner’s Dilemma is coded. A strategy to play the game is defined by the class Strategy. The main loop lets two strategies play each other for 100 rounds (it is not hard to create a main loop that lets more than two strategies play each other in pairs, but that increases the size of the code quite a bit and is not important for the exercise). Strategy has not implemented the choice() method. To create a strategy, you inherit a new class from Strategy, and code the choice() method. Optionally you can also implement the lastmove() method, and extend the __init__() method.

Implement the following strategies:

• Random just plays COOPERATE or DEFECT at random.

• AlwaysDefect always plays DEFECT.

• TitForTat starts with COOPERATE, then plays what the opponent played on the previous move (the lastmove() method gets to see what the opponent played after a choice has been made).

• TitForTwoTats starts with two COOPERATEs, then plays DEFECT if the opponent played DEFECT on both the previous two moves, otherwise COOPERATEs.

• Majority starts with COOPERATE, then plays what the opponent played on the majority of the previous moves.

If you want to implement more strategies, be my guest. Test out some of the strategies against each other by filling in the assignments for strategy1 and strategy2 (do not forget to give them a name between the parentheses).

Note that the shorthand way that I use in this code to write a simple condition (with a statement like 3 if c1 == COOPERATE else 1), which looks like a list comprehension (see Chapter 13¹), is just to save some space and make the code a bit more readable. It would be just as well to write the 4 lines of code that would be needed to do this with a regular if statement.

# Iterated Prisoner's Dilemma
COOPERATE = 'C'
DEFECT = 'D'
ROUNDS = 100

class Strategy:
    def __init__( self, name="" ):
        self.name = name
        self.score = 0
    def choice( self ):
        # Should return COOPERATE or DEFECT
        return NotImplemented
    def lastmove( self, mymove, opponentmove ):
        # Gets passed the last move made, after a call of choice()
        pass
    def incscore( self, n ):
        self.score += n

strategy1 = Strategy()
strategy2 = Strategy()

for i in range( ROUNDS ):
    c1 = strategy1.choice()
    c2 = strategy2.choice()
    if c1 == c2:
        strategy1.incscore( 3 if c1 == COOPERATE else 1 )
        strategy2.incscore( 3 if c2 == COOPERATE else 1 )
    else:
        strategy1.incscore( 0 if c1 == COOPERATE else 6 )
        strategy2.incscore( 0 if c2 == COOPERATE else 6 )
    strategy1.lastmove( c1, c2 )
    strategy2.lastmove( c2, c1 )

print( "End score of", strategy1.name, "is", strategy1.score )
print( "End score of", strategy2.name, "is", strategy2.score )