Solution of 6.6.3, Axelrod 1980



In [1]:

    
import numpy as np

Implement the five strategies



In [2]:

    
# We are going to implement five strategies. 
# Each strategy takes as input the history of the turns played so far
# and returns 1 for cooperation and 0 for defection.

# 1) Always defect
def always_defect(previous_steps):
    return 0

# 2) Always cooperate
def always_cooperate(previous_steps):
    return 1

# 3) Purely random, with probability of defecting 0.5
def random(previous_steps):
    if np.random.random(1) > 0.5:
        return 1
    return 0

# 4) Tit for tat
def tit_for_tat(previous_steps):
    if len(previous_steps) == 0:
        return 1
    return previous_steps[-1]

# 5) Tit for two tat
def tit_for_two_tat(previous_steps):
    if len(previous_steps) < 2:
        return 1
    # if the other player defected twice
    if sum(previous_steps[-2:]) == 0:
        # retaliate
        return 0
    return 1

Write a function that accepts the name of two strategies and competes them in a game of iterated prisoner's dilemma for a given number of turns.

You could implement a series of if elif that plays each strategy against the other. Here, we present a more advanced approach that matches a string such as "strategy_1" with a name of a corresponding function. The call globals()[strategy_1] does just that. Now pl1 is an "alias" that calls the function that corresponds to the chosen strategy.



In [3]:

    
def play_strategies(strategy_1, strategy_2, nsteps = 200):
    pl1 = globals()[strategy_1]
    pl2 = globals()[strategy_2]    
    # We create two vectors to store the moves of the players
    steps_pl1 = []
    steps_pl2 = []
    # and two variables for keeping the scores. 
    # (because we said these are numbers of years in prison, we 
    # use negative payoffs, with less negative being better)
    points_pl1 = 0
    points_pl2 = 0
    # Iterate over the number of steps
    for i in range(nsteps):
        # decide strategy:
        # player 1 chooses using the history of the moves by player 2
        last_pl1 = pl1(steps_pl2) 
        # and vice versa
        last_pl2 = pl2(steps_pl1)
        # calculate payoff
        if last_pl1 == 1 and last_pl2 == 1:
            # both cooperate -> -1 point each
            points_pl1 = points_pl1 - 1
            points_pl2 = points_pl2 - 1
        elif last_pl1 == 0 and last_pl2 == 1:
            # pl2 lose
            points_pl1 = points_pl1 - 0
            points_pl2 = points_pl2 - 3
        elif last_pl1 == 1 and last_pl2 == 0:
            # pl1 lose
            points_pl1 = points_pl1 - 3
            points_pl2 = points_pl2 - 0
        else:
            # both defect
            points_pl1 = points_pl1 - 2
            points_pl2 = points_pl2 - 2
        # add the moves to the history
        steps_pl1.append(last_pl1)
        steps_pl2.append(last_pl2)
    # return the final scores
    return((points_pl1, points_pl2))



In [4]:

    
# Your numbers will differ given the involved randomness
play_strategies("random", "always_defect")









    Out[4]:





(-506, -188)

Implement a round-robin tournament, in which each strategy is played against every other (including against itself) for 10 rounds of 1000 turns each.



In [5]:

    
def round_robin(strategies, nround, nstep):
    nstrategies = len(strategies)
    # initialize list for results
    strategies_points = [0] * nstrategies
    # for each pair
    for i in range(nstrategies):
        for j in range(i, nstrategies):
            print("Playing", strategies[i], "vs.", strategies[j])
            for k in range(nround):
                res = play_strategies(strategies[i], 
                                      strategies[j], 
                                      nstep)
                # print(res)
                strategies_points[i] = strategies_points[i] + res[0]
                strategies_points[j] = strategies_points[j] + res[1]
    print("\nThe final results are:")
    for i in range(nstrategies):
        print(strategies[i] + ":", strategies_points[i])
    print("\nand the winner is....")
    print(strategies[strategies_points.index(max(strategies_points))])



In [6]:

    
my_strategies = ["always_defect",
                 "always_cooperate", 
                 "random", 
                 "tit_for_tat", 
                 "tit_for_two_tat"]



In [7]:

    
# Your numbers will differ slightly given the involved randomness
round_robin(my_strategies, 10, 1000)









    



Playing always_defect vs. always_defect
Playing always_defect vs. always_cooperate
Playing always_defect vs. random
Playing always_defect vs. tit_for_tat
Playing always_defect vs. tit_for_two_tat
Playing always_cooperate vs. always_cooperate
Playing always_cooperate vs. random
Playing always_cooperate vs. tit_for_tat
Playing always_cooperate vs. tit_for_two_tat
Playing random vs. random
Playing random vs. tit_for_tat
Playing random vs. tit_for_two_tat
Playing tit_for_tat vs. tit_for_tat
Playing tit_for_tat vs. tit_for_two_tat
Playing tit_for_two_tat vs. tit_for_two_tat

The final results are:
always_defect: -89916
always_cooperate: -90012
random: -85013
tit_for_tat: -74957
tit_for_two_tat: -77465

and the winner is....
tit_for_tat