囚徒困境合作、背叛、以牙还牙、随机python代码
时间: 2023-07-10 14:39:30 浏览: 77
关于囚徒困境,合作和背叛是两种策略。如果两个囚犯都合作,那么他们都可以获得最小的刑期。如果两个囚犯都背叛,那么他们都会受到较重的刑期。但是如果一个囚犯背叛而另一个囚犯合作,那么背叛的囚犯将获得自由,而合作的囚犯将受到最重的刑期。因此,在这种情况下,背叛是一个更有利的策略。
以牙还牙是一种策略,即在第一轮中合作,然后在后续的每一轮中,采取对前一轮对手的行为进行回应的方式。如果对手合作,则合作;如果对手背叛,则背叛。这种策略可以避免被对手不断背叛,同时也可以鼓励对手合作。
下面是一个随机生成Python代码的示例:
```python
import random
for i in range(10):
num = random.randint(1, 100)
if num % 2 == 0:
print(num, "is even")
else:
print(num, "is odd")
```
这段代码会生成10个1到100之间的随机整数,并判断它们是奇数还是偶数。然后将结果输出到控制台。
相关问题
囚徒困境合作、背叛、以牙还牙、随机策略python代码
以下是一个实现囚徒困境游戏的Python代码示例,其中包括合作、背叛、以牙还牙和随机策略:
```python
import random
# 定义囚徒困境游戏
def prisoner_dilemma(p1, p2):
if p1 == 'C' and p2 == 'C':
return (3, 3)
elif p1 == 'C' and p2 == 'D':
return (0, 5)
elif p1 == 'D' and p2 == 'C':
return (5, 0)
elif p1 == 'D' and p2 == 'D':
return (1, 1)
# 定义策略函数
def cooperate_strategy(opponent_moves):
return 'C'
def defect_strategy(opponent_moves):
return 'D'
def tit_for_tat_strategy(opponent_moves):
if opponent_moves:
return opponent_moves[-1]
else:
return 'C'
def random_strategy(opponent_moves):
return random.choice(['C', 'D'])
# 测试策略函数
strategies = [cooperate_strategy, defect_strategy, tit_for_tat_strategy, random_strategy]
for strategy in strategies:
score = 0
for i in range(10):
my_move = strategy([])
opponent_move = random_strategy([])
my_score, opponent_score = prisoner_dilemma(my_move, opponent_move)
score += my_score
print(f"Strategy {strategy.__name__} scores {score}")
```
在这个示例中,我们首先定义了囚徒困境游戏的规则,然后定义了四种不同的策略函数:合作、背叛、以牙还牙和随机策略。最后,我们对每一个策略函数进行了测试,并输出了它的得分。
囚徒困境以牙还牙python代码
下面是一个实现囚徒困境以牙还牙策略的 Python 代码示例:
```python
# 策略:以牙还牙
def tit_for_tat(last_opponent_action):
if last_opponent_action == 'cooperate':
return 'cooperate'
else:
return 'defect'
# 主函数
def play_game(player1, player2, num_rounds):
p1_cooperated = True
p2_cooperated = True
p1_score = 0
p2_score = 0
for i in range(num_rounds):
p1_action = player1('cooperate' if p1_cooperated else 'defect')
p2_action = player2('cooperate' if p2_cooperated else 'defect')
if p1_action == 'cooperate' and p2_action == 'cooperate':
p1_score += 3
p2_score += 3
p1_cooperated = True
p2_cooperated = True
elif p1_action == 'cooperate' and p2_action == 'defect':
p1_score += 0
p2_score += 5
p1_cooperated = True
p2_cooperated = False
elif p1_action == 'defect' and p2_action == 'cooperate':
p1_score += 5
p2_score += 0
p1_cooperated = False
p2_cooperated = True
else:
p1_score += 1
p2_score += 1
p1_cooperated = False
p2_cooperated = False
return p1_score, p2_score
# 测试代码
if __name__ == '__main__':
player1 = tit_for_tat
player2 = tit_for_tat
num_rounds = 5
p1_score, p2_score = play_game(player1, player2, num_rounds)
print(f'Player 1 score: {p1_score}')
print(f'Player 2 score: {p2_score}')
```
这个代码实现了一个简单的囚徒困境游戏,其中两个玩家使用相同的以牙还牙策略来做出决策。在每一轮中,两个玩家都会做出决策,然后根据它们的决策来更新分数和合作状态。最后,输出两个玩家的得分。
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