囚徒困境顺序激活机制代码+结果
时间: 2024-03-02 08:03:41 浏览: 14
下面是一个基于顺序激活机制的囚徒困境Python代码示例:
```python
# 两个囚徒的策略
def prisoner_1(history):
if not history:
return "C"
elif history[-1] == "C":
return "C"
else:
return "D"
def prisoner_2(history):
if not history:
return "C"
elif history[-1] == "C":
return "D"
else:
return "C"
# 顺序激活机制
def sequence_activation(player_1, player_2, iterations):
history_1 = []
history_2 = []
score_1 = 0
score_2 = 0
# 迭代次数
for i in range(iterations):
if i % 2 == 0:
move = player_1(history_2)
history_1.append(move)
score_1 += 3 if move == "C" and history_2[-1] == "C" else 0
score_1 += 5 if move == "D" and history_2[-1] == "C" else 0
score_1 += 1 if move == "C" and history_2[-1] == "D" else 0
score_1 += 1 if move == "D" and history_2[-1] == "D" else 0
else:
move = player_2(history_1)
history_2.append(move)
score_2 += 3 if move == "C" and history_1[-1] == "C" else 0
score_2 += 5 if move == "D" and history_1[-1] == "C" else 0
score_2 += 1 if move == "C" and history_1[-1] == "D" else 0
score_2 += 1 if move == "D" and history_1[-1] == "D" else 0
return (score_1, score_2)
# 运行游戏
print(sequence_activation(prisoner_1, prisoner_2, 10))
```
这个代码中,我们定义了两个囚徒的策略函数,`prisoner_1` 和 `prisoner_2`,以及一个顺序激活机制的函数 `sequence_activation`。在这个顺序激活机制中,每个囚徒在每一轮交互中轮流成为先手和后手。在每一轮中,先手玩家采取自己的策略,后手玩家采取自己的策略,并根据双方的决策计算得分。注意,先手玩家的策略函数使用后手玩家的历史记录,后手玩家的策略函数使用先手玩家的历史记录。
在这个示例中,我们运行了 10 局游戏并打印了最终分数。由于每个囚徒的策略函数都是确定性的,因此在不同的游戏中得分可能会有所不同。