修道士和野人问题状态空间
时间: 2024-06-12 18:10:04 浏览: 13
修道士和野人问题的状态空间可以用(Nm,Nc,B)来表示,其中Nm和Nc分别表示左岸上的修道士和野人数量,B表示船的位置,B=0表示船在左岸,B=1表示船在右岸。初始状态为(3,3,0),目标状态为(0,0,1)。中间状态包括(2,2,0)、(1,1,0)、(0,2,0)、(0,1,0)、(0,3,1)、(1,1,1)、(2,2,1)、(3,0,1)、(3,1,1)、(3,2,1)等。
相关问题
状态空间法野人与修道士Python
好的,您想了解有关状态空间法解决野人与修道士问题的 Python 实现是吗?
首先,您需要了解这个问题的规则:
在一条河上,有三个野人和三个修道士,他们想要过河。他们只有一个小船,这个小船最多能容纳两个人。如果河岸上的野人数量大于修道士的数量,那么修道士就会被野人杀死。如何才能使所有人都安全过河?
状态空间法是一种搜索算法,用于在有限状态空间中搜索目标状态。对于野人与修道士问题,状态空间法可以通过以下步骤来解决问题:
1. 定义状态:将野人和修道士的数量、船的位置等信息定义为状态。
2. 定义操作:将船上的人数、船的位置等信息定义为操作。
3. 确定初始状态:将所有人和船都放在起始岸。
4. 确定目标状态:将所有人和船都放在对岸。
5. 实现搜索算法:通过遍历状态空间,找到从初始状态到目标状态的路径。
以下是一个简单的 Python 实现,使用 BFS 算法来搜索状态空间,找到从初始状态到目标状态的路径:
```python
from collections import deque
def is_valid_state(state):
left_bank = state[0]
right_bank = state[1]
if left_bank[0] < left_bank[1] and left_bank[0] != 0:
return False
if right_bank[0] < right_bank[1] and right_bank[0] != 0:
return False
return True
def get_possible_states(state):
possible_states = []
left_bank = state[0]
right_bank = state[1]
if left_bank[2] == 1:
for i in range(3):
for j in range(3):
if i + j == 0 or i + j > 2:
continue
new_left_bank = (left_bank[0] - i, left_bank[1] - j, 0)
new_right_bank = (right_bank[0] + i, right_bank[1] + j, 1)
new_state = (new_left_bank, new_right_bank)
if is_valid_state(new_state):
possible_states.append(new_state)
else:
for i in range(3):
for j in range(3):
if i + j == 0 or i + j > 2:
continue
new_left_bank = (left_bank[0] + i, left_bank[1] + j, 1)
new_right_bank = (right_bank[0] - i, right_bank[1] - j, 0)
new_state = (new_left_bank, new_right_bank)
if is_valid_state(new_state):
possible_states.append(new_state)
return possible_states
def bfs(initial_state, goal_state):
visited = set()
queue = deque([(initial_state, [])])
while queue:
state, path = queue.popleft()
if state == goal_state:
return path
if state in visited:
continue
visited.add(state)
for next_state in get_possible_states(state):
queue.append((next_state, path + [next_state]))
initial_state = ((3, 3, 1), (0, 0, 0))
goal_state = ((0, 0, 0), (3, 3, 1))
path = bfs(initial_state, goal_state)
print(path)
```
该实现中,`is_valid_state` 函数用于检查状态是否合法,即左岸和右岸的修道士数量都大于等于野人数量,或者修道士数量为零。`get_possible_states` 函数用于获取当前状态下的所有可能状态,即将 1 或 2 个人从左岸或右岸运到另一岸。`bfs` 函数使用 BFS 算法来遍历状态空间,找到从初始状态到目标状态的路径。最后,打印出路径,即可得到所有安全过河方案的步骤。
状态空间法野人与修道士PYTHON
下面是使用 Python 实现状态空间法求解野人与修道士问题的代码:
```python
# 定义状态类
class State:
def __init__(self, cannibal_left, missionary_left, boat, cannibal_right, missionary_right):
self.cannibal_left = cannibal_left
self.missionary_left = missionary_left
self.boat = boat
self.cannibal_right = cannibal_right
self.missionary_right = missionary_right
self.parent = None
# 判断当前状态是否合法
def is_valid(self):
if self.missionary_left >= 0 and self.missionary_right >= 0 \
and self.cannibal_left >= 0 and self.cannibal_right >= 0 \
and (self.missionary_left == 0 or self.missionary_left >= self.cannibal_left) \
and (self.missionary_right == 0 or self.missionary_right >= self.cannibal_right):
return True
else:
return False
# 判断当前状态是否为目标状态
def is_goal(self):
if self.cannibal_left == 0 and self.missionary_left == 0 \
and self.cannibal_right == 3 and self.missionary_right == 3:
return True
else:
return False
# 生成当前状态的所有可能下一步状态
def generate_next_states(self):
next_states = []
if self.boat == 'left':
# 一名修道士过河
new_state = State(self.cannibal_left, self.missionary_left - 1, 'right',
self.cannibal_right, self.missionary_right + 1)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
# 一名野人过河
new_state = State(self.cannibal_left - 1, self.missionary_left, 'right',
self.cannibal_right + 1, self.missionary_right)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
# 一名修道士和一名野人过河
new_state = State(self.cannibal_left - 1, self.missionary_left - 1, 'right',
self.cannibal_right + 1, self.missionary_right + 1)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
else:
# 一名修道士过河
new_state = State(self.cannibal_left, self.missionary_left + 1, 'left',
self.cannibal_right, self.missionary_right - 1)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
# 一名野人过河
new_state = State(self.cannibal_left + 1, self.missionary_left, 'left',
self.cannibal_right - 1, self.missionary_right)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
# 一名修道士和一名野人过河
new_state = State(self.cannibal_left + 1, self.missionary_left + 1, 'left',
self.cannibal_right - 1, self.missionary_right - 1)
if new_state.is_valid():
new_state.parent = self
next_states.append(new_state)
return next_states
# 广度优先搜索算法
def bfs(start_state):
queue = [start_state]
visited = set()
while queue:
current_state = queue.pop(0)
if current_state.is_goal():
return current_state
visited.add(current_state)
next_states = current_state.generate_next_states()
for state in next_states:
if state not in visited:
queue.append(state)
return None
# 输出结果
def print_solution(solution):
path = []
current_state = solution
while current_state:
path.append(current_state)
current_state = current_state.parent
path.reverse()
for state in path:
print("(" + str(state.cannibal_left) + "," + str(state.missionary_left) + "," + state.boat \
+ "," + str(state.cannibal_right) + "," + str(state.missionary_right) + ")")
return
# 测试代码
start_state = State(3, 3, 'left', 0, 0)
solution = bfs(start_state)
print_solution(solution)
```
运行以上代码,将输出所有人都安全过河的方案。
相关推荐
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)
![text/x-c](https://img-home.csdnimg.cn/images/20210720083646.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)