用A*算法实现八数码难题
时间: 2023-07-10 15:12:14 浏览: 49
下面是用A*算法实现八数码难题的伪代码:
```
function Astar_Search(initial_state, goal_state):
open_list = [initial_state] // 开放列表,初始状态加入其中
closed_list = [] // 封闭列表,不包含初始状态
g_score = {} // g值,表示从初始状态到当前状态的代价
g_score[initial_state] = 0 // 初始状态的代价为0
f_score = {} // f值,表示从初始状态到目标状态的估计代价
f_score[initial_state] = heuristic(initial_state, goal_state) // 初始状态的估计代价
parent = {} // 父节点,用于回溯
while open_list is not empty:
current = state in open_list with the lowest f_score // 从开放列表中选取f值最小的状态
open_list.remove(current)
closed_list.append(current)
if current == goal_state: // 找到目标状态,回溯路径
path = []
while current in parent:
path.append(current)
current = parent[current]
path.append(initial_state)
path.reverse()
return path
for neighbor in get_neighbors(current): // 扩展当前状态的邻居节点
if neighbor in closed_list:
continue // 跳过封闭列表中的节点
tentative_g_score = g_score[current] + 1 // 计算邻居节点的代价,每次移动一个数字或空格算作一步
if neighbor not in open_list or tentative_g_score < g_score[neighbor]:
parent[neighbor] = current
g_score[neighbor] = tentative_g_score
f_score[neighbor] = g_score[neighbor] + heuristic(neighbor, goal_state)
if neighbor not in open_list:
open_list.append(neighbor) // 加入开放列表,如果邻居节点不在开放列表中
return [] // 无解
function heuristic(state, goal_state):
// 启发函数,计算从当前状态到目标状态的估计代价,使用曼哈顿距离
distance = 0
for i in range(1, 9):
current_pos = get_position(state, i)
goal_pos = get_position(goal_state, i)
distance += abs(current_pos[0] - goal_pos[0]) + abs(current_pos[1] - goal_pos[1])
return distance
function get_position(state, value):
// 获取数字value在状态state中的位置
for i in range(3):
for j in range(3):
if state[i][j] == value:
return (i, j)
function get_neighbors(state):
// 获取当前状态的邻居节点
neighbors = []
empty_pos = get_position(state, 0)
for move in [(0, -1), (-1, 0), (0, 1), (1, 0)]:
new_pos = (empty_pos[0] + move[0], empty_pos[1] + move[1])
if 0 <= new_pos[0] < 3 and 0 <= new_pos[1] < 3:
new_state = swap(state, empty_pos, new_pos)
neighbors.append(new_state)
return neighbors
function swap(state, pos1, pos2):
// 交换状态state中位置pos1和位置pos2上的数字或空格
new_state = [row[:] for row in state]
new_state[pos1[0]][pos1[1]] = state[pos2[0]][pos2[1]]
new_state[pos2[0]][pos2[1]] = state[pos1[0]][pos1[1]]
return new_state
```
在实现A*算法时,需要注意以下几点:
1. 需要使用优先队列来维护开放列表,确保每次选取f值最小的状态进行扩展。
2. 启发函数和代价函数的设计很关键,可以影响算法的效率和正确性。对于八数码难题,可以使用曼哈顿距离作为启发函数,每次移动一个数字或空格算作一步。
3. 在计算估计代价时,需要使用从起始状态到当前状态的代价加上从当前状态到目标状态的启发函数值。
4. 通过回溯,可以找到从起始状态到目标状态的最优路径。