python - 为什么最佳优先搜索 Python 实现没有给出正确的输出

Question

我尝试在 8 个难题上实现最佳优先搜索算法。但是无论我采用什么矩阵，我都会得到与 A* 代码相同的路径。另外，有人可以帮我打印每个矩阵下的启发式吗？我在输出中只得到“1”。

最佳首次搜索代码-

from copy import deepcopy
from collections import deque

class Node:
    def __init__(self, state=None, parent=None, cost=0, depth=0, children=[]):
        self.state = state
        self.parent = parent
        self.cost = cost
        self.depth = depth
        self.children = children

    def is_goal(self, goal_state):
        return is_goal_state(self.state, goal_state)

    def expand(self):
        new_states = operator(self.state)
        self.children = []
        for state in new_states:
            self.children.append(Node(state, self, self.cost + 1, self.depth + 1))

    def parents(self):
        current_node = self
        while current_node.parent:
            yield current_node.parent
            current_node = current_node.parent
    def gn(self):
        costs = self.cost
        for parent in self.parents():
            costs += parent.cost
        return costs


def is_goal_state(state, goal_state):
    for i in range(len(state)):
        for j in range(len(state)):
            if state[i][j] != goal_state[i][j]:
                return False
    return True


def operator(state):
    states = []

    zero_i = None
    zero_j = None

    for i in range(len(state)):
        for j in range(len(state)):
            if state[i][j] == 0:
                zero_i = i
                zero_j = j
                break

    def add_swap(i, j):
        new_state = deepcopy(state)
        new_state[i][j], new_state[zero_i][zero_j] = new_state[zero_i][zero_j], new_state[i][j]
        states.append(new_state)

    if zero_i != 0:
        add_swap(zero_i - 1, zero_j)

    if zero_j != 0:
        add_swap(zero_i, zero_j - 1)

    if zero_i != len(state) - 1:
        add_swap(zero_i + 1, zero_j)

    if zero_j != len(state) - 1:
        add_swap(zero_i, zero_j + 1)

    return states


R = int(input("Enter the number of rows:")) 
C = int(input("Enter the number of columns:")) 

# Initialize matrix 
inital = [] 
print("Enter the entries rowwise:") 

# For user input 
for i in range(R):          # A for loop for row entries 
    a =[] 
    for j in range(C):      # A for loop for column entries 
         a.append(int(input())) 
    inital.append(a) 

# For printing the matrix 
for i in range(R): 
    for j in range(C): 
        print(inital[i][j], end = " ") 
    print() 

R = int(input("Enter the number of rows:")) 
C = int(input("Enter the number of columns:")) 

# Initialize matrix 
final = [] 
print("Enter the entries rowwise:") 

# For user input 
for i in range(R):          # A for loop for row entries 
    a =[] 
    for j in range(C):      # A for loop for column entries 
         a.append(int(input())) 
    final.append(a) 

# For printing the matrix 
for i in range(R): 
    for j in range(C): 
        print(final[i][j], end = " ") 
    print() 





def search(state, goal_state):

    def gn(node):
        return node.gn()

    tiles_places = []
    for i in range(len(goal_state)):
        for j in range(len(goal_state)):
            heapq.heappush(tiles_places, (goal_state[i][j], (i, j)))

    def hn(node):
        cost = 0
        for i in range(len(node.state)):
            for j in range(len(node.state)):
                tile_i, tile_j = tiles_places[node.state[i][j]][1]
                if i != tile_i or j != tile_j:
                    cost += abs(tile_i - i) + abs(tile_j - j)
        return cost

    def fn(node):
        return 1

    return bfs_search(state, goal_state, fn)

def bfs_search(state, goal_state, fn):
    queue = []
    entrance = 0
    node = Node(state)
    while not node.is_goal(goal_state):
        node.expand()
        for child in node.children:
            #print(child)
            #print(fn(child))
            queue_item = (fn(child), entrance, child)
            heapq.heappush(queue, queue_item)
            entrance += 1
        node = heapq.heappop(queue)[2]
    output = []
    output.append(node.state)
    for parent in node.parents():
        output.append(parent.state)
    output.reverse()
    return (output,fn)
l , n = search(inital,final)
for i in l:
    for j in  i:
        print(j)
    print(n(Node(i)))
    print("\n")

这是输出-

Enter the number of columns:3
Enter the entries rowwise:
2
8
3
1
6
4
7
0
5
2 8 3 
1 6 4 
7 0 5 
Enter the number of rows:3
Enter the number of columns:3
Enter the entries rowwise:
8
0
3
2
6
4
1
7
5
8 0 3 
2 6 4 
1 7 5 
[2, 8, 3]
[1, 6, 4]
[7, 0, 5]
1


[2, 8, 3]
[1, 6, 4]
[0, 7, 5]
1


[2, 8, 3]
[0, 6, 4]
[1, 7, 5]
1


[0, 8, 3]
[2, 6, 4]
[1, 7, 5]
1


[8, 0, 3]
[2, 6, 4]
[1, 7, 5]
1

尽管我通过所有中间步骤到达了正确的目标节点，但我无法理解它考虑的启发法的基础是什么。

Answer 1

一般来说，A* 和贪婪最佳优先搜索不能保证达到相同的解决方案。如果使用可接受的启发式，A* 将始终找到最佳解决方案，而对于最佳优先搜索则不然。

有关 A* 和最佳优先搜索将给出不同结果的示例，请参阅 this answer .