Я пытался реализовать алгоритм Best First Search для 8 задач. Но я получаю тот же путь, что и в коде A *, независимо от того, какую матрицу я выберу.Кроме того, кто-то может помочь мне распечатать эвристику под каждой матрицей?Я получаю только «1» на выходе.
Лучший код первого поиска -
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")
Вот вывод -
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
Хотя я достигаю правильного целевого узласо всеми промежуточными шагами я не могу понять, на каком основании эвристики это рассматривает.