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2D_Astar.py
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"""
Purpose: Given a binary matrix of N*M order where 0 is the wall and 1 is way.
Find the shortest distance from a source cell to a destination cell,
traversing through limited cells only. Also you can move only
up, down, left and right. If found then print the distance and
path in separate lines, else return -1.
"""
from heapq import heappop, heappush
# Manhattan Distance for heuristic functionn
def heuristic_function(p1, p2):
x1, y1 = p1
x2, y2 = p2
return abs(x1 - x2) + abs(y1 - y2)
def Astar(maze, src, des, way=1):
# Base Case: If there is no way from the source, returnn False
if(maze[src[0]][src[1]] != 1):
return False
# Dimention of the maze
n = len(maze)
m = len(maze[0])
hp = []
count = 0
x, y = src
# To keep a track of visited nodes, also mark source as visited
visited = [[False] * m for i in range(n)]
visited[x][y] = True
# All possible moves from a cell
moves = {(1, 0): 'D', (-1, 0): 'U', (0, 1): 'R', (0, -1): 'L'}
parent = {}
# Initilize the heap with the source cell
heappush(hp, [0, count, src])
# Initilze the G_score for each node to infinity
# And G_score of source is 0
g_score = [[float('inf')] * m for i in range(n)]
g_score[x][y] = 0
# Initilze the F_score for each node to infinity
# And F score of source is 1
# F_source = G_score + heuristic function
f_score = [[float('inf')] * m for i in range(n)]
f_score[x][y] = 1
while hp:
cur_pos = heappop(hp)[2]
# print(cur_pos)
xx, yy = cur_pos
if cur_pos == des:
path = ''
# Calculate the path by backtracking with the parent dict
while cur_pos != src:
print(cur_pos)
prev_move = parent[cur_pos]
m = (cur_pos[0] - prev_move[0], cur_pos[1] - prev_move[1])
path += moves[m]
cur_pos = prev_move
# Return the distance and path
return len(path), path[::-1]
for i in moves.keys():
r = xx + i[0]
c = yy + i[1]
# If the next node inside the maze , has a way and not yet visited
# then mark it visited and push it in the queue
if 0 <= r < n and 0 <= c < m and maze[r][c] == way and not visited[r][c]:
temp = g_score[xx][yy] + 1
if temp < g_score[r][c]:
parent[(r, c)] = (xx, yy)
g_score[r][c] = temp
# F_source = G_score + heuristic function
f_score[r][c] = temp + heuristic_function(des, (r, c))
count += 1
heappush(hp, [f_score[r][c], count, (r, c)])
visited[r][c] = 1
return False
# --------------------------------DRIVER CODE ---------------------------------
if __name__ == "__main__":
N, M = map(int, input("Enter the Dimension of the maze:- ").split())
print("Enter the Maze: ")
maze = []
# Input the Maze
for _ in range(N):
maze.append([int(i) for i in input().split()])
src = tuple(map(int, input("Enter the Source cell: ").split()))
des = tuple(map(int, input("Enter the Destination cell: ").split()))
ans = Astar(maze, src, des)
# If ans is false, i.e. no way is possible, else print distance and path
if ans is False:
print("No Path exists between", src, "and", des)
else:
dist = ans[0]
path = ans[1]
print("Disance= ", dist)
print("Path: ", path)
"""
Time Compexity: O(N*M)
Space Complexity: O(N*M)
Sample Input / Output
Enter the Dimension of the maze:- 5 5
Enter the Maze:
1 0 1 1 1
1 0 1 0 1
1 0 1 0 1
1 0 0 0 1
1 1 1 1 1
Enter the Source cell: 0 0
Enter the Destination cell: 4 4
Disance= 8
Path: DDDDRRRR
Enter the Dimension of the maze:- 5 5
Enter the Maze:
1 0 1 1 1
1 0 1 0 1
1 0 0 0 1
1 0 1 0 1
1 1 1 0 1
Enter the Source cell: 0 0
Enter the Destination cell: 4 4
No Path exists between (0, 0) and (4, 4)
Enter the Dimension of the maze:- 5 8
Enter the Maze:
1 0 1 1 1 1 1 1
1 0 1 0 0 0 0 1
1 0 1 1 1 1 0 1
1 0 0 0 0 1 0 1
1 1 1 1 1 1 0 1
Enter the Source cell: 0 0
Enter the Destination cell: 4 7
Disance= 25
Path: DDDDRRRRRUULLLUURRRRRDDDD
"""