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2D_BFS.py
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2D_BFS.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 collections import deque
# Main BFS function to find the path
def BFS(maze, src, des, way=1):
# Base case: is source has no way, return false
if maze[src[0]][src[1]] != way:
return False
# Dimeention of the matrix and a queue is initlized with the source vertex
n = len(maze)
m = len(maze[0])
queue = deque([(src[0], src[1], 0)])
# To keep a track of the visdited nodes and parents
visited = [[False] * m for i in range(n)]
parent = {}
# All possible moves from a cell
move = {(1, 0): 'D', (-1, 0): 'U', (0, 1): 'R', (0, -1): 'L'}
while queue:
# Extract the left most node from the queue
v = queue.popleft()
# If the node is destination, calculate the path and return
if (v[0], v[1]) == des:
path = ''
dis = v[2]
cur = (v[0], v[1])
# Calculate the path by backtracking with the parent dict
while cur != src:
prev_move = parent[cur]
m = (cur[0] - prev_move[0], cur[1] - prev_move[1])
path += move[m]
cur = prev_move
# Return the distance and path
return dis, path[::-1]
# For a given node check each possible move
for i in move.keys():
r = v[0] + i[0]
c = v[1] + 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]:
visited[r][c] = True
parent[(r, c)] = (v[0], v[1])
queue.append((r, c, v[2] + 1))
# If the queue is empty, there is no way possible, return False
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 = BFS(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 1 0 1
1 1 1 0 1
Enter the Source cell: 0 0
Enter the Destination cell: 4 4
Disance= 16
Path: DDDDRRUUUURRDDDD
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
"""