Using breadth First Search for unweighted edges in a 2d matrix

Question

Problem:

A 2D maze(of size RxC) has R rows and C columns. Each empty cell is marked by ‘O’ (letter O) and the one with a barrier is marked by ‘#’. Two coordinates from the Maze(a,b) and (x,y). Currently, the starting point is at (a,b) and the destination is (x,y).

Move only in the following directions.

L: move to left cell from the current cell
R: move to right cell from the current cell
U: move to upper cell from the current cell
D: move to the lower cell from the current cell

Example:

Maze= [
                ['O', 'O', 'O', 'O', 'O'],

                ['O', 'O', '#', 'O', 'O'],

                ['O', 'O', 'O', 'O', 'O'],

                ['#', 'O', '#', '#', 'O'],

                ['O', '#', 'O', 'O', 'O']

            ]

Example #1

Input: Maze, (0,2), (2,2)

Output: ([(0, 2), (0, 1), (1, 1), (2, 1), (2, 2)], ‘LDDR’)

Example #2

Input: Maze, (0,0), (4,4)

Output: ([(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 4), (2, 4), (3, 4), (4, 4)], ‘RRRRDDDD’)

Example #3

Input: Maze, (0,0), (4,0)

Output: None

Not an empty list, but the None object.

So I decided to do BSF for this problem but now I’m stuck. I don’t know when should I insert the indices of each position in the path. Since I have to output a list of tuples(starting with the source and ending with the destination) as well as return a set of possible
directions as well in the form of a string. Any guidance would be appreciated, thank you.

Here is my code:

def puzzle(Board, Source, Destination):
#    if Source is None or Source not in Board:
#         return Source 

    dx = [-1, +1, 0, 0]
    dy = [0, 0 , +1, -1]
    visited = [[False]*cols for i in range(rows)]
    distances = {vertex : float('inf') for vertex in range(rows)}

    reached_end = False
    distances[Source] = 0
 
    queue = deque()
    queue.append((currentRow,currentCol))
    visited[currentRow][currentCol] = True

    while queue:
        r,c = queue.popleft()
        path = (r,c)
        if(path == Destination):
            reached_end = True    
            return
        for neighbor in range(0,4):
                rr = r + dx[neighbor]
                cc = c + dy[neighbor]
                if (rr < 0 or cc < 0):
                    continue
                if (rr >= rows or cc >= cols):
                    continue
                if visited[rr][cc]:
                    continue
                if Board[rr][cc] == '#':
                    continue
                queue.append((rr,cc))
                visited[rr][cc] = True

Asked By: vmilk-

||

Source

Answer 1

One way to do this, is to add the list of moves ("LDDR…") to the tuple that is queued. Then when the destination is found, you can build the coordinates from that.

Here is your code with those adaptations:

def puzzle(Board, Source, Destination):
    if Source is None or Source not in Board:
        return
    cols = len(Board[0])
    rows = len(Board)
    currentRow,currentCol = Source

    dx = [-1, +1, 0, 0]
    dy = [0, 0 , +1, -1]
    dirs = "UDRL"
    visited = [[False]*cols for i in range(rows)]
    distances = {vertex : float('inf') for vertex in range(rows)}

    distances[Source] = 0
 
    queue = deque()
    queue.append((currentRow,currentCol, ""))
    visited[currentRow][currentCol] = True

    while queue:
        r,c,moves = queue.popleft()
        path = (r,c)
        if path == Destination:
            path = [Source]
            r,c = Source
            for move in moves:
                i = dirs.index(move)
                r += dx[i]
                c += dy[i]
                path.append((r,c))
            return path, moves
        for neighbor in range(0,4):
            rr = r + dx[neighbor]
            cc = c + dy[neighbor]
            if (rr < 0 or cc < 0):
                continue
            if (rr >= rows or cc >= cols):
                continue
            if visited[rr][cc]:
                continue
            if Board[rr][cc] == '#':
                continue
            queue.append((rr,cc,moves+dirs[neighbor]))
            visited[rr][cc] = True

Answered By: trincot

Using breadth First Search for unweighted edges in a 2d matrix

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