CODEWITHSUNDEEP
pythonrecursiongraph
bard
bard

Reputation: 3052

Count the number of possible paths through a grid

There is a robot at the top-left corner of an N*M grid. The robot can move up, down, left and right, but cannot visit the same cell more than once in each traversal. How do I find the total number of ways the robot can reach the bottom-right corner?

(the robot does not need to visit every cell for a path to be valid)

I think there is a recursive solution to this but I can't get it somehow.

Here's what I've got so far:

def initialize(row, cols):
    grid = [ [ 0 for c in range(cols) ] for r in range(rows) ]
    pos_r, pos_c = 0, 0
    grid[pos_r][pos_c] = 1 # set start cell as visited
    dst_x, dst_y = len(grid)-1, len(grid[0])-1 # coords of bottom-right corner
    print move_robot(grid, dst_x, dst_y, pos_r, pos_c)

def move_robot(grid, dst_x, dst_y, pos_r, pos_c, prev_r=None, prev_c=None):
    num_ways = 0
    if reached_dst(dst_x, dst_y, pos_r, pos_c):
        undo_move(grid, pos_r, pos_c)
        undo_move(grid, prev_r, prev_c)
        return 1
    else:
        moves = get_moves(grid, pos_r, pos_c)
        if len(moves) == 0:
            undo_move(grid, prev_r, prev_c)
            return 0
        for move in moves:
            prev_r = pos_r
            prev_c = pos_c
            pos_r = move[0]
            pos_c = move[1]
            update_grid(grid, pos_r, pos_c)
            num_ways += move_robot(grid, dst_x, dst_y, pos_r, pos_c, prev_r, prev_c)
    return num_ways

if __name__ == '__main__':
    initialize(4, 4)

I left out some function definitions for brevity. Get_moves retrieves the all legal moves, checking whether each move would still be on the board and whether the cell has already been visited. Update_grid sets the specified cell to '1', which means visited. Undo_move does the opposite, setting the specified cell to '0'.

I get the right answer for the simplest possible case (2*2 grid), but for larger grids the output is always too low. What's wrong with my code, and is there a simpler way of doing this?

Upvotes: 4

Views: 4127

Answers (1)

Nir Alfasi
Nir Alfasi

Reputation: 53525

The recursion is pretty straight forward but one should be careful to create copies of the matrix while recursing in order to receive good results:

from copy import copy, deepcopy
def calc(i, j, mat):
    if i < 0 or j < 0 or i >= len(mat) or j >= len(mat[0]):
        return 0 # out of borders
    elif mat[i][j] == 1:
        return 0 # this cell has already been visited
    elif i == len(mat)-1 and j == len(mat[0])-1:
        return 1 # reached destination (last cell)
    else:
        mat[i][j] = 1 # mark as visited
        # create copies of the matrix for the recursion calls
        m1 = deepcopy(mat)
        m2 = deepcopy(mat)
        m3 = deepcopy(mat)
        m4 = deepcopy(mat)
        # return the sum of results of the calls to the cells: 
        # down + up + right + left 
        return calc(i+1, j, m1) + calc(i-1, j, m2) + calc(i, j+1, m3) + calc(i, j-1, m4)

def do_the_robot_thing(m, n):
    # an un-visited cell will be marked with "0"
    mat = [[0]*n for x in xrange(m)]
    return calc(0, 0, mat)


print(do_the_robot_thing(3, 3))

OUTPUT:

12

Upvotes: 4

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