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Program to check if number of compass usage to get out of a maze is enough in Python
Suppose, we are playing a game where we are trapped in a maze. We have to find our way out of the maze. The maze can be presented as an x m matrix, where n is the number of rows and m is the number of columns. Each cell/element of the matrix contains any of the symbols 'O','D', 'S', or '-'. 'O' means that the path is blocked, 'D' is the way out from the maze, 'S' is our starting position, and '-' means we can move through the path. We can move freely through any of the '-' marked cells. Now, we also have a compass using which we can find the exit path ('D' cell) from the maze. When we have to find a direction, we have to use the compass. But, we can use the compass at most k times. Given the maze as a matrix and the number of times we can use the compass; we have to find out if we can get out of the maze having used the compass that many times only. If it is possible we return True, or else we return False.
So, if the input is like grid =
| - | O | - | O | - | - | - | - | - | - | O |
| - | O | D | - | O | - | O | O | O | - | O |
| - | O | O | - | O | - | O | O | O | - | O |
| - | O | O | - | O | - | O | S | - | - | - |
| - | - | - | - | - | - | O | O | O | O | - |
n = 4, m = 11, k = 3; then the output will be True.
To solve this, we will follow these steps −
Define a function path_search() . This will take curr_pos, grid, total_rows, total_cols, k, predecessor
x:= x value of curr_pos
y:= y value of curr_pos
if grid[x, y] is same as "*", then
if k is same as 0, then
return True
otherwise,
return False
otherwise,
parent := predecessor[curr_pos]
succ_pos := a new list from the return values of succesor_positions(curr_pos, grid, total_rows, total_cols, parent)
use_compass := True if size of succ_pos > 1
for each position in succ_pos, do
predecessor[position] := curr_pos
if use_compass is non-zero, then
path_search(position, grid, total_rows, total_cols, k - 1, predecessor)
- otherwise,
path_search(position, grid, total_rows, total_cols,k, predecessor)
Define a function succesor_positions() . This will take curr_pos, grid, total_rows, total_cols, parent
x:= x value of curr_pos
y:= y value of curr_pos
succ_pos := a new list
o if y > 0, then
left := x, y - 1
insert left at the end of succ_pos
if y < total_cols - 1, then
right := x, y + 1
insert right at the end of succ_pos
if x > 0, then
up := x - 1, y
insert up at the end of succ_pos
if x < total_rows - 1, then
down := x + 1, y
insert down at the end of succ_pos
if for every position in succ_pos, grid[position[0], pos[1]] is
not same as "X" and pos is not same as parent, then −
return the elements in succ_pos satisfying the condition
Now perform the following −
curr_pos := a new empty pair
for each index i, item row in grid, do
for each index j, item element in row, do
if element is same as 'M', then
curr_pos := a new pair containing i, j
predecessor := a new map where the curr_pos := Null initially
path_search(curr_pos, grid, n, m, k, predecessor)
Source Code (Python)
Let us see the following implementation to get better understanding −
def path_search(curr_pos, grid, total_rows, total_cols, k, predecessor):
x, y = curr_pos
if grid[x][y] == "D":
if k == 0:
print('True')
else:
print('False')
else:
parent = predecessor[curr_pos]
succ_pos = list(succesor_positions(curr_pos, grid, total_rows, total_cols, parent))
use_compass = len(succ_pos) > 1
for position in succ_pos:
predecessor[position] = curr_pos
if use_compass:
path_search(position, grid, total_rows, total_cols, k - 1, predecessor)
else:
path_search(position, grid, total_rows, total_cols, k, predecessor)
def succesor_positions(curr_pos, grid, total_rows, total_cols, pred):
x, y = curr_pos
succ_pos = []
if y > 0:
left = (x, y - 1)
succ_pos.append(left)
if y < total_cols - 1:
right = (x, y + 1)
succ_pos.append(right)
if x > 0:
up = (x - 1, y)
succ_pos.append(up)
if x < total_rows - 1:
down = (x + 1, y)
succ_pos.append(down)
return filter(lambda pos: grid[pos[0]][pos[1]] != "O" and pos != pred, succ_pos)
def solve(grid, n, m, k):
curr_pos = ()
for i, row in enumerate(grid):
for j, element in enumerate(row):
if element == 'S':
curr_pos = (i, j)
path_search(curr_pos, grid, n, m, k, predecessor = {curr_pos: None})
grid = [['-', 'O', '-', 'O', '-', '-', '-', '-', '-', '-', 'O'],
['-', 'O', 'D', '-', 'O', '-', 'O', 'O', 'O', '-', 'O'],
['-', 'O', 'O', '-', 'O', '-', 'O', 'S', '-', '-', '-'],
['-', '-', '-', '-', '-', '-', 'O', 'O', 'O', 'O', '-']]
solve(grid, 4, 11, 3)Input
grid = [['-', 'O', '-', 'O', '-', '-', '-', '-', '-', '-', 'O'], ['-', 'O', 'D', '-', 'O', '-', 'O', 'O', 'O', '-', 'O'], ['-', 'O', 'O', '-', 'O', '-', 'O', 'S', '-', '-', '-'], ['-', '-', '-', '-', '-', '-', 'O', 'O', 'O', 'O', '-']] , 4, 11, 3
Output
True
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