Maze

maze.py

from sys import setrecursionlimit

maze = [
    ['#', '#', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#'],  # 0
    ['#', '#', '#', '#', '#', '#', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '.', '#', '#', '#'],  # 1
    ['#', '#', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '#', '#', '#', '#', '.', '#', '#', '#'],  # 2
    ['#', '#', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '#', '.', '.', '.', '.', '#', '#', '#'],  # 3
    ['#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '.', '#', '#', '.', '#', '#', '#'],  # 4
    ['#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '#'],  # 5
    ['#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '#'],  # 6
    ['#', '#', '.', '#', '#', '#', '#', '.', '#', '#', '#', '#', '#', '.', '#', '#', '.', '#', '#', '#'],  # 7
    # 0,   1,  2,   3,   4,   5,   6,   7,  8,     9,   10,  11,  12,  13,  14,  15,  16, 17,  18,  19
]


def is_exit_here(x, y):
    """
    Function will check if the coordinates match the coordinates of the exit
    :param x: list coordinate
    :param y: item coordinate in the list
    :return: locations dict
    """
    # possible exit positions for x or y
    if any([x == main_list_length - 1, y == 0, x == 0, y == internal_list_length - 1]):
    # if x == main_list_length - 1 or y == 0 or x == 0 or y == internal_list_length - 1:
        # Skip entry coordinates
        if locations["enter"] != [x, y]:
            # There may be more than one exit, so let's add all of them
            locations["exit"].append([x, y])
    return locations


def solve(maze, x, y, visited):
    """
    Finding an exit of the maze
    :param maze: NxN matrix
    :param x (int): list coordinate
    :param y (int): item coordinate in the list
    :param visited: size duplicate maze
    """
    if maze[x][y] == ".":
        # mark the current cell as visible
        visited[x][y] = True

        # check exit
        is_exit_here(x, y)
        # down
        if x + 1 < main_list_length and not visited[x + 1][y]:
            solve(maze, x + 1, y, visited)
        # up
        if x - 1 >= 0 and not visited[x - 1][y]:
            solve(maze, x - 1, y, visited)
        # right
        if y + 1 < internal_list_length and not visited[x][y + 1]:
            solve(maze, x, y + 1, visited)
        # left
        if y - 1 >= 0 and not visited[x][y - 1]:
            solve(maze, x, y - 1, visited)

    return visited


if __name__ == "__main__":
    # Set the maze entrance coordinates
    row = 0
    column = 6
    enter = [row, column]

    # Main list length
    main_list_length = len(maze)
    # Internal list length
    internal_list_length = len(maze[0])

    # Different variants of maze are being considered.
    # It is necessary to prevent RecursionError.
    # I don't think making recursion infinite is a good idea,
    # so I'll try increasing it as follows:
    setrecursionlimit(1000 + main_list_length * main_list_length)

    # matrix - a duplicate of the maze to follow the path.
    visited = [
        [False for x in range(internal_list_length)] for y in range(main_list_length)
    ]

    # Save the entrance and exit to the maze in dict
    locations = {"enter": enter, "exit": []}

    # Find a solution
    solve(maze, enter[0], enter[1], visited)

    if not locations["exit"]:
        print(
            "Buddy, it's really bad. This maze has no way out. "
            "I hope you're ready for the walk back 😞"
        )
    else:
        print(
            "Yay! It was a challenge, but we did it!'🎉\n" "Exit location:",
            *locations["exit"],
            sep="\n"
        )
    print(*visited, sep="\n")
    assert [7, 13] == locations["exit"][0]

Maze solution