backtracking/n_queens.py - master - ksushaksusha/Python

backtracking/n_queens.py
109 строк3 KB
Mozart Maia
add some doctests to algos in backtracking (#11911)
13 мар 2026, 07:40
13 мар 2026, 07:407e4b60b
"""

The nqueens problem is of placing N queens on a N * N
chess board such that no queen can attack any other queens placed
on that chess board.
This means that one queen cannot have any other queen on its horizontal, vertical and
diagonal lines.

"""

from __future__ import annotations

solution = []


def is_safe(board: list[list[int]], row: int, column: int) -> bool:
    """
    This function returns a boolean value True if it is safe to place a queen there
    considering the current state of the board.

    Parameters:
    board (2D matrix): The chessboard
    row, column: Coordinates of the cell on the board

    Returns:
    Boolean Value

    >>> is_safe([[0, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
    True
    >>> is_safe([[0, 1, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
    False
    >>> is_safe([[1, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
    False
    >>> is_safe([[0, 0, 1], [0, 0, 0], [0, 0, 0]], 1, 1)
    False
    >>> is_safe([[1, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 2)
    True
    >>> is_safe([[1, 0, 0], [0, 0, 0], [0, 0, 0]], 2, 1)
    True
    >>> is_safe([[0, 0, 0], [1, 0, 0], [0, 0, 0]], 0, 2)
    True
    >>> is_safe([[0, 0, 0], [1, 0, 0], [0, 0, 0]], 2, 2)
    True
    """

    n = len(board)  # Size of the board

    # Check if there is any queen in the same upper column,
    # left upper diagonal and right upper diagonal
    return (
        all(board[i][j] != 1 for i, j in zip(range(row), [column] * row))
        and all(
            board[i][j] != 1
            for i, j in zip(range(row - 1, -1, -1), range(column - 1, -1, -1))
        )
        and all(
            board[i][j] != 1
            for i, j in zip(range(row - 1, -1, -1), range(column + 1, n))
        )
    )


def solve(board: list[list[int]], row: int) -> bool:
    """
    This function creates a state space tree and calls the safe function until it
    receives a False Boolean and terminates that branch and backtracks to the next
    possible solution branch.
    """
    if row >= len(board):
        """
        If the row number exceeds N, we have a board with a successful combination
        and that combination is appended to the solution list and the board is printed.
        """
        solution.append(board)
        printboard(board)
        print()
        return True
    for i in range(len(board)):
        """
        For every row, it iterates through each column to check if it is feasible to
        place a queen there.
        If all the combinations for that particular branch are successful, the board is
        reinitialized for the next possible combination.
        """
        if is_safe(board, row, i):
            board[row][i] = 1
            solve(board, row + 1)
            board[row][i] = 0
    return False


def printboard(board: list[list[int]]) -> None:
    """
    Prints the boards that have a successful combination.
    """
    for i in range(len(board)):
        for j in range(len(board)):
            if board[i][j] == 1:
                print("Q", end=" ")  # Queen is present
            else:
                print(".", end=" ")  # Empty cell
        print()


# Number of queens (e.g., n=8 for an 8x8 board)
n = 8
board = [[0 for i in range(n)] for j in range(n)]
solve(board, 0)
print("The total number of solutions are:", len(solution))