Python/backtracking/n_queens.py
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Python

"""
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([[1, 0, 0], [0, 0, 0], [0, 0, 0]], 1, 1)
False
"""
n = len(board) # Size of the board
# Check if there is any queen in the same row, column,
# left upper diagonal, and right upper diagonal
return (
all(board[i][j] != 1 for i, j in zip(range(row, -1, -1), range(column, n)))
and all(
board[i][j] != 1 for i, j in zip(range(row, -1, -1), range(column, -1, -1))
)
and all(board[i][j] != 1 for i, j in zip(range(row, n), range(column, n)))
and all(board[i][j] != 1 for i, j in zip(range(row, n), range(column, -1, -1)))
)
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))