n queens Algorithm
The eight queens puzzle is an example of the more general N queens problem of put N non-attacking queens on an n×n chessboard, for which solutions exist for all natural numbers N with the exception of N = 2 and N = 3.The eight queens puzzle is the problem of put eight chess queens on an 8×8 chessboard so that no two queens threaten each other; thus, a solution necessitates that no two queens share the same row, column, or diagonal. Chess composer Max Bezzel published the eight queens puzzle in 1848.Nauck also extended the puzzle to the N queens problem, with N queens on a chessboard of n×n squares. In 1874, S. Gunther proposed a method use determinants to find solutions.
"""
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.
"""
solution = []
def isSafe(board, row, column):
"""
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) : board
row ,column : coordinates of the cell on a board
Returns :
Boolean Value
"""
for i in range(len(board)):
if board[row][i] == 1:
return False
for i in range(len(board)):
if board[i][column] == 1:
return False
for i, j in zip(range(row, -1, -1), range(column, -1, -1)):
if board[i][j] == 1:
return False
for i, j in zip(range(row, -1, -1), range(column, len(board))):
if board[i][j] == 1:
return False
return True
def solve(board, row):
"""
It 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 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
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 isSafe(board, row, i):
board[row][i] = 1
solve(board, row + 1)
board[row][i] = 0
return False
def printboard(board):
"""
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=" ")
else:
print(".", end=" ")
print()
# n=int(input("The no. of queens"))
n = 8
board = [[0 for i in range(n)] for j in range(n)]
solve(board, 0)
print("The total no. of solutions are :", len(solution))
