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Program to check whether we can get N queens solution or not in Python
Suppose we have a binary matrix where 0 is representing empty cell and 1 is representing a chess queen at that cell. We have to check whether we can fill this board and get a valid nqueen solution or not. As we know the n queens puzzle asks to place n queens on an n × n chessboard so that no two chess queens can attack each other.
So, if the input is like
| 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
then the output will be True, as one solution is like −
| 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 |
To solve this, we will follow these steps −
Define a function isSafe() . This will take board, i, j
for r in range 0 to size of board, do
if r is not same as i and board[r, j] is same as 1, then
return False
r := i + 1, c := j + 1
while r < row size of board and c < column size of board, do
if board[r, c] is same as 1, then
return False
r := r + 1, c := c + 1
r:= i + 1, c := j - 1
while r < row size of board and c >= 0, do
if board[r, c] is same as 1, then
return False
r := r + 1, c := c - 1
r := i - 1, c := j + 1
while r >= 0 and c < column size of board, do
if board[r, c] is same as 1, then
return False
r := r - 1, c := c + 1
r := i - 1, c := j - 1
while r >= 0 and c >= 0, do
if board[r, c] is same as 1, then
return False
r := r - 1, c := c - 1
return True
From the main method do the following −
r := 0, c := 0
stack := a new stack
while r < row size of board, do
if 1 is in board[r], then
r := r + 1
go for next iteration
otherwise,
found := False
while c < column size of board, do
if isSafe(board, r, c) is true, then
board[r, c] := 1
insert [r, c] into stack
found := True
come out from the loop
c := c + 1
if found is true, then
c := 0, r := r + 1
otherwise,
if stack is empty, then
return False
m := delete top element from stack
r := m[0], c := m[1] + 1
board[r, c - 1] := 0
return True
Example
Let us see the following implementation to get better understanding −
class Solution: def solve(self, board): def isSafe(board, i, j): for r in range(len(board)): if r != i and board[r][j] == 1: return False r, c = i + 1, j + 1 while r < len(board) and c < len(board[0]): if board[r][c] == 1: return False r += 1 c += 1 r, c = i + 1, j - 1 while r < len(board) and c >= 0: if board[r][c] == 1: return False r += 1 c -= 1 r, c = i - 1, j + 1 while r >= 0 and c < len(board[0]): if board[r][c] == 1: return False r -= 1 c += 1 r, c = i - 1, j - 1 while r >= 0 and c >= 0: if board[r][c] == 1: return False r -= 1 c -= 1 return True r = c = 0 stack = [] while r < len(board): if 1 in board[r]: r += 1 continue else: found = False while c < len(board[0]): if isSafe(board, r, c): board[r][c] = 1 stack.append([r, c]) found = True break c += 1 if found: c = 0 r += 1 else: if not stack: return False m = stack.pop() r, c = m[0], m[1] + 1 board[r][c - 1] = 0 return True ob = Solution() matrix = [ [1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 1], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0] ] print(ob.solve(matrix))
Input
[ [1, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 1], [0, 0, 0, 0, 0], [0, 0, 0, 1, 0] ]
Output
True
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