The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, following is a solution for 4 Queen problem.

The N Queen is the problem of placing N chess queens on an N×N chessboard so that no two queens attack each other. For example, following are two solutions for 4 Queen problem.

In previous post, we have discussed an approach that prints only one possible solution, so now in this post the task is to print all solutions in N-Queen Problem. The solution discussed here is an extension of same approach.

**Backtracking Algorithm**

The idea is to place queens one by one in different columns, starting from the leftmost column. When we place a queen in a column, we check for clashes with already placed queens. In the current column, if we find a row for which there is no clash, we mark this row and column as part of the solution. If we do not find such a row due to clashes then we backtrack and return false.

1) Start in the leftmost column 2) If all queens are placed return true 3) Try all rows in the current column. Do following for every tried row. a) If the queen can be placed safely in this row then mark this [row, column] as part of the solution and recursively check if placing queen here leads to a solution. b) If placing queen in [row, column] leads to a solution then return true. c) If placing queen doesn't lead to a solution then unmark this [row, column] (Backtrack) and go to step (a) to try other rows. 3) If all rows have been tried and nothing worked, return false to trigger backtracking.

There is only a slight modification in above algorithm that is highlighted in the code.

/* C/C++ program to solve N Queen Problem using backtracking */ #define N 4 #include<stdio.h> /* A utility function to print solution */ void printSolution(int board[N][N]) { static int k = 1; printf("%d-\n",k++); for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) printf(" %d ", board[i][j]); printf("\n"); } printf("\n"); } /* A utility function to check if a queen can be placed on board[row][col]. Note that this function is called when "col" queens are already placed in columns from 0 to col -1. So we need to check only left side for attacking queens */ bool isSafe(int board[N][N], int row, int col) { int i, j; /* Check this row on left side */ for (i = 0; i < col; i++) if (board[row][i]) return false; /* Check upper diagonal on left side */ for (i=row, j=col; i>=0 && j>=0; i--, j--) if (board[i][j]) return false; /* Check lower diagonal on left side */ for (i=row, j=col; j>=0 && i<N; i++, j--) if (board[i][j]) return false; return true; } /* A recursive utility function to solve N Queen problem */ bool solveNQUtil(int board[N][N], int col) { /* base case: If all queens are placed then return true */ if (col == N ) { printSolution(board); return true; } /* Consider this column and try placing this queen in all rows one by one */ for (int i = 0; i < N; i++) { /* Check if queen can be placed on board[i][col] */ if ( isSafe(board, i, col) ) { /* Place this queen in board[i][col] */ board[i][col] = 1; /* recur to place rest of the queens */ solveNQUtil(board, col + 1) ; // below commented statement is replaced // by above one /* if ( solveNQUtil(board, col + 1) ) return true;*/ /* If placing queen in board[i][col] doesn't lead to a solution, then remove queen from board[i][col] */ board[i][col] = 0; // BACKTRACK } } /* If queen can not be place in any row in this column col then return false */ return false; } /* This function solves the N Queen problem using Backtracking. It mainly uses solveNQUtil() to solve the problem. It returns false if queens cannot be placed, otherwise return true and prints placement of queens in the form of 1s. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/ void solveNQ() { int board[N][N] = { {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, }; if (solveNQUtil(board, 0)) { printf("Solution does not exist"); return ; } return ; } // driver program to test above function int main() { solveNQ(); return 0; }

Output:

1- 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 2- 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0

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