N Queen Problem

  • Let us discuss N Queen problem that can be solved using Backtracking.
  • 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 the solution for 4 Queen problem.
 N Queen Problem
  • Binary matrix which has 1’s for the blocks where queens are placed. For example, following is the output matrix for above 4 queen solution.
{ 0,  1,  0,  0 }
{ 0, 0, 0, 1 }
{ 1, 0, 0, 0 }
{ 0, 0, 1, 0 }

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 can checkout to 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 from 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 the queen in [row, column] leads to a solution then return
true.
c) If placing queen does not lead to a solution then umark this [row,
column] (Backtrack) and go to step (a) to try other rows.
4) If all rows have been tried and nothing worked, return false to trigger
backtracking.

Sample Code in C

/* C/C++ program to solve N Queen Problem using 
backtracking */
#define N 4
#include<stdio.h>
#include<stdbool.h>

/* A utility function to print solution */
void printSolution(int board[N][N])
{
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
printf(" %d ", board[i][j]);
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)
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 the 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 */
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 the queen cannot be placed in any row in
this colum 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.*/
bool 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) == false )
{
printf("Solution does not exist");
return false;
}

printSolution(board);
return true;
}

// driver program to test above function
int main()
{
solveNQ();
return 0;
}

Output

0  0  1  0 
1 0 0 0
0 0 0 1
0 1 0 0

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