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 expected output is a binary matrix which has 1s 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}
Naive Algorithm
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.
Generate all possible configurations of queens on board and print a configuration that satisfies the given constraints.
while there are untried conflagrations
{
generate the next
configuration
if queens don't attack in
this configuration then
{
print this
configuration;
}
}
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.
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 umark 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.
Implementation of Backtracking solution
/* 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])
{
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 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 queen
can not be place 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;
}
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