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N-Queen4(SSF).cpp
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N-Queen4(SSF).cpp
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//MY Program to solve the N-Queen problem
//grid[][] is represent the 2-d array with value(0 and 1) for grid[i][j]=1 means queen i are placed at j column.
//we can take any number of queen , for this time we take the atmost 10 queen (grid[10][10]).
#include<iostream>
using namespace std;
int grid[10][10];
//print the solution
void print(int n) {
for (int i = 0;i <= n-1; i++) {
for (int j = 0;j <= n-1; j++) {
cout <<grid[i][j]<< " ";
}
cout<<endl;
}
cout<<endl;
cout<<endl;
}
//function for check the position is safe or not
//row is indicates the queen no. and col represents the possible positions
bool isSafe(int col, int row, int n) {
//check for same column
for (int i = 0; i < row; i++) {
if (grid[i][col]) {
return false;
}
}
//check for upper left diagonal
for (int i = row,j = col;i >= 0 && j >= 0; i--,j--) {
if (grid[i][j]) {
return false;
}
}
//check for upper right diagonal
for (int i = row, j = col; i >= 0 && j < n; j++, i--) {
if (grid[i][j]) {
return false;
}
}
return true;
}
//function to find the position for each queen
//row is indicates the queen no. and col represents the possible positions
bool solve (int n, int row) {
if (n == row) {
print(n);
return true;
}
//variable res is use for possible backtracking
bool res = false;
for (int i = 0;i <=n-1;i++) {
if (isSafe(i, row, n)) {
grid[row][i] = 1;
//recursive call solve(n, row+1) for next queen (row+1)
res = solve(n, row+1) || res;//if res ==false then backtracking will occur
//by assigning the grid[row][i] = 0
grid[row][i] = 0;
}
}
return res;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cout<<"Enter the number of queen"<<endl;
cin >> n;
for (int i = 0;i < n;i++) {
for (int j = 0;j < n;j++) {
grid[i][j] = 0;
}
}
bool res = solve(n, 0);
if(res == false) {
cout << -1 << endl; //if there is no possible solution
} else {
cout << endl;
}
return 0;
}