How to solve 8 queens 1D array with brute force? - c++

Question

I was given an assignment to modify an 8 Queens program to use a 1D array and to use brute force (already did backtracking). I've come up with the following code:
#include <cmath>
#include <iostream>
using namespace std;
bool ok(int board[8]){
for(int j = 0; j <= 7; j++){ //check for repeating digits
cout << "ok loop 1"<<endl;
for (int k = 0; k <= 7; k++)
{
cout << "ok loop 2"<<endl;
if (board[k] = board[j]){ return false; }
}
}
for(int c = 7; c >= 0; c--){ //check if position is safe
cout << "ok loop 3"<<endl;
//int r = 0;
for(int i = 1; i <= c; i++){
cout << "ok loop 4"<<endl;
if(board[c-i] == c)
return false;
else if ((board[c]-i)>0 && board[c-i]-i == 1)
return false;
else if ((board[c]+i)<=7 && board[c-i]+i == 1)
return false;
} // for loop
} // for loop
return true;
} // ok
void print(int board[8], int c){
cout << "Solution " << c << ": " << endl;
for(int i = 0; i < 8; i++){
{
cout << board[i] <<" ";
}
}
cout << endl;
}
int main ()
{
int b[8]={0}; //initialize the array
int count = 0;
for(b[0]=0; b[0]<8; b[0]++)
for(b[1]=0; b[1]<8; b[1]++)
for(b[2]=0; b[2]<8; b[2]++)
for(b[3]=0 ; b[3]<8; b[3]++)
for(b[4]=0; b[4]<8; b[4]++)
for(b[5]=0; b[5]<8; b[5]++)
for(b[6]=0; b[6]<8; b[6]++)
for(b[7]=0; b[7]<8; b[7]++)
if(ok(b))
{
count++;
print(b, count);
}
system("PAUSE");
return 0;
}
It keeps looping forever and I am not sure why. Would anyone mind helping me?

Answer 1

There's a few things that could be improved:
If you passed a reference to a constant array of eight chars to ok() instead of just a pointer to non-const ints, the compiler could have told you about one of the issues.
How many different positions can a queen have? I'd say 64, though your code suggests eight. I'd start with documenting the actual meaning of variables and constants throughout your code, because you seem to be confused there yourself.
You check if board[x] is board[y], but with x and y being equal, and from that you claim that there are repeating digits.
You make a difference between the different queens. In other words, your program will find all permutations of how the queens could be positioned on the same eight positions. This is not incorrect, but inefficient. If you fix the number of positions, that will make a noticeable difference.