Cheking the surrounding indexes around an index in 2d matrix

Question

I'm working on the Conway's game of life program and I'm at the point where the dead/live cell checks it's surrounding neighbors and counts the amount of live neighbors surrounding it. Right now, I'm working on having [0][0] checked. The problem I'm having is that [0][0] is being checked along with the surrounding indexes. I thought if I were to put " if (k!=o && l!=p)", it would exclude [0][0], but it doesn't.

public class Life {

public static void main(String[] args)
{
    int N = 5;
    boolean[][] b = new boolean[N][N];
    double cellmaker = Math.random();

    int i = 0;
    int j = 0;


    int o=0;
    int p=0;
    int livecnt = 0; //keeps track of the alive cells surrounding cell

     System.out.println("First Generation:");
     // Makes the first batch of cells
     for ( i = 0; i < N ; i++)
     {

         for ( j = 0; j< N; j++)
         {
              cellmaker = Math.random();


             if (cellmaker > 0.5) // * = alive; - = dead
             {
                 b[i][j]=true;

                 System.out.print( "* ");

             }


             if (cellmaker < 0.5)
            { b[i][j] = false;


             System.out.print("- ");

            }

         }
         System.out.println();
     }       

     // Checks for the amount of "*" surrounding (o,p)
     for (int k=(o-1); k <= o+1; k++ )
        {

            for (int l =(p-1); l <=p+1; l++)
            {
                 if ( k >= 0 && k < N && l >= 0 && l < N) //for the border indexes.
                 { 
                     //if (k!=o && l!=p)
                    {
                        if (b[k][l] == true)
                            {

                                livecnt++;

                            }
                    }

                 }

            }
        } 
    System.out.println(livecnt++);
}

}

Answer 1

You want to check the surrounding of (o,p), try something like this:

if (!(k== o && l==p))

instead of:

if (k!=o && l!=p)

because with the above condition you are not checking the coordinates (k,p-1), (k,p+1), (k-1,p) and (k+1,p)

Answer 2

Your code is nearly correct. You just need to change

if (k!=o && l!=p)

into

if (k!=o || l!=p)

You only want to count a field if the coordinates (k,l) are not equal to (o,p) In other words: !(k==o && l==p) Remember (k!=o || l!=p) is equal to !(k==o && l==p) according to De Morgan's laws.