There's something wrong in my Magic Square Algorithm

Question

Hi I created a Magic Square program in java It works fine if you input a number 3 but if i input 5 and so on there's a problem occurring.. The pattern becomes wrong.

Please help me to find out what's wrong in my code:

Here's my code:

    Scanner input = new Scanner(System.in);

    System.out.print("Enter a number: ");
    int num = input.nextInt();

    // Number must be ODD and not less than or equals to one to continue
    while ((num % 2 == 0) || (num <= 1)) {
        System.out.println("Enter a valid number: ");
        num = input.nextInt();
    }

    int[][] magic = new int[num][num];

    int row = 0;
    int col = num / 2;
    magic[row][col] = 1;

    for (int i = 2; i <= num * num; i++) {
        if (magic[(row + 5) % num][(col + 2) % num] == 0) {
            row = (row + 5) % num;
            col = (col + 2) % num;
        } else {
            row = (row + 1 + num) % num;
        }
        magic[row][col] = i;
    }

    for (int x = 0; x < num; x++) {
        for (int j = 0; j < num; j++) {
            System.out.print(magic[x][j] + "\t");
        }
        System.out.println();
    }

}

It's correct when i Input 3, here's the output:

enter image description here

But when i type a number like 5: It becomes:

enter image description here

UPDATED!

sorry for incomplete information. So i edited and now this is my new question — repsajznav, Dec 10 '14 at 16:31
Why move row + 5 and col + 2? move both +1 and if the value is 0 move to the previous row and same col. — vzamanillo, Dec 10 '14 at 18:02

score 1 · Answer 1 · answered Dec 10 '14 at 16:54

You seem to be trying to implement the Method for constructing a magic square of odd order.

The method prescribes starting in the central column of the first row with the number 1.

You seem to have that right.

After that, the fundamental movement for filling the squares is diagonally up and right, one step at a time.

I would code this as:

int newRow = row - 1;
int newCol = col + 1;

When an "up and to the right" move would leave the square, it is wrapped around to the last row or first column, respectively.

This is clearly:

if ( newRow < 0 ) {
  newRow = num - 1;
}
if ( newCol > num - 1 ) {
  newCol = 0;
}

If a filled square is encountered, one moves vertically down one square instead, then continues as before.

if (magic[newRow][newCol] != 0) {
  newRow = row + 1;
  newCol = col;
}

I know this doesn't actually solve your problem but I hope it gets you somewhere towards your goal. See how I take the words of the algorithm and match them as accurately as possible with code.

Thanks for answering my question I would try to test this but is there a way that i could just edit my work because i think there's something wrong in my loop, i think my formula was wrong, i need a new formula for that — repsajznav, Dec 10 '14 at 17:37
@repsajznav The code I posted would be a good replacement of what you have in your loop — OldCurmudgeon, Dec 10 '14 at 20:05

There's something wrong in my Magic Square Algorithm

1 Answers1