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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Forming a Magic Square
  5. Discussions

Forming a Magic Square

  • zettix1
     + 0 comments

    I first precomputed all magic squares. There's only one: 4 rotations and 4 mirrors, hence 8 representations of the same magic square. I then printed them and hard coded them.

    public static int formingMagicSquare(List<List<Integer>> s) {
// Write your code here
// precomputed all magic squares.
    int[][] allmagic = new int[][] {
    {2, 7, 6, 9, 5, 1, 4, 3, 8}, 
    {2, 9, 4, 7, 5, 3, 6, 1, 8}, 
    {4, 3, 8, 9, 5, 1, 2, 7, 6}, 
    {4, 9, 2, 3, 5, 7, 8, 1, 6}, 
    {6, 1, 8, 7, 5, 3, 2, 9, 4}, 
    {6, 7, 2, 1, 5, 9, 8, 3, 4}, 
    {8, 1, 6, 3, 5, 7, 4, 9, 2}, 
    {8, 3, 4, 1, 5, 9, 6, 7, 2}
    };
    int[] sint = new int[9];
    for (int y = 0; y < 3; y++) {
        for (int x = 0; x < 3; x++) {
            sint[x + y * 3] = s.get(y).get(x);
        }
    }
    int count = 9 * 9;
    for (int i = 0; i < allmagic.length;i++) {
        int mycount = 0;
        for (int j = 0; j < 9; j++) {
            mycount += Math.abs(allmagic[i][j] - sint[j]);
        }
        if (mycount < count) {
            count = mycount;
        }
    }
    return count;
}