Magic Square

This is another typical constraint satisfaction problem. There is a grid board. We are required to assign each grid a different digit from 1 to N such that all rows, all columns, and both of the diagonals have the same total. This example illustrates how to use the sum function.

  
Figure 5.4:  A solution to the magic square problem.

The following shows the program.

main class Magic { 
     public static final int N = 3; 
     component Cell cells[N][N];	
     attribute int total;	

     for (i in 0..N-1)   //columns 
          sum(cells[i][j].value, j in 0..N-1)==total; 
     for (j in 0..N-1)   //rows 
          sum(cells[i][j].value, i in 0..N-1)==total; 
     sum(cells[i][i].value,i in 0..N-1) == total; 
                         //left-upper-down diagonal
     sum(cells[i][j].value, 
         i in 0..N-1, j in 0..N-1, i+j==N-1) == total;
                         // left-bottom-up diagonal 
     for (i1 in 0..N-1, j1 in 0..N-1, 
          i2 in 0..N-1, j2 in 0..N-1)
          (i1!=i2 || j1!=j2) -> cells[i1][j1].value != cells[i2][j2].value; 
                         // all different
     grid(cells); // constrain the position
}

class Cell {
      component Label lb;	
      attribute int value in 1..Magic.N*Magic.N;
      lb.text == String(value);
}