Comparing 2 elements of a 3x3 matrix to see if element 2 is adjacent to element 1.

Question

Basically, I'm creating a puzzle where you can swap pieces. And I want to make sure that when swapping 2 elements, the selection is valid.

Since the puzzle is only 9 pieces (3x3), I am currently using the code:

  function valid_selection(p1, p2) {
   if (p1 == 1 && (p2 == 2 || p2 == 4)) return true;
   if (p1 == 2 && (p2 == 1 || p2 == 3 || p2 == 5)) return true;
   if (p1 == 3 && (p2 == 2 || p2 == 6)) return true;
   if (p1 == 4 && (p2 == 1 || p2 == 5 || p2 == 7)) return true;
   if (p1 == 5 && (p2 == 2 || p2 == 4 || p2 == 6 || p2 == 8)) return true;
   if (p1 == 6 && (p2 == 3 || p2 == 5 || p2 == 9)) return true;
   if (p1 == 7 && (p2 == 4 || p2 == 8)) return true;
   if (p1 == 8 && (p2 == 5 || p2 == 7 || p2 == 9)) return true;
   if (p1 == 9 && (p2 == 6 || p2 == 8)) return true;

   return false;
  }

But, can I do this programatically? Anyone know of such an algorithm?

Any help is appreciated.

Answer 1

Assuming your matrix has positions like so:

1 2 3
4 5 6
7 8 9

You should be able to do the following:

if ( abs(p2-p1) == 3 // test for vertical connectedness
        || ( abs(p2-p1) == 1 // test for horizontal connectedness
        && ( p1+p2 != 7 && p1+p2 != 13) ) ) // except for edge cases (3,4 and 6,7)
    return true;

Answer 2

You could also convert each piece on the grid into coordinate form.

ie:

1 is (0,0), 2 is (0,1), 3 is (0,2), 4 is (1,0), etc

So, given that the coordinate of p1 is (x_p1, y_p1) and p2 is (x_p2, y_p2) then your function would return true if:

( abs(x_p2 - x_p1) + abs(y_p2 - y_p1) ) == 1

I think...? Haven't actually tried it.

And this should work regardless of grid size.

Answer 3

Assuming this is JavaScript:

var N = 3;  // size of matrix

var x1 = p1 % N, y1 = Math.floor(p1 / N);
var x2 = p2 % N, y2 = Math.floor(p2 / N);

return (x1 == x2 && Math.abs(y2 - y1) == 1) ||
       (y1 == y2 && Math.abs(x2 - x1) == 1);