e   0 1 2 3
   -------
0| 0 1 0 1
1| 1 0 1 0
2| 0 1 0 1
3| 1 0 1 0 

f(2,1) = 1 (y is either 1 or 3 | x is either 0 or 2)

p1 sends 0

   0 1 2 3
0| 0 1 0 1
2| 0 1 0 1

p2 sends 1

  1 3
  ---
0| 1 1 
2| 1 1

Why AND Fails:

   0 1 2 3
   --------
0| 0 0 0 0
1| 0 1 0 1
2| 0 0 1 1   
3| 0 1 1 1

f(1,2) = 1

p1 sends....there is no way to send information without revealling out input...the matrix is not partitionable.