fieldnotes/content/random-notes

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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.