1. Using Reduction for the Game of NIM

2. At each turn, a player chooses one pile and removes some sticks. The player who takes the last stick wins. Problem: Is there a move that guarantees a win for the current player? Nim

3. Nim ● Obvious approach: search the space of possible moves. ● Reduction to an XOR computation problem: 100110 101101 010011 011

4. Nim ● Obvious approach: search the space of possible moves. ● Reduction to an XOR computation problem: 100110 101101 010011 011 ● XOR them together: ♦ 0 + means state is losing for current player ♦ otherwise current player can win by making a move that makes the XOR 0.

5. Correctness Proof of Nim Reduction If all 0’s, there is no move that preserves all 0’s: If not all 0’s, there is a move that makes it all 0’s:

6. Correctness Proof of Nim Reduction If all 0’s, there is no move that preserves all 0’s: Suppose there were such a move. Then it would change some row n. Since it must subtract at least 1 from the number in row n, it would necessarily change a 1 in some column j to a 0. Since the XOR of that column was 0, it now becomes 1. The only way to return it to 0 would be to alter some other row in column j. But a given move can only alter a single row. So the new XOR will contain at least one 1. 1 1 0 0 1 0

7. Correctness Proof of Nim Reduction 1.Scan from the left to find the first column, j, that contains a 1 in the XOR row. 2.Choose a row, i, that contains a 1 in column j. Change the 1 to 0. 3.Continue moving from the left. If the XOR row contains a 1 in column j, flip bit (i, j). 1 0 1 1 1 1 0 0 1 0 1 1 If not all 0’s, there is a move that makes it all 0’s: At the end, the XOR row is all 0’s because we flipped exactly one bit in every column that had been 1 and no bits in columns that had been 0. The flipping corresponds to a legal move because it changes only one row and it creates a row that represents a smaller number than the original one.