Nim - Proof of The Winning Formula

Proof of The Winning Formula

The soundness of the optimal strategy described above was demonstrated by C. Bouton.

Theorem. In a normal Nim game, the player making the first move has a winning strategy if and only if the nim-sum of the sizes of the heaps is nonzero. Otherwise, the second player has a winning strategy.

Proof: Notice that the nim-sum (⊕) obeys the usual associative and commutative laws of addition (+), and also satisfies an additional property, x ⊕ x = 0 (technically speaking, the nonnegative integers under ⊕ form an Abelian group of exponent 2).

Let x₁, ..., x_n be the sizes of the heaps before a move, and y₁, ..., y_n the corresponding sizes after a move. Let s = x₁ ⊕ ... ⊕ x_n and t = y₁ ⊕ ... ⊕ y_n. If the move was in heap k, we have x_i = y_i for all i ≠ k, and x_k > y_k. By the properties of ⊕ mentioned above, we have

t = 0 ⊕ t = s ⊕ s ⊕ t = s ⊕ (x₁ ⊕ ... ⊕ x_n) ⊕ (y₁ ⊕ ... ⊕ y_n) = s ⊕ (x₁ ⊕ y₁) ⊕ ... ⊕ (x_n ⊕ y_n) = s ⊕ 0 ⊕ ... ⊕ 0 ⊕ (x_k ⊕ y_k) ⊕ 0 ⊕ ... ⊕ 0 = s ⊕ x_k ⊕ y_k (*) t = s ⊕ x_k ⊕ y_k.

The theorem follows by induction on the length of the game from these two lemmas.

Lemma 1. If s = 0, then t ≠ 0 no matter what move is made.

Proof: If there is no possible move, then the lemma is vacuously true (and the first player loses the normal play game by definition). Otherwise, any move in heap k will produce t = x_k ⊕ y_k from (*). This number is nonzero, since x_k ≠ y_k.

Lemma 2. If s ≠ 0, it is possible to make a move so that t = 0.

Proof: Let d be the position of the leftmost (most significant) nonzero bit in the binary representation of s, and choose k such that the dth bit of x_k is also nonzero. (Such a k must exist, since otherwise the dth bit of s would be 0.) Then letting y_k = s ⊕ x_k, we claim that y_k < x_k: all bits to the left of d are the same in x_k and y_k, bit d decreases from 1 to 0 (decreasing the value by 2d), and any change in the remaining bits will amount to at most 2d−1. The first player can thus make a move by taking x_k − y_k objects from heap k, then

t = s ⊕ x_k ⊕ y_k (by (*)) = s ⊕ x_k ⊕ (s ⊕ x_k) = 0.

The modification for misère play is demonstrated by noting that the modification first arises in a position that has only one heap of size 2 or more. Notice that in such a position s ≠ 0, therefore this situation has to arise when it is the turn of the player following the winning strategy. The normal play strategy is for the player to reduce this to size 0 or 1, leaving an even number of heaps with size 1, and the misère strategy is to do the opposite. From that point on, all moves are forced.