# More than Nim (combinatorics problem)

A two-player game is played with two piles of stones, with sizes m,n. On a player's turn, that player can remove any positive number of stones from one pile, or the same positive number of stones from each pile. A player loses when they are unable to take a stone. If 1≤m,n≤30, for how many of the 30×30=900 starting positions does the first player have a winning strategy?

• I.e., the winner takes the last stones. If there are the same number of stones on both piles, the next player can take all, so the strategy must be to keep them different and none empty (one pile empty means the other player can take it all). – vonbrand Mar 23 '13 at 20:32
• This is known as Wythoff's game. It has a very interesting theory. – MJD Mar 24 '13 at 13:38

## 1 Answer

As pointed out by MJD, this game is known as Wythoff's game and has a very interesting theory. As discovered by Wythoff, the losing positions are the pairs $(m_k,n_k)$ and $(n_k,m_k)$ with

$$n_k=\left\lfloor k\phi\right\rfloor\;,\\ m_k=\left\lfloor k\phi^2\right\rfloor\;,$$

where $k\in\mathbb N$ and $\phi=\frac{1+\sqrt5}2$ is the golden ratio. Solving $k\phi^2\lt31$ for $k$ yields $k\lt31/\phi^2\approx 11.84$, so there are $2\cdot11=22$ losing positions and thus $900-22=878$ winning positions with $1\le m,n\le30$.