I'm asking that question because I still cannot figure out the solution after hours of thinking.
You are given two towers where first has exactly n stones and second has exactly m stones. You are also given a number k. During each round you can perform one of the actions listed below:
- take exactly k stones from the first tower,
- take exactly k stones from the second tower,
- take exactly k stones from both towers.
There are two players: A and B. Player A starts. The winner is the player who has performed last possible action. How to determine after given n, m, k numbers who has winning strategy (who will win independently from opponent moves).
You can divide through by $k$, equvalently making $k=1$.
Let $m=xk+a, n=yk+b$, with both $a$ and $b$ between $0$ and $ k-1$. Then the $a$ and $b$ stones left over play no part in the game. Every round, you reduce either $x$, $y$ or both by $1$.
So let $k=1$. Which positions with $y=0$ are wins, and which are losses ? If $y=1$, can you put the position into a lost position for your opponent ? So which positions are wins when $y=1$? Then do $y=2$, and so on.