# Game theory problem - two towers

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).

• @MohammadZuhairKhan the link you attached is for problem where you should take k stones at turn, but there's no such requirement at all there to take that amount at every turn May 12 '19 at 11:24
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.