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).

  • $\begingroup$ @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 $\endgroup$ May 12 '19 at 11:24
  • $\begingroup$ My bad, I should have read it better. en.wikipedia.org/wiki/Wythoff%27s_game may help you. $\endgroup$ May 12 '19 at 11:26
  • $\begingroup$ No problem. Thanks for help. $\endgroup$ May 12 '19 at 11:29

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.


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