the game is structured like this:

  • two players
  • the players alternate moves
  • 4 heaps $h_1,h_2,h_3,h_4$ with sizes $n_1,n_2,n_3,n_4$
  • at each move, the player can either remove one or two elements from any of the heaps (meaning that if the player takes two elements, those two can either be from the same heap or from two different heaps, as long as the total number of removed elements per turn is 2).
  • the game stops when the player (who loses the game) is left with 3 heaps of size $0$ and one heap with size greater than $0$.

Can anyone suggest a winning strategy for this game?

ty.thank-you

  • just to clarify, only one or two elements can be removed from the heap at every move by a player? – user3483902 Jan 29 at 8:26
  • @user3483902 from the heapS. You can remove two elements from two different heaps. And yes, for each turn one can remove at most 2 elements in total. – Simone Procaccia Feb 1 at 20:07
  • What happens in the scenario where your ending condition never occurs? So $\{1, 1, 0, 0\} \to \{0, 0, 0, 0\}$ in the final turn. – orlp Feb 2 at 12:07
  • @orlp I would say it is a losing position too. You can't make any move from $\{0,0,0,0\}$, so I believe it is a losing position. – Simone Procaccia Feb 2 at 20:07

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