# Winning strategy

Is there a winning strategy for player one or two in the following scenario: The game begins with the number 2012. In one turn, a player can subtract from the current number any natural number less than or equal to it that is a power of 2. The player who reaches 0 wins.

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This is an impartial game, which is defined as one where each player has the same moves from a given position. The en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem shows that each position is equivalent to a Nim heap and how to calculate which one. There is a good discussion in vol 1 of Winning Ways amazon.com/Winning-Ways-Your-Mathematical-Plays/dp/1568811306/… The Wikipedia article also cites math.ucla.edu/~tom/Game_Theory/comb.pdf which looks very good to me. – Ross Millikan Jun 14 '12 at 21:15

so for a slow winning game subtract $1$ or $2$ each time and for a fast winning game subtract $2^{\lfloor \log_2 n \rfloor}$ or $2^{\lfloor \log_2 n \rfloor - 1}$ each time – Henry Jun 14 '12 at 19:45