Let me be more pointed about what troubles me. Suppose we play Grabbing Cubes with a 2x2x2 block. Number the cubes (i,j,k) with i,j,k each equal to 1 or 2. If the first player grabs (1,1,1) (removing also (1,1,2) and (1,2,1) and (2,1,1)) and the second player grabs (2,2,2) (removing also (1,2,2) and (2,1,2) and (2,2,1)), then the game ends after two moves. On the other hand, if the first player grabs (1,1,1) and the second player grabs (1,2,2), then the game lasts more than two moves. Am I missing something here? Jim Propp On Thu, Mar 10, 2016 at 12:11 PM, James Propp <jamespropp@gmail.com> wrote:
Am I the only one who is dissatisfied with the solution to Problem 327 in the Playground feature of the February 2016 issue of "Math Horizons" (see http://digital.ipcprintservices.com/article/The_Playground/2387298/289441/ar... , page 31)?
To show that a certain cube-grabbing game always ends in 65 moves, the authors first show that an obvious way to play the game ends in 65 moves, and then show that "the algorithm for removing the maximum number of cubes at each steps takes 65 steps".
But I don't see why it follows that EVERY way to play the game ends in 65 moves.
Question #1: Am I right to be dissatisfied?
Question #2 (assuming that the answer to Question #1 is "Yes"): Can someone provide a proof that will satisfy me?
Jim Propp