I agree that the explanation in Math Horizons doesn't prove what it claims to prove. It also sloppily omits the dimensions of the intended box: 5 x 11 x 13 (= 2015). But it also says that each player must play in such a way as to remove, in each move, the maximum number of cubes they can. The 2nd example of the second player picking (1, 2, 2) can't remove the maximum number of remaining cubes, or it would end the game. —Dan
On Mar 10, 2016, at 12:41 PM, James Propp <jamespropp@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun