Thinking about this some more, I realize that you can, in fact, arbitrarily permute the coordinates of any given axis, at any point in the game, without affecting game play. So rather than think of the coordinates as ordered, you can think of them as unordered sets of indices. I think James Propp may have made this observation in an earlier post? With this observation, I believe the second move can be limited to 4 possibilities. You may preserve at most one coordinate index with respect to the initial move (to obtain a maximal number of cubes), so that leaves the following possibilities for the second move: (1) preserve x (2) preserve y (3) preserve z (4) preserve none. I'm not sure if this helps, but I thought I'd throw it out there. Tom Tom Karzes writes:
Here's an observation: The edges of the box actually serve no purpose, other than to provide a frame of reference. Specifically, I believe you could rotate the cubes along any of the 3 axes without affecting the game play. This implies, for instance, that all initial moves are equivalent.
Perhaps it would be better to think of the box as the surface of a 3D torus (i.e., a torus with a 3D surface), with no edges at all? The lines can then be thought of as circles through the torus.
Tom
Allan Wechsler writes:
On third thought, all of these rules just amount to arranging the 2015 cubes in some order. It might be worth assigning a random order to them to see how well that works. Alternatively, just pick one of the maximal-yield cubes at random.
On Fri, Mar 11, 2016 at 4:53 PM, Allan Wechsler <acwacw@gmail.com> wrote:
What if you just break ties by x + 5y + 65z from the outset?
On Fri, Mar 11, 2016 at 4:48 PM, Mike Beeler <mikebeeler@verizon.net> wrote:
I tried 4 tie breaking rules, applied when more than one candidate cube has the same (maximal) size of family that would be removed. If the rule leaves more than one candidate, the one with smallest index x+5y+65z is chosen, where box dimensions in x, y, z are 5, 13, 31 and x, y, z begin at 0.
A) Remove cube closest to any corner of the box. B) Remove cube closest to center of any edge of the box. C) Remove cube closest to center of any side of the box. D) Remove cube closest to center of the box.
I was surprised that all four rules make games that end after 65 moves. Maybe it’s time to look for a proof.
However, the number of cubes Alice and Bob end with differ:
A) Alice 1023, Bob 992 (like Taylor and Jamie found) B) same as A C) Alice 1025, Bob 990 D) Alice 1021, Bob 994
Perhaps Dmitry (Alice 1024, Bob 991) had yet another tie breaking rule.
— Mike
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