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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