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This question is inspired by IBM's "Ponder This" for this month. August 2013. So, perhaps without too much help, could someone let me know if I would be on the right track to try to create a 3 "layer" Hasse of a 9 cube. Because a 9 cube has 512 vertices, I wanted to make sure this wasn't futile before I ventured down this path. Thanks for any help. The original question is here.

Thanks again. -court

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By three layer, do you mean going down 3 dimensions to a 6-cube, and thinking of the 9-cube as a Hasse involving first drawing the 6-cubes (based on the last three entries in the 9-tuple being taken from $\{0,1\}$ in each of the eight possible ways? If so that may be a good thing to try, since one can more easily visualize the connecting up process, hopefully arriving at a solution. –  coffeemath Aug 3 '13 at 3:51
EXACTLY! I just wanted to make sure I wasn't going on a wild chase. Thanks so much... –  court Aug 3 '13 at 3:58

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