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