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Apr 7, 2014 at 8:08 comment added nsrt In the 3D case, $n=p$, the first two steps can be done similarly. One obtains a placement where one must still switch all numbers on one edge of the cube (instead of one corner in the 2D case). This should be possible for $p>3$, but I didn't see a "clean" way how to do it for all such primes $p$.
Apr 3, 2014 at 12:38 history answered nsrt CC BY-SA 3.0