Message #1277

From: Andrew Gould <>
Subject: 3^4 one 4C left to orient
Date: Fri, 03 Dec 2010 13:41:05 -0600

I’m having trouble proving something for the 4D cubes. In either case, ALL
but one 4C piece is perfectly solved. I can see that there are at least 4
possible states for this remaining 4C piece (the solved state and three
states where 2 independent pairs of stickers are permuted).and although I
haven’t seen it, I can’t prove that there are more possible states. Anyone
have an argument as to why you can’t have all stickers solved except a
3-cycle of stickers on the remaining 4C piece?

I see in 5 dimensions (and higher) you can have a single 3-cycle on the
remaining 5C piece (and higher).

I also had similar trouble with the original 3D cubes: trying to prove why
you can’t have everything solved except one 3C piece (where its stickers
would be in an unsolved 3-cycle). My sketchy argument there had to define
what it means to ‘permute 3C pieces without disorienting them’.I defined it
as yellow/white corner stickers must be facing the yellow/white face (yellow
is opposite white on my cube). I then argued that a simple twist always
orients the SUM of all 3C pieces by a multiple of 360 degrees. I don’t know
if a similar argument will work in 4D.