# Message #1277

From: Andrew Gould <agould@uwm.edu>

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.

–

Andy