Message #1278

From: Andrey <>
Subject: Re: 3^4 one 4C left to orient
Date: Mon, 06 Dec 2010 08:02:29 -0000

what we need there is to map orientations of 4C piece to the group Z_3={-1,0,1} in such way that: (1) changing of the code of some piece orientation during a twist depends only on the piece position (not on its orientation before twist!) and is additive (i.e. m’=m+f(t,p) mod 3, where m is code of orientation "before", m’ is code of orientation of the same piece "after", t is twist description, p is piece position) and (2) that sum(f(t,p)) for all positions p is zero for every twist.
I think that the following construction will work:
Let faces of the cube be {a,b,c,d,A,B,C,D} ("A" is opposite to "a" and so on). Divide this faces to 2 sets (tubes): one set is {a,b,A,B} and another is {c,d,C,D}. Call a piece well-oriented if its stickers that initially belong to one tube (say, "a" and "B") are laying in faces that also belong to one tube (e.g. sticker "a" is in the face "C" and sticker "B" is in the face "D"). This orientation has code=0. If sticker in not well-oriented do the following. Enumerate faces of the cube that contains this sticker now such that first is "a" or "A", second is "b" or "B" and two last faces give positive (right-handed,…) orientation of the vertex (like abcd, aBDC, AbcD etc.) Take the sticker S of the piece that lays in the first face and look for sticker S’ that initially was in the same tube with S. Now it can lay in second, third or fourth faces wrt our enumeration. The code of the orientation will be 0,1 and -1 respectively.
The rest is to prove properties (1) and (2) for this mapping. I didn’t do it yet but hope that they both are true.
Even if this construction works, it’s not easy to generalize it to other figures - shallow-cut simplex and 120-cell. But somehow I think that the invariant shoud be true for them too.

And I think that couple of years ago there was long post about the same problem for 120-cell. Don’t remember when was it and how was the author.


— In, "Andrew Gould" <agould@…> wrote:
> 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