Message #1282

From: Andrew Gould <>
Subject: RE: [MC4D] Re: 3^4 one 4C left to orient
Date: Tue, 07 Dec 2010 01:21:55 -0600

Thanks Andrey,

I never realized that if ‘a’ and ‘b’ start in the same tube, then they
stay in the same tube in the klein 4 subgroup:
{{a,b,c,d},{b,a,d,c},{c,d,a,b},{d,c,b,a}}. Interesting and neat.

Your construction seems to be true for (1): either your code stays the
same for all 4C pieces, or else for the 4C pieces that do get twisted: the
code of the 4C pieces in the CUBE positions with C (or c) before D (or d)
does that opposite of the code of the 4C pieces in the CUBE positions with D
(or d) before C (or c). That is, if you travel between 4C pieces via the
1-dimensional edges only, every other 4C piece (that gets twisted) increases
it’s code by 1(mod3)–the code of the remaining twisted pieces decreases by
1(mod3). This is true for 90-degree 3D face twists, 120-degree twists, and
180 degree twists.and even my 90-degree 2D face twists. Since the number of
pieces that increase in code = the number of pieces that decrease in code,
sum(f(t,p)) = 0(mod3). Hence it’s true for (2).

I’m eager to check the links out when I find time.


From: [] On Behalf
Of Andrey
Sent: Monday, December 06, 2010 2:10
Subject: [MC4D] Re: 3^4 one 4C left to orient

Here is the message about 120cell:
Less than 2 years ago, but already almost in the middle of the archive!

— In <> ,
"Andrey" <andreyastrelin@…> wrote:
> Andy,
> 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.
> Andrey
> — In <> ,
"Andrew Gould" <agould@> wrote:
> >
> > I’m having trouble proving something for the 4D cubes. In either case,
> > but one 4C piece is perfectly solved. I can see that there are at least
> > possible states for this remaining 4C piece (the solved state and three
> > states where 2 independent pairs of stickers are permuted).and although
> > haven’t seen it, I can’t prove that there are more possible states.
> > 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
> > 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
> > what it means to ‘permute 3C pieces without disorienting them’.I defined
> > as yellow/white corner stickers must be facing the yellow/white face
> > 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
> > if a similar argument will work in 4D.
> >
> >
> >
> > –
> >
> > Andy
> >