Message #1283

From: Roice Nelson <roice3@gmail.com>
Subject: Re: [MC4D] Re: 3^4 one 4C left to orient
Date: Tue, 07 Dec 2010 17:53:12 -0600

Hi Andrew,

I’m chiming in a bit late here, but I thought I’d mention that David Smith
based the portion of his argument for M120C corner orientations on material
from a paper called "The Rubik Tesseract <http://helm.lu/cube/tesseract.pdf>"
by Kamack/Keane, and there you can find the proof you’re seeking for the
hypercube puzzle.

Btw, Don gave an intuitive explanation for why you can twirl a corner in
isolation on the 5D cube in message
243<http://games.groups.yahoo.com/group/4D_Cubing/message/243>
.

> This is fairly easy to see if you think about it–
> it’s exactly the same reasoning that lets you
> twirl a single 3-sticker cubie on the 3^4 puzzle.
> The crux of the matter is that
> in >=5 dimensions you can’t tell the difference
> between a cycle and an anti-cycle (of 3 stickers on a corner cubie);
> that is, one can be rotated to the other.
>

The topic of twirled corners is also highly interesting in the context of an
analogy to quarks discussed in Hofstadter’s Metamagical
Themas<http://www.amazon.com/gp/product/0465045669?ie=UTF8&tag=gravit-20&linkCode=as2&camp=1789&creative=390957&creativeASIN=0465045669>.
I’ve never investigated, but I wonder if the particle physics analogies
could be improved with our higher dimensional variants.

Cheers,
Roice

On Mon, Dec 6, 2010 at 2:10 AM, Andrey <andreyastrelin@yahoo.com> wrote:

> Here is the message about 120cell:
> http://games.groups.yahoo.com/group/4D_Cubing/message/658
> Less than 2 years ago, but already almost in the middle of the archive!
>
>
> — In 4D_Cubing@yahoogroups.com, "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 4D_Cubing@yahoogroups.com, "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
> > >
> >
>
>