# 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

> > >

> >

>

>