Message #1463

From: Roice Nelson <roice3@gmail.com>
Subject: Re: [MC4D] Social dream
Date: Mon, 28 Feb 2011 18:18:51 -0600

Hi Andy/Andrey,

I clearly missed the issue Andy raised earlier, so apologies for my
previous email in this thread, which probably appeared as both obvious and a
bit of a non sequitur.

Anyway, I can’t help myself from feeling like this is getting quite
complicated, which feels a bit unfortunate. I do like that the new
suggestion looks to handle the 5D case now, but I found my mind trying to
find a way around enumerating the full set of rotational planes in the
file (especially in the 4D case). Those will all have to be generated and
ordered in a specified manner, like the axes. (This left me wondering if it
would be simpler to specify how to algorithmically generate/order stickers.)

I am uncertain about something else now too (in the 4D case). Early on,
Andrey wrote:

> For every axis we have number and positions of cutting planes.

The spaces orthogonal to the 4D symmetry axes are 3D. So it seems a cutting
plane position will need to contain orientation information as well (not
just a single number of how deep the cut is on the axis). Is that right?
If so, maybe the rotational plane and cutting plane data both reference the
same underlying set?

Best,
Roice

On Mon, Feb 28, 2011 at 4:32 AM, Andrey <andreyastrelin@yahoo.com> wrote:

> Hi Andy,
> I see 12 more rotational planes: they have fixed bivectors
> (1,1,0,0)x(0,0,1,1) with sign changes and coordinate permutations.
> These planes can’t be used in 3D-centered twists (because bi-axis doesn’t
> contain coordinate vectors), but there is a couple of 180-deg twists with
> them:
> 2D-centered twist is here: <img src="
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1184359940/pic/527704145/view"
> />
> and 0D-centered is here: <img src="
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1184359940/pic/579107863/view"
> />
>
> So if you have some rotation in 4D around the plane [P,Q] where P and Q are
> some vectors, you may select any cutting plane perpendicular to vector R
> from [P,Q] and then rotate the slice around some other vector R1 from [P,Q].
>
> For example, in the most simple situation when bi-axis is
> (1,0,0,0)x(0,1,0,0), one can use it to rotate any of 4 dufferent 3D cells
> (R=(1,0,0,0) or (0,1,0,0)), or any of 4 2D ridges (R=(1,1,0,0) or
> R=(1,-1,0,0)).
>
> When I wrote my description of twist, I thought that if we select vector R1
> perpendicular to R, then it will be an axis of the figure. But now I see
> that it’s wrong: if we take R=(1,1,1,1) (0D-centered twist) and perform
> 120-degree rotation, then actual axis R1 will be something like (-3,1,1,1)
> that is not any axis of the tesseract. Of course, I can write in log that
> R1=(0,1,1,1) or (1,0,0,0) (so that [R,R1] gives the rotation bi-axis), but
> such selection will be arbitrary, not canonical, and I’m not sure that the
> rotational bi-axis will always contain two axes of the figure.
>
> It looks like we should enumerate all rotational planes as well as axes,
> and use orthogonal pairs (axis,plane) in the log file.
>
> This method will work for all finite puzzles where twists can be defined in
> 2D way (with D-2 dimensional multy-axis), but there will be a lot of
> external data for the axes set definitions (list of axes, and list of planes
> - and there are some infinite families of sets…) So format will be not as
> elegant as it supposed to be.
>
> For 3^5 there is a restricted set with only 5 axes and 10 rotational planes
> - and only 30 axis/plane pairs may appear in log. For 3^7 there are 7 axes
> and 21 planes (105 possible combinations) an for 5D simplex we have 6 axes
> and 20 planes (or 10? not sure) that give us 120 deg-twists of faces.
>
>
> Andrey
>
> — In 4D_Cubing@yahoogroups.com, "Andrew Gould" <agould@…> wrote:
> >
> > I’m seeing 46 rotational planes for the tesseract (6 planes for 90-degree
> > rotations, 24 for 180-degree, 16 for 120-degree), now I’m trying to
> > translate into Andrey’s 40 axes–I got confused.
> >
> >
> >
> > I’m keeping my so-called "2D twists" in mind. They seem to use the same
> > axes/rotational planes that are already used.
> >
> > http://groups.yahoo.com/group/4D_Cubing/photos/album/1774759718/pic/list
> >
> >
> >
> > The deal in 5D is that you can twist 4D slices, 3D slices, or 2D
> > slices…but again I’m seeing that they all get twisted about
> > axes/rotational planes that are already used for twisting 4D slices.
> >
> >
> >
> > –
> >
> > Andy
>