Message #1464

From: Andrew James Gould <>
Subject: Re: [MC4D] Social dream
Date: Mon, 28 Feb 2011 21:44:32 -0600

Hi guys,
No problem, Roice. It opened my mind a bit. I like 4D geometry once I get the lingo.

Speaking of which, what’s a cutting plane? In the 2^4 cube, my guess is that there are four: the 3D hyperplanes between slices of cubies, namely, w=0, x=0, y=0, and z=0. If so, then I agree with Andrey, you may select any cutting plane perpendicular to any vector R from [P,Q]. You can select multiple such cutting planes for a single twist to get my 2D slices. Technically, one could either take the union or intersection of the subsets created by multiple cutting planes. I agree with Roice, this does get complex.

Andrey, this afternoon I did a little calculus to verify that your twelve 180-degree rotations do work, but if you cut the tesseract up like MC4D does, you can only apply them to the whole tesseract. If you try them on a 4C cubie alone, the corners (and 2D faces, etc.) of that cubie run into other cubies–same with a 2D slice: cubies run into cubies. Now that I think about it, your pic does have diagonal cuts. …(analyzing)…they are correct. Each only uses one cutting plane.

For your simple bi-axis (1,0,0,0)x(0,1,0,0), you can also rotate four internal 2D slices on the 3^4 (between R=(1,1,0,0) and R=(1,-1,0,0), etc.), so I count eight 2D slices not including mask layers.

For your 120-degree twist about R=(1,1,1,1): If it were me, I would write R1=(1,0,0,0) in the log–namely, the center of the face that gets rotated.


—– Original Message —–
From: "Roice Nelson" <>
To: "4D Cubing" <>
Sent: Monday, February 28, 2011 6:18:51 PM
Subject: Re: [MC4D] Social dream

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?


On Mon, Feb 28, 2011 at 4:32 AM, Andrey < > 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=" " />
and 0D-centered is here: <img src=" " />

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.


— In , "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.
> 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