Regular compounds in 4D

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Re: Regular compounds in 4D

Postby username5243 » Tue Feb 11, 2020 10:05 pm

For convex hulls, I'm worried some of the 3-tes compounds might just be variants of icoic polychora. (Maybe not all of them, though.) I say this because Klitzing's site gives representations of many icoic polychora as tegum sums of three tessic polychora, though that's when considered under demitessic subsymmetry, so I'm not sure if they'd all coincide with variants of known polychora. The 2-pens and 2-icoes (and consequently, the 6-tesses) would be new though.

Tere's also some compounds of 2 tessic polychora if they remain uniform under demitessic symmetry, but there are some that won't be truly "uniform". These are like haddet (2-hex compound), but not all tessic uniforms can form these. For instance, there's a compound of 2 tesses (in fact, the dual of haddet), but I'm fairly sure that 16 of the vertices will coincide by 2 (providing a so shaped verf at those 8 combined points) while the other 16 don't coincide (tet shaped verf). So this 2-tes compound is not uniform. Not sure if that happens to other tessic polychora as well...
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Re: Regular compounds in 4D

Postby username5243 » Tue Feb 11, 2020 10:07 pm

Mecejide wrote:
Mercurial, the Spectre wrote:
Klitzing wrote:
username5243 wrote:Hmm wonder if Bowers (or anyone else) has names thought up for these yet...


I contacted him privately and he answered me:

I've already played around with these as well as 2-tes and 2-ico versions. Here are the short names for them:

<list snipped>

I had these written in one of my spirals from several years ago.

These will be in categories C4 (tupens and threetesses) and C5 (tuicoes and sixtesses), C4 will also contain the 2-tesses. I'm making an attempt to list compound categories in 4-D, they'll be on my next website update. I also named all of the tupens and even done renders of them.

Jonathan B.


I missed a few on my list, I accidently skipped past the thex ones. I also noticed that I called 2-ico stic on that list, I'll need to check my notes to see where I called it stoc and where I called it stic:

<more list snipped>

Jonathan B.

--- rk


I would be interested in their convex hulls.


Me too, especially coquatapdic.


I'm not even sure what the convex hull of iquatoc itself is, though... How do you even find the convex hull of a general non-convex Wythoffian uniform polychoron, anyway?
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Re: Regular compounds in 4D

Postby ndl » Wed Feb 12, 2020 2:09 am

username5243 wrote:I'm not even sure what the convex hull of iquatoc itself is, though... How do you even find the convex hull of a general non-convex Wythoffian uniform polychoron, anyway?

I use stella4D to find convex hulls.

Iquatoc is w3x4o3x.

The 2 compound has the extended symmetry and doesn't have a good way to represent the vertex arrangement.
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Re: Regular compounds in 4D

Postby Klitzing » Wed Feb 12, 2020 11:32 am

ndl wrote:The 2 compound has the extended symmetry and doesn't have a good way to represent the vertex arrangement.

At least for pennic and icoic 2-compounds you might try with a3b3b3a resp. a3b4b3a for hulls.
--- rk
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Re: Regular compounds in 4D

Postby ndl » Wed Feb 12, 2020 9:22 pm

Klitzing wrote:At least for pennic and icoic 2-compounds you might try with a3b3b3a resp. a3b4b3a for hulls.
--- rk


I'm not understanding what the "a" and "b" represent.
If I wanted, for example, to describe the convex hull of the 6sidpith compound mentioned above. I know it has a cube with w edges and square antiprism with w and k edges, along with 2 different tets. So how would that be written in inline notation?
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Re: Regular compounds in 4D

Postby Klitzing » Wed Feb 12, 2020 9:38 pm

a and b are free variables (of corresponding edge size). - But keep in mind that I intensionally spoke of "pennic" (pentachoron symmetric) resp. "icoic" (icositetrachoron symmetric), refering to the being used components, and considering corresponding 2 componds for sure.
--- rk
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Re: Regular compounds in 4D

Postby username5243 » Wed Feb 12, 2020 10:19 pm

You need to use the "&#zx" tegum sum notation to represnet hulls of these hulls, since many of them aren't uniform or even variants of uniforms that remain vertex transitive with different edge lengths.

The 2-pens and 2-icoes have non-uniform hulls in the tegum sum notation, represented by things like oo3xo3ox3oo&#zy (for 2-rap) and xo3ox3xo3ox&#zy (2-srip), where y is some other edge length chosen to put the components on the same 4-D hyperplane. The same thing happens for the 2-hex and its relatives, but you need to use demitessic ymmetry - and I believe must of those hull are mere vraiants of tessic uniform polychora. i'm less sure what the hulls of the 3-tes and 6-tes relatives - maybe icoic uniform variants for the 3-tesses and variants of the hulls of the 2-ices for the 6-tesses? - but it's harder to write hulls of 3 or 6 polychora than two in tegum sum notation, generally. (You might have to break the tessics into demi-tessic symmetry, or something...)
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Re: Regular compounds in 4D

Postby ndl » Thu Feb 27, 2020 2:59 pm

Looking at the 6-sidpith compound, 9 ondips can be created using sets of those vertices, with 3 ondips at each vertex. Is that compound uniform? Is it a valid compound at all?
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Re: Regular compounds in 4D

Postby ndl » Thu Feb 27, 2020 3:24 pm

ndl wrote:Looking at the 6-sidpith compound, 9 ondips can be created using sets of those vertices, with 3 ondips at each vertex. Is that compound uniform? Is it a valid compound at all?


After playing around in stella with the faceting of the combined vertex figures, it appears I have created a new scaliform!
384 vertices

288 + 96 tets
192 trips
288 cubes
288 digon cupola blends

Can anyone confirm this?

ondip scaliform.JPG
ondip scaliform.JPG (34.47 KiB) Viewed 5926 times
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Re: Regular compounds in 4D

Postby Mecejide » Thu Feb 27, 2020 6:45 pm

ndl wrote:Yes, that's what I had in mind.

Here's the 6-sidpith compound. It has 3+1 Cube components, the tets are all so, and the trips remain lonely.

6sidpith.JPG


The 3-rit compound turns out to be a faccetting of rico.

Still examining more of these. I wonder if I can make a uniform 3-ondip compound.

How did you make it?
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Re: Regular compounds in 4D

Postby Mecejide » Thu Feb 27, 2020 9:25 pm

ndl wrote:
ndl wrote:Looking at the 6-sidpith compound, 9 ondips can be created using sets of those vertices, with 3 ondips at each vertex. Is that compound uniform? Is it a valid compound at all?


After playing around in stella with the faceting of the combined vertex figures, it appears I have created a new scaliform!
384 vertices

288 + 96 tets
192 trips
288 cubes
288 digon cupola blends

Can anyone confirm this?

ondip scaliform.JPG

That is the blend of 24 sidpiths, in category S12.
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Re: Regular compounds in 4D

Postby ndl » Fri Feb 28, 2020 4:31 am

Mecejide wrote:That is the blend of 24 sidpiths, in category S12.


Yes, it appears I didn't read that earlier post by J.B. fully.

Mecejide wrote:How did you make it?


Which are you referring to?
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Re: Regular compounds in 4D

Postby Mecejide » Fri Feb 28, 2020 6:50 pm

ndl wrote:
Mecejide wrote:That is the blend of 24 sidpiths, in category S12.


Yes, it appears I didn't read that earlier post by J.B. fully.

Mecejide wrote:How did you make it?


Which are you referring to?

How did you make the compound of 6 sidpiths?
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Re: Regular compounds in 4D

Postby ndl » Sun Mar 01, 2020 1:34 am

Mecejide wrote:
ndl wrote:
Mecejide wrote:That is the blend of 24 sidpiths, in category S12.


Yes, it appears I didn't read that earlier post by J.B. fully.

Mecejide wrote:How did you make it?


Which are you referring to?

How did you make the compound of 6 sidpiths?


I took the compound of 3 sidpith with convex hull of x3w4o3o and rotated it along the icoic extended symmetry axis and put them together. As mentioned above this can be done with any polychoron with icoic symmetry.

Here's the list of vertices:

Code: Select all
#1

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#2

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#3

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#4

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0.70710678118654752   1.70710678118654752   0.70710678118654752   0.70710678118654752

#5

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#6

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Re: Regular compounds in 4D

Postby Mecejide » Sat Mar 07, 2020 5:41 pm

Each of the icoic snubs have three 2-compounds (one with full icoic symmetry and two with ionic contic symmetry) as well as a 4-compound (with full contic symmetry). Do these have names yet?
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Re: Regular compounds in 4D

Postby username5243 » Sat Mar 07, 2020 10:49 pm

I think these should work, the ones with full ico symmetry will be like the 3D "disnub" compounds. THe ikes (or gikes, for the starry cases) might well combine into their respective uniform 2-compounds here. Some of the tets might, too.
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Re: Regular compounds in 4D

Postby username5243 » Sat Mar 07, 2020 11:16 pm

Moving back to the initial topic, I thought I'd list all hte truly "regular" compounds that I can think of.

There's the 2-pen (sted), 2-hex (haddet), and 2-ico (stoc).

In the ico army we have 3-tes (gico) and 3-hex (stico) - I suspect most compounds of icoes can have ico replaced with either of these compounds. For example, from the 2-ico we get the 6-hex (sistic) and 6-tes (gistic).

Then there's dox (25-ico) in the ex army and sishi regiment. You can also get a 75-tes compound (dac) and a 75-hex compound (which needs a name still...any ideas? I started calling it "stidox" for stellated dox, but I'm not sure.) There's also a chiral 5-ico compound (chi) related to dox. I think chiral 15-tes and chiral 15-hex should exist, though I'm not sure.

Then there's the 10-ex (called "sody"), and it's relatives - many regulars (ex, fix, gohi, gahi, sishi, gaghi, gishi, gashi, gofix, and gax) seem to form regular compounds of 10 of them, that don't seem to be named yet. I also know there seems to be a chiral 5-ex formed from "halving" sody, so there's probably chiral 5 versions of the others too (for example a chiral 5-gaghi). The 10-sishi regiment would have a compound of icoes in it too - I would think 250, but Klitzing has it as 225, any clue where the difference comes from? Regardless, that ico compound will have its inscribed tes and hex compounds too.

Finally Klitzing says there's apparently a 120-pen compound (mix) and a 720-pen compound.

That seems to be all the regulars known so far, a whopping total of 35... any that I missed, or that I counted but shouldn't have?
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Re: Regular compounds in 4D

Postby wendy » Sun Mar 08, 2020 1:28 pm

The regular compounds that preserve all of the symmetry of the elements, are {4,3+3}, {3+3,4} and {3,3,3}_120.

The first is x3o3o4o *a4/2*b, the second is the dual thereof.

Two more preserve the symmetry, but are not regular figures, being the compound of two pens, and two icos, ie xo3ooXoo3oox for X=3, 4

The number in brackets eg (120) represent a set of vertices or directions of the centre, to the different regular polychora. If it is given as (120) the same set is used for the vertices and faces, eg {5/2, 5,3}. (24,24) represent the verticies of dual icos, which point in different directions.

The compounds of the ico figures in [3,3,5], are as i list them in the first post, some 124 different examples (or so). The trick is that the compound of 5(120) in (600), goes as (600)[5 (120,120)] 5 (120). This means, for example, the compound of 5 icos in {3,3,5} goes (120) [5 (24, 24)] (120) = (120,120). Because of this, we can insert 1, 2, 3, 4, 5 of this 5(24,24) cell, represented by eg 1, 2, 3, 4, 5 octahedra of the compound of 5 octahedra that make up the faces of the 600-vertex compound.

Since (600)[5 (120,120)] 5(120) is itself chiral, we can put its reflection in as 2(600)[10(120,120)]10(120). This is represented as five rows and five columns of a 5*5 grid. The whole grid represents the (600), the rows or columns are the (120), and the cells are (24). In the dual, the grid represents (120), each row or column represents (24), and each cell represents hexagons.

In the (600), since each row and column represents (120), we can replace each row/column with 1, 2, 3, 4, 5 sets of 5 (24). But in each case, there is a 'special' compound, represented by the cells of the grid itself. The other four are different fillings of the row/column by 5 different sets of 24ch. What we're doing is in effect, taking (600)[25(24,24)](120), and squeezing this into the (120,120) thing.

So to each cell, one can have four left-chiral 24ch, and four right-chiral 24ch. Any distinct pattern does, so we get for LSR (ie left, special, right)

100, 200, 300, 400, 101, 201, 301, 401, 202, 302, 402, 303, 403, 404
110, 210, 310, 410, 111, 211, 311, 411, 212, 312, 412, 313, 413, 414.

where n = L+S+R, we get compounds of n(600)[25n(24,24)]

The reciprocal 24's point in different directions (a total of 5400 vertices), but a single (24) represents both the compounds of 3 16ch and 3 tesseracts.

Each number in the table above represents four different compounds. 414 is the biggest, has a density of 225 for the 24ch and 675 for cubes and cross polychora,

In the s/f group, or 600n(120n(5,5))600n.

We note that the middle ring of the grand antiprism is 100 tetrahedra, the vertices of 120c at these points, form a bi-decagon prism. The six hundred vertices of this figure form sets of six bidecagon-prisms, which are the more remote vertices of a chain of ten dodecahedra. This forms a six-by-six array, since there are six left-chiral and six right-chiral clifford-parallels. In total there are 18 inscribed bi-decagonal prisms.

Now, of course, a bidecagon prism contains 100 vertices, which fall into 20 left-chiral pens, and 20 right-chiral pens. A set of 6 clifford-parallel bi-decagon prisms, contains therefore all 600 vertices, which we can choose to divide into 20 left or 20 right cliffords. If we restrict ourselves to left-cliffords, each row is 120 pens, and six rows make 720 pens. That is, there is a solution for the above of n=1,2,3,4,5,6. This includes all of the pens, so the right-chirals are left-chirals in another row. So while we can not get both the rows and column trick, we do know that we can select 20 left and 20 right, and these will for a single case, give the 120 pens twice over. This is different to '2 rows', and of course, the complement exists (6 minus 20 left - 20 right). So 2 and 4 can be doubled.

There is the metestar, the compound of 120 fully symmetric figures. In ring 18 of the twelfty-cell, there are 28 vertices, 24 as above, and four forming a symmetric tetrahedron. This is the mete star. It is adds to each of the members of the above, as well as being an isolate. This makes a compound of 840 pens in the (600).
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Re: Regular compounds in 4D

Postby Mecejide » Sun Mar 08, 2020 1:59 pm

wendy wrote:Two more preserve the symmetry, but are not regular figures, being the compound of two pens, and two icos, ie xo3ooXoo3oox for X=3, 4

Why do you not consider sted (the compound of 2 pens) and stoc (the compound of 2 icoes) to be regular?
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Re: Regular compounds in 4D

Postby wendy » Mon Mar 09, 2020 11:56 am

In order to be regular, either the hull or the core must be regular.

In the case of two pen or two icos, the core is o3x3x3o and o3x4x3o.
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Re: Regular compounds in 4D

Postby URL » Wed Apr 22, 2020 12:53 pm

wendy wrote:In order to be regular, either the hull or the core must be regular.


This seems needlessly restrictive. As I see it, being regular is about symmetry, not about convex hulls or cores.
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Re: Regular compounds in 4D

Postby wendy » Wed Apr 22, 2020 1:13 pm

Every polytope that derives from the [3,3,5] or the [3,3,5] symmetry contains vertices that belong to a {3,3,5}, such that one (or two) such exist at each vertex. That means that among the vertices of say, x3x3x5x, which has 14400 (twe 1,00,00) vertices, you can have 120 copies of {3,3,5}, and thus 120 copies of whatever compounds (120, gives. Since i think this numbers something like 33 + 12 + 12 = 57 compounds, you have 57.00 regilar compounds in this figure, and then 28.60 in any of the four of x3x3x5o, x3x3o5o, x3o3x5x and o3x3x5x, 28.00 in x3o3x5o, o3x3o5x, o3x3x5o, 18.90 in x3o3o5x, o3o3x5x, and all the way down to 57.

Most numbers are in base 120, it's too late to translate them.

But take the argument up with Don Coxeter.
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Re: Regular compounds in 4D

Postby Klitzing » Wed Apr 22, 2020 8:38 pm

The thing about regularity of compounds is (just as the notation of those) from Coxeter, and as such not arguable.

(Fully) regular compounds then are the ones of the form aP[bQ]cR, where
  • P, Q, R all are valide Schläfli symbols, i.e. regular polytopes, and
  • a, b, c are positive integers ("1"s usually are omitted).
That code string then has to be dechiffered as follows: the (fully) regular compound under consideration (and such described) ...
  • ... uses b components Q;
  • its hull is P;
  • it has a components per hull vertex (or given in other words: the vertex figure of the compound itself is a subdimensional compound with a components);
  • its kernel is R;
  • it supports every facet hyperplane of R by c facet hyperplanes of the compound (or stated otherwise: the facet figure of the compound itself is a subdimensional compound with c components).

He further defined also vertex regular compounds. Those then can be described at least by aP[bQ]. Similarily facet regular compounds are describable as [bQ]cR. Thus fully regular compounds are the ones, which are both, vertex regular and facet regular.

Wendy's statement, which was questioned by URL, put simply was that any of these types of regularities either asks for a regular hull and/or for a regular kernel. But clearly, if being reduced to that only, it becomes wrong: the components in addition have to be regular also!

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Re: Regular compounds in 4D

Postby username5243 » Wed Apr 22, 2020 11:13 pm

Aha, I see now why the 2-pen and 2-ico compounds wouldn't be considered to be "regular" by that definition, even if they appear to have only one type of each element themselves; their hulls aren't regular (or even uniform)... It just seems like they should be considered regular though, though I think I get the reasoning.
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Re: Regular compounds in 4D

Postby URL » Thu Apr 23, 2020 12:20 am

So, it's a "by-definition" restriction. I get it.
Still though. The definition of regular polytopes I've most often seen is that all of its flags (every set of a vertex on an edge on a face... on the polytope) must be identical under symmetry. If we took a pentachoron and its reflection through its center, wouldn't that satisfy that property? Sorry to differ with Coxeter, but if that's the case, I believe that this deserves to be classified as regular.
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Re: Regular compounds in 4D

Postby Klitzing » Thu Apr 23, 2020 1:42 pm

Haha, yes they could be written as [2{3,3,3}]{3;3;3} and [2{3,4,3}]{3;4;3} respectively.
Here both description (the kernels) somehow mimic the Schläfli symbols, but in fact they use Wendy's extension on them, i.e. allow for decorated nodes (the ";"-signs), where the usual Schläfli symbol then would read {p,q,r} = {;p,q,r}.
So in fact these 2 compounds are nothing but stellations of deca and cont respectively.

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Re: Regular compounds in 4D

Postby Mecejide » Thu Apr 23, 2020 4:29 pm

I don't care if that was Coxeter's definition. It's bad and I'm using a better one. There is 1 regular compound polyhedron (2 tetrahedra) and there are 7 regular compound polychora (2 pentachora, 3 hexadecachora, 6 hexadecachora, 120 pentachora, 3 tesseracts, 2 icositetrachora, and 6 tesseracts).
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Re: Regular compounds in 4D

Postby username5243 » Thu Apr 23, 2020 5:18 pm

What definition are you using? Why wouldn't, say, the 5 cubes count?

Also, are there even any truly regular compounds (using the Coxeter definition) in 5+ dimensions? I don't think there are...
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Re: Regular compounds in 4D

Postby Mecejide » Thu Apr 23, 2020 6:07 pm

The compound of 5 cubes isn't flag-transitive.
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Re: Regular compounds in 4D

Postby username5243 » Thu Apr 23, 2020 6:34 pm

So by your definition the only regular compounds in 5+ dimensions are those of 2 simplexes? Which wouldn't even count under Coxeter's definition (I don't think there are any that do) in 5D)?
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