Bilbirothawroids (D4.3 to D4.9)

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

Re: Johnsonian Polytopes

Postby Keiji » Tue Feb 11, 2014 7:19 am

Just a quick thought about higher-dimensional CRFs (this thread is not titled "Johnsonian Polychora", after all!) - the Cartesian product of any two regular polygons is a duoprism and thus CRF (and uniform). Would it also be the case that the Cartesian product of any two CRF polytopes gives another CRF polytope? e.g. a Johnson solid and a regular polygon giving a 5D CRF?

Assuming that is the case, I presume 6D Cartesian products of two 3D CRFs would not have any 4D elements that are not CRFs we already know about (duoprisms, or prisms of 3D CRFs), but what about taking such a 6D figure and diminishing it? Could this be an easy way to find new 4D CRFs?
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Re: Johnsonian Polytopes

Postby wendy » Tue Feb 11, 2014 7:42 am

It is true that the cartesian product of two CRF figures is CRF, but normally the list is restricted to primitives. That is, prisms are discounted. It starts to be a problem in 4D, where each of J1 to J92 gives a slab- or line-prism.

The CRF in the outer extent includes uniforms, and the prism product. What is generally given by J1 to J92 is the OCRF that are neither prism-products nor uniforms, just as the archemedian are uniforms that are not platonic.
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Re: Johnsonian Polytopes

Postby Keiji » Tue Feb 11, 2014 8:18 am

Johnson may have excluded prisms and uniforms from his list, but I personally prefer to see them as a series of sets of increasing scope, i.e. Regulars < Uniforms < CRFs. (that's why I talk of CRF polytopes, not Johnsonian polytopes.) That way, the existence of prisms of Johnson solids is not a problem either.

I also had another thought. It looks like all the 3D crown jewels are related to the rhodomorphs, as opposed to the stauromorphs. This makes me wonder whether there will be any 4D "stauromorphic" crown jewels - and if there are not, there may be no crown jewels at all in 5D and up, due to the lack of any rhodomorphs above 4D!

So if we are to find higher dimensional crown jewels, either they would have to be stauromorphic, or somehow based upon the symmetry of rhodomorphic prisms, without being prisms themselves.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 11, 2014 1:30 pm

Well, Keiji, it highly depends on what you are considering a crown jewel to be.

Look e.g. at my set of convex segmentochora. There are several ones, which have different symmetries for bottom and top layer, even if we neglect all those diminishings etc., i.e. restrict the bases to full higher symmetry groups only. Then you will have to consider the common subgroup of those. That one thus would be the true axial symmetry, Independent on what the base polyhedra would show up themselves. Most often that common subgroup would be itself a known Coxeter group. Then both base polyhedra can be given as Dynkin symbol with respect to that one, and you'll get the lace prism notation for free. Just for the cube||ike that common subsymmetry happens to be the non-Coxeter pyrithohedral group. This is what makes this beasty so special.

But also tet||oct, tet||co, oct||tut, tut||co, and tut||toe all show up different symmetries in their bases (here: tetrahedral vs. cubical). But the common subgroup here is the tetrahedral group itself.

The same holds true in higher dimensions as well. You surely could stack quite often hypercubical and demihypercubical polytopes. But the common subsymmetry then again is the demihypercubical one. - Sure the amount of non-trivial subgroups most generally would get higher the larger a group is. This is why the hyic (hecatonicosachoral) symmetry allows for more jewels than the tessic (tesseractic) group. And as already mentioned: here the demitessic subgroup already has to be discarded...

Finally regarding the icoic group. But, as you know, both, a tesseract and a hexadecachoron could be vertex inscribed to it. (In fact, using the remaining vertices each.) Therefore the common subgroup here generally happens to be either the tessic or demitessic one. But again both are Coxeter groups.

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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 11, 2014 3:38 pm

Keiji wrote:Johnson may have excluded prisms and uniforms from his list, but I personally prefer to see them as a series of sets of increasing scope, i.e. Regulars < Uniforms < CRFs. (that's why I talk of CRF polytopes, not Johnsonian polytopes.) That way, the existence of prisms of Johnson solids is not a problem either.

I also had another thought. It looks like all the 3D crown jewels are related to the rhodomorphs, as opposed to the stauromorphs. This makes me wonder whether there will be any 4D "stauromorphic" crown jewels - and if there are not, there may be no crown jewels at all in 5D and up, due to the lack of any rhodomorphs above 4D!

Probably, the reason we have only found crown jewels involving rhodomorphs is because they are the easiest to find? :) Also, both J91 and J92 are directly related to the 3D rhodomorphs, and we have only searched for crown jewels containing them as cells so far. Marek et al have suggested searching for crown jewels containing the snub disphenoid -- I haven't started looking in that direction yet (still looking at other J92 possibilities), but it may well be that crown jewels exist in that direction too. Absence of proof is not the proof of absence!

So if we are to find higher dimensional crown jewels, either they would have to be stauromorphic, or somehow based upon the symmetry of rhodomorphic prisms, without being prisms themselves.

We have already proven that ursatopes exist in all dimensions, and they aren't based on stauromorphic or rhodomorphic prisms. ;)
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Re: Johnsonian Polytopes

Postby quickfur » Tue Feb 11, 2014 3:45 pm

Klitzing wrote:Well, Keiji, it highly depends on what you are considering a crown jewel to be.

Look e.g. at my set of convex segmentochora. There are several ones, which have different symmetries for bottom and top layer, even if we neglect all those diminishings etc., i.e. restrict the bases to full higher symmetry groups only. Then you will have to consider the common subgroup of those. That one thus would be the true axial symmetry, Independent on what the base polyhedra would show up themselves. Most often that common subgroup would be itself a known Coxeter group. Then both base polyhedra can be given as Dynkin symbol with respect to that one, and you'll get the lace prism notation for free. Just for the cube||ike that common subsymmetry happens to be the non-Coxeter pyrithohedral group. This is what makes this beasty so special.

But also tet||oct, tet||co, oct||tut, tut||co, and tut||toe all show up different symmetries in their bases (here: tetrahedral vs. cubical). But the common subgroup here is the tetrahedral group itself.

The cuboctahedron, while usually considered to have cubic symmetry, can also be regarded as the tetrahedral equivalent of the rhombi- polyhedra (i.e., x3o3x), and similarly the truncated octahedron can be regarded as the omnitruncated tetrahedron (x3x3x). So in that sense these segmentochora just belong to the tetrahedral family.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Feb 11, 2014 6:10 pm

quickfur wrote:
Klitzing wrote:Well, Keiji, it highly depends on what you are considering a crown jewel to be.

Look e.g. at my set of convex segmentochora. There are several ones, which have different symmetries for bottom and top layer, even if we neglect all those diminishings etc., i.e. restrict the bases to full higher symmetry groups only. Then you will have to consider the common subgroup of those. That one thus would be the true axial symmetry, Independent on what the base polyhedra would show up themselves. Most often that common subgroup would be itself a known Coxeter group. Then both base polyhedra can be given as Dynkin symbol with respect to that one, and you'll get the lace prism notation for free. Just for the cube||ike that common subsymmetry happens to be the non-Coxeter pyrithohedral group. This is what makes this beasty so special.

But also tet||oct, tet||co, oct||tut, tut||co, and tut||toe all show up different symmetries in their bases (here: tetrahedral vs. cubical). But the common subgroup here is the tetrahedral group itself.

The cuboctahedron, while usually considered to have cubic symmetry, can also be regarded as the tetrahedral equivalent of the rhombi- polyhedra (i.e., x3o3x), and similarly the truncated octahedron can be regarded as the omnitruncated tetrahedron (x3x3x). So in that sense these segmentochora just belong to the tetrahedral family.

Haha, quickfur, this was exactly what I was proclaiming. There does exist a subsymmetry which itself is a Coxeter group. And so the corresponding base in most of the relevant cases can be given as a Wythoffian polytope with resp. to that symmetry as well.

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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 12, 2014 4:13 pm

I think I may have found another possible CRF construction involving J91. Take a look at the following projection of the rectified 120-cell o5x3o3o:

Image

Look at where two o5x3o's touch each other. There are two pentagons straddled by two tetrahedral gaps, which, if you fill in the tetrahedra, results in unit edges from the pentagons to each other. The outline of the two pentagons and two triangles thus introduced corresponds with the outline of a J91. Furthermore, the ends of the J91 outline coincide exactly with where the o5x3o's can be bisected into pentagonal rotunda. So it would appear that we have a CRF fragment here consisting of 2 pentagonal rotundae girded by 5 J91's. I'm not sure yet what would need to be inserted on the far side to close it up, but the construction sure looks rather promising! My initial thought is to put a reflected copy of the same fragment on the other side, in which case we have here another rhombochoron-like construction. :)

A more remote possibility, which I have to think about more carefully, is if you look at the o5x3o's in this projection and note that their outward-facing face is a triangle, that is, this is the part of their surface that matches that of J92. If we bisect the polychoron with a hyperplane that bisects these cells parallel to the triangular faces, we will get a non-CRF polystratic cup of the o5x3o3o, but hexagons may be inserted to turn the non-CRF bisected o5x3o's into J92's, and maybe some other CRF fragments can then be added to make the result CRF. The gaps where the exposed cyan cells lie, are where another 12 o5x3o's lie, and these are in just the right orientation to be bisected into pentagonal rotundae. So perhaps it is possible to make a CRF cup of o5x3o3o that contains 12 pentagonal rotundae and 20 J92's?
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 12, 2014 4:56 pm

Woohoo, 91st post!! :D
quickfur wrote:I think I may have found another possible CRF construction involving J91. Take a look at the following projection of the rectified 120-cell o5x3o3o:

Image

Look at where two o5x3o's touch each other. There are two pentagons straddled by two tetrahedral gaps, which, if you fill in the tetrahedra, results in unit edges from the pentagons to each other. The outline of the two pentagons and two triangles thus introduced corresponds with the outline of a J91. Furthermore, the ends of the J91 outline coincide exactly with where the o5x3o's can be bisected into pentagonal rotunda. So it would appear that we have a CRF fragment here consisting of 2 pentagonal rotundae girded by 5 J91's. I'm not sure yet what would need to be inserted on the far side to close it up, but the construction sure looks rather promising! My initial thought is to put a reflected copy of the same fragment on the other side, in which case we have here another rhombochoron-like construction. :)

interesting!
it's lace-city would look like this:
Code: Select all
       x5x

    f5o   f5A

  o5x        o5B

    f5o   f5A

       x5x

you can compare that with the lace-city of a bilbro:
Code: Select all
       x

    f     f

  o          o

    f     f

       x
when you add ..5? to everything, you get what you have. Basically this means you place 5 bilbro's on the vertices of a pentagon. Unfortunately this also means A and B aren't equal to o resp. x. It might be that A=x though, would be very cool if it is.
EDIT: A=x; B=u should work
Last edited by student91 on Wed Feb 12, 2014 5:41 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 12, 2014 5:37 pm

Conversation with Robert Webb (author of Great Stella) told me that there's an easy method to measure dichoral angles in the program. This means that if I get *.off files of the polychora we have so far, I can check them for possible augmentations.

Take icosahedral rotunda, for example. Great Stella claims that the dichoral angles are:

pentagonal pyramid/truncated tetrahedron (triangle): 157.761
pentagonal pyramid/truncated icosahedron (pentagon): 36
truncated tetrahedron/icosahedron (triangle): 157.761
truncated tetrahedron/truncated icosahedron (hexagon): 22.2388

Note the very low values for truncated icosahedron, showing that this rotunda is in fact quite shallow (though not as shallow to serve as augmentation of o5x3x3o or o5x3x3x).

Now let's look at something more interesting: the J92-choron.

It has following cells:

12 tetrahedra, each adjoining 1 triangular prisms (172.239), 2 pentagonal pyramids (142.239) and 1 metabidiminished icosahedron (142.239).
12 square pyramids A, each adjoining a triangular prism through its square face (155.905) and 2 pentagonal pyramids (157.761), 1 square pyramid B (167.379) and 1 metabidiminished icosahedron (82.2388) through triangular faces.
12 square pyramids B, each adjoining a J92 through its square face (135) and 1 J92 (82.2388), 2 pentagonal pyramids (157.761) and 1 square pyramid A (167.379) through triangular faces.
6 square pyramids C, each adjoining a triangular prism through its square face (155.905) and 2 metabidiminished icosahedra (157.761) and 2 J92's (142.239) through triangular faces.
6 triangular prisms, each adjoining a two square pyramids A (155.905) and one square pyramid C (155.905) through its square faces and two tetrahedra (172.239) through its triangular faces.
24 pentagonal pyramids, each adjoining another pentagonal pyramid through its pentagonal face (144) and 1 tetrahedron (142.239), 1 square pyramid A (157.761), 1 square pyramid B (157.761), 1 metabidiminished icosahedron (120) and 1 J92 (120) through its triangular faces.
6 metabidiminished icosahedra, each adjoining two J92's through its pentagonal faces (144) and 2 tetrahedra (142.239), 2 square pyramids A (82.2388), 2 square pyramids C (157.761) and 4 pentagonal pyramids (120) through its triangular faces.
4 J92's, each adjoining one J92 through its hexagonal face (60), 3 metabidiminished icosahedra through its pentagonal faces (144), 3 square pyramids B through its square faces (135) and 3 square pyramids C (142.239), 6 pentagonal pyramids (120) and 1 J92 (120) through its triangular faces.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 12, 2014 6:04 pm

Marek14 wrote:Conversation with Robert Webb (author of Great Stella) told me that there's an easy method to measure dichoral angles in the program. This means that if I get *.off files of the polychora we have so far, I can check them for possible augmentations.

Take icosahedral rotunda, for example. Great Stella claims that the dichoral angles are:

pentagonal pyramid/truncated tetrahedron (triangle): 157.761
pentagonal pyramid/truncated icosahedron (pentagon): 36
truncated tetrahedron/icosahedron (triangle): 157.761
truncated tetrahedron/truncated icosahedron (hexagon): 22.2388

Note the very low values for truncated icosahedron, showing that this rotunda is in fact quite shallow (though not as shallow to serve as augmentation of o5x3x3o or o5x3x3x).

Sounds about right, you can see this rotunda / polystratic cup on the o5o3x3x in this projection:

Image

Especially obvious is the side-view on the upper right side of this image, which clearly shows a slice that corresponds to the icosahedral rotunda / cup.

Now what's interesting, is that all the 120-cell family polychora share the same underlying hyperplanes corresponding to the 120-cell/600-cell symmetry group, meaning that there is a correspondence between the icosahedron of this cup with a dodecahedron of the 120-cell, and the 12 pentagonal pyramids with the 12 dodecahedra surrounding this dodecahedron. Therefore, the pentagonal pyramids lie in hyperplanes parallel to the hyperplanes of 12 dodecahedra, which are also parallel to the hyperplanes of the elements corresponding with the 120-cell's cells lie in, in any of the other 120-cell family uniforms, for example, the icosidodecahedra in o5x3o3o, the icosahedra in o5o3x3o, etc.. So if you can join these pentagonal pyramids to the corresponding elements of a 120-cell uniform, they would be coplanar, and the pentagonal pyramids would become augmentations of those cells, which, if the cells admit CRF augmentations with pentagonal pyramids, would result in a CRF augmentation of the entire polytope.

Now let's look at something more interesting: the J92-choron.

It has following cells:

12 tetrahedra, each adjoining 1 triangular prisms (172.239), 2 pentagonal pyramids (142.239) and 1 metabidiminished icosahedron (142.239).
12 square pyramids A, each adjoining a triangular prism through its square face (155.905) and 2 pentagonal pyramids (157.761), 1 square pyramid B (167.379) and 1 metabidiminished icosahedron (82.2388) through triangular faces.
12 square pyramids B, each adjoining a J92 through its square face (135) and 1 J92 (82.2388), 2 pentagonal pyramids (157.761) and 1 square pyramid A (167.379) through triangular faces.
6 square pyramids C, each adjoining a triangular prism through its square face (155.905) and 2 metabidiminished icosahedra (157.761) and 2 J92's (142.239) through triangular faces.
6 triangular prisms, each adjoining a two square pyramids A (155.905) and one square pyramid C (155.905) through its square faces and two tetrahedra (172.239) through its triangular faces.
24 pentagonal pyramids, each adjoining another pentagonal pyramid through its pentagonal face (144) and 1 tetrahedron (142.239), 1 square pyramid A (157.761), 1 square pyramid B (157.761), 1 metabidiminished icosahedron (120) and 1 J92 (120) through its triangular faces.
6 metabidiminished icosahedra, each adjoining two J92's through its pentagonal faces (144) and 2 tetrahedra (142.239), 2 square pyramids A (82.2388), 2 square pyramids C (157.761) and 4 pentagonal pyramids (120) through its triangular faces.
4 J92's, each adjoining one J92 through its hexagonal face (60), 3 metabidiminished icosahedra through its pentagonal faces (144), 3 square pyramids B through its square faces (135) and 3 square pyramids C (142.239), 6 pentagonal pyramids (120) and 1 J92 (120) through its triangular faces.

Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 12, 2014 7:19 pm

quickfur wrote:Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.


What do you use to look into questions like that?
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 12, 2014 8:35 pm

Marek14 wrote:
quickfur wrote:Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.


What do you use to look into questions like that?

Right now, I don't really have very advanced tools for this. I just try to analytically determine the structure as much as possible, then compute (partial) coordinates for it and run it through my polytope viewer's convex hull algorithm to see what comes out.

I've discovered that Wendy's lace cities are very useful for computing coordinates: once you analyze a possible structure in terms of a lace city, you can then compute the first 2 coordinates just from the labels themselves (the AnB's are just polygons, after all, so coordinates should be easy). Then you use the unit edge length requirement to compute the distances between each lace tower (which corresponds with cells). For simplicity, you could compute each cell in 3D, for example, for the J92 rhombochoron, I first computed the 3D coordinates of the J92 using the lace tower:

x3o
f3x
o3F
x3x

Then I add the gyrated J92 beneath the hexagon x3x:

x3o
f3x
o3F
x3x
F3o
x3f
o3x

This produces coordinates for two coplanar J92's joined at their hexagonal faces, so when you add the 4th coordinate, the vertices will have the form (x,y,z,0). Since we're going to bend these two cells into 4D at the hexagonal face, it makes things easiest to keep the hexagon fixed, so we fix the x3x's coordinates to the form (x,y,0,0) and adjust the z values of the other points accordingly.

Next, we introduce the dichoral angle between the two cells by transforming the coordinates of the first J92 into (x,y,z*sin A, z*cos A), and the coordinates of the second J92 into (x,y,-z*sin A, -z*cos A), where A is the free variable we're trying to solve for. At this point, you can either experiment with randomly chosen values for A to see what comes out of the convex hull algorithm, or you can try to fit certain cells between the J92's, say for example you see the respective squares and triangles of the J92's being possibly connected by a square pyramid. This would imply that the distance between the vertices of o3F and the vertices of x3f should be unit length. Since those vertices have triangular symmetry, it is enough to pick two representative points that would be connected by the new unit edge, so you have two points P1=(x1,y1,z1*sin A, z1*cos A) and P2=(x2,y2,z2*sin A, z2*cos A), and the requirement that |P1 - P2| = 1. (Or, in my case, |P1 - P2|=2 because by convention I always work with coordinates that give edge length 2, it simplifies a number of commonly-encountered values like <±1,±1,±1> for the cube rather than <±1/2, ±1/2, ±1/2>.) Then you solve for the value of A, and plug it into the coordinates of the points, then run it through the convex hull algorithm to see what comes out.

This is how I arrived at my first J92-choron attempt, in fact. I got lucky when by chance I picked A=30° (equiv. dichoral angle 60°) and noticed that the convex hull algorithm spitted out something with lots of phi-scaled edge lengths, implying that it could be CRF-ed by introducing pentagons at the right places.

When working this way, I usually keep the 4D viewpoint for the renders at <0,0,0,-5>, that is, looking at the fixed hexagonal ridge so that it's easier to see what's happening as the value of A is varied. After I found that A=30° is giving me a lot of phi-scaled edges, I did a render from <0,0,0,5> (i.e. from the back of the polychoron) to see what's going on, and saw a phi-elongated octahedron surrounded by what looked like incomplete (non-CRF) diminished icosahedra. That was what gave me the idea to insert an antipodal point, which, sure enough, made the far side CRF by turning the elongated octahedron into two tetrahedra, and the other cells into CRF tridiminished icosahedra. Unfortunately, I could not get rid of 6 of the phi-scaled edges that lie on the near side of the polychoron, and while trying to "complete the pentagon" around those edges, I made a wrong calculation that convinced me that it's not CRF-able.

But after student91 pointed out the mistake, I redid the calculations, and managed to get the near side to be CRF (and in the process the apex that introduced the two tetrahedra become coplanar with a far-side cell, so it vanished). Then looking at the far side again, I saw something again with phi-scaled edges but this time an odd arrangement of cells that I wasn't sure how to make CRF. After thinking for a long time about what to do next, I suddenly noticed that the new vertices introduced based on student91's idea were coplanar with the top and bottom triangles of the J92's, which implied that the front half of the polychoron was CRF, even though the back wasn't. So this suggested that if I glued two copies of this polychoron back-to-back, those new vertices from either copy would coincide, which means the back side would become CRF as well, leaving only the equatorial cells to deal with, which is a far easier problem.

So that's what I tried next, and lo and behold, all edges came out as unit length (turns out there was no need to do anything more for the equatorial cells), and the J92 rhombochoron was born. :)

As you can see, there's a fair amount of geometric intuition going on here in deciding how to proceed at each step, and a fair amount of 4D visualization to understand how the shape is progressing as you add points to it. And also a lot of manual calculations, much of which I hope to automate with the new version of my polytope viewer (but unfortunately that's going to take quite a bit of time to develop).
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 12, 2014 9:05 pm

Hm, looks very interesting. So far, I can only generate off files by hand :) I checked whether square biantiprismatic ring can be augmented by a cube prism (apparently, it can).

EDIT: Yup.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 12, 2014 9:16 pm

quickfur wrote:[...]Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.

I thought of this as well. What happens is that you get an x3o in the middle of the hexagon, surrounded by bigger %3%'s. this means there is some sort of pit in the middle of the polytope (like a red blood cell). you could try to fill this pit up, but that means you have to change the interior of the hexagon, and that probably means there aren't any J92-outlines anymore in the middle of the polytope, so what you're left with is just a hexagon of J92's if you try to fill it my way.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 12, 2014 9:41 pm

student91 wrote:
quickfur wrote:[...]Since the J92 rhombochoron has a 120°-60° rhombus shape, I'm thinking that it should be possible to glue 3 of them together to make a hexagon. Since the hexagons and triangles line up nicely in such a gluing, it's possible that the result can be made CRF by inserting a few more cells. If this is true, then it would be a CRF containing 6 J92's in a hexagonal ring. This would give enough space for interesting things to happen in the orthogonal ring; it might be possible to get some kind of 6,6-duoprism equivalent here.

I thought of this as well. What happens is that you get an x3o in the middle of the hexagon, surrounded by bigger %3%'s. this means there is some sort of pit in the middle of the polytope (like a red blood cell). you could try to fill this pit up, but that means you have to change the interior of the hexagon, and that probably means there aren't any J92-outlines anymore in the middle of the polytope.

Well, yes, once you join two J92-rhombochora together, two of the J92's would become internal to the polychoron and would no longer appear on the surface. So it's already expected that only six of the J92's would lie on the surface. The concave area around the x3o would need to be filled up by something in order to be CRF, so it's expected that no J92's will be present in the "middle" of the polytope (i.e., on the orthogonal ring to the 6 (J92's). At least, none from the original 3 rhombochora; the thought did occur to me that we might be able to use new J92's to close up the orthogonal ring. If this is possible, we might get a duo-hexagonal-J92-choron. :D
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 12, 2014 10:09 pm

quickfur wrote:[...]
Well, yes, once you join two J92-rhombochora together, two of the J92's would become internal to the polychoron and would no longer appear on the surface. So it's already expected that only six of the J92's would lie on the surface. The concave area around the x3o would need to be filled up by something in order to be CRF, so it's expected that no J92's will be present in the "middle" of the polytope (i.e., on the orthogonal ring to the 6 (J92's). At least, none from the original 3 rhombochora; the thought did occur to me that we might be able to use new J92's to close up the orthogonal ring. If this is possible, we might get a duo-hexagonal-J92-choron. :D

I understand the J92's themselves won't occur on the surface, but that was not what I tried to say. I tried to sa that the x3o itself would become internal, so when you make it convex, the vertices of the x3o can't be used anymore. This isn't very nice, as most of the J92-rhomb's structure will vanish when the x3o is deleted :( .
Your other idea, the duo-hexagonal J92-choron, does seem interesting, I hope you/we will be able to get a nice CRF out of that. :)
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 12, 2014 10:15 pm

BTW, I tried to research the augments of square biantiprismatic ring. I'd appreciate if someone checked it, but with dichoral angles from Great Stella, it looks like the cube and both square antiprisms can be independently augmented with pyramids. If both square antiprisms are augmented, two of the pyramids merge into octahedron.

This leads to 2 augmented, 2 biaugmented and 1 triaugmented square biantiprismatic ring.

Here's off file for cube-augmented ring: I'd need some program to properly make the others...

Code: Select all
4OFF
# NumVertices, NumFaces, NumEdges, NumCells
13 43 40 16

# Vertices
1.0                     1.0                    1.0  0.0
1.0                     1.0                   -1.0  0.0
1.0                    -1.0                    1.0  0.0
1.0                    -1.0                   -1.0  0.0
-1.0                     1.0                    1.0  0.0
-1.0                     1.0                   -1.0  0.0
-1.0                    -1.0                    1.0  0.0
-1.0                    -1.0                   -1.0  0.0
1.41421356237309504880  0.0                    0.0  1.35219344945395667954
-1.41421356237309504880  0.0                    0.0  1.35219344945395667954
0.0                     1.41421356237309504880 0.0  1.35219344945395667954
0.0                    -1.41421356237309504880 0.0  1.35219344945395667954
0.0                     0.0                    0.0 -1.0

# Faces
4 0 1 3 2
4 0 1 5 4
4 0 2 6 4
4 1 3 7 5
4 2 3 7 6
4 4 5 7 6
4 8 10 9 11
3 0 1 8
3 0 1 10
3 2 3 8
3 2 3 11
3 4 5 9
3 4 5 10
3 6 7 9
3 6 7 11
3 0 2 8
3 1 3 8
3 0 4 10
3 1 5 10
3 2 6 11
3 3 7 11
3 4 6 9
3 5 7 9
3 0 8 10
3 1 8 10
3 2 8 11
3 3 8 11
3 4 9 10
3 5 9 10
3 6 9 11
3 7 9 11
3 0 1 12
3 0 2 12
3 0 4 12
3 1 3 12
3 1 5 12
3 2 3 12
3 2 6 12
3 3 7 12
3 4 5 12
3 4 6 12
3 5 7 12
3 6 7 12

# Cells
10 2 6 15 17 19 21 23 25 27 29  255 0 0
10 3 6 16 18 20 22 24 26 28 30  255 0 0
4 7 8 23 24  0 0 255
4 9 10 25 26  0 0 255
4 11 12 27 28  0 0 255
4 13 14 29 30  0 0 255
5 0 7 9 15 16  127 127 0
5 1 8 12 17 18  127 127 0
5 4 10 14 19 20  127 127 0
5 5 11 13 21 22  127 127 0
5 0 31 32 34 36  0 255 0
5 1 31 33 35 39  0 255 0
5 2 32 33 37 40  0 255 0
5 3 34 35 38 41  0 255 0
5 4 36 37 38 42  0 255 0
5 5 39 40 41 42  0 255 0
# The end.
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Re: Johnsonian Polytopes

Postby student91 » Wed Feb 12, 2014 10:32 pm

Marek14 wrote:BTW, I tried to research the augments of square biantiprismatic ring. I'd appreciate if someone checked it, but with dichoral angles from Great Stella, it looks like the cube and both square antiprisms can be independently augmented with pyramids. If both square antiprisms are augmented, two of the pyramids merge into octahedron.

This leads to 2 augmented, 2 biaugmented and 1 triaugmented square biantiprismatic ring.

According to my calculations, the 4-biantipr. ring has dichoral angle 53.5156 between cube and antipr., and 72.96875 between two antipr. Dichoral angle of cube-pyr at square is 53.5156 too45, and for antipr.-pyr it is 69.295 53.5156. This means the cube and the antipr. can be independently augmented. The two antiprisms however, can't be similtaiously augmented while retaining convexity. :) :| :( EDIT: it can
I based my conclusions on my calculated list of dichoral angles of the segmentochora, here.
I would really like to know whether you have the same angles, I don't want my list to have more wrong numbers :\
Last edited by student91 on Thu Feb 13, 2014 2:50 pm, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Feb 12, 2014 10:43 pm

student91 wrote:
quickfur wrote:[...]
Well, yes, once you join two J92-rhombochora together, two of the J92's would become internal to the polychoron and would no longer appear on the surface. So it's already expected that only six of the J92's would lie on the surface. The concave area around the x3o would need to be filled up by something in order to be CRF, so it's expected that no J92's will be present in the "middle" of the polytope (i.e., on the orthogonal ring to the 6 (J92's). At least, none from the original 3 rhombochora; the thought did occur to me that we might be able to use new J92's to close up the orthogonal ring. If this is possible, we might get a duo-hexagonal-J92-choron. :D

I understand the J92's themselves won't occur on the surface, but that was not what I tried to say. I tried to sa that the x3o itself would become internal, so when you make it convex, the vertices of the x3o can't be used anymore. This isn't very nice, as most of the J92-rhomb's structure will vanish when the x3o is deleted :( .
Your other idea, the duo-hexagonal J92-choron, does seem interesting, I hope you/we will be able to get a nice CRF out of that. :)

What I have in mind is something like this:
Code: Select all
            x3x
        F3o     F3o
    x3f             x3f
            o3F
o3x                     o3x
    F3x     f3x     F3x
x3f                     x3f
        x3F     x3F
            x3o
F3o     f3x     f3x     F3o
    o3F             o3F
            x3F
x3x                     x3x
            F3x
    F3o             F3o
        x3f     x3f
            o3x

While it is true that the x3o in the middle of the lace city will vanish once we add more things to make it convex (and probably same goes for the f3x's and x3F's neighbouring it), the other 3 x3o's still lie on the surface of the polychoron, so this still makes a ring of six J92's.

I'm not as sure what will happen to the o3F's and F3x's, as those protrude quite a bit out from the hyperplane of the lace city, so they might still remain intact, which would imply that the pentagonal pyramids and triangular prism + square pyramids + tetrahedra combos might still exist in the result.

One interesting feature of this lace city is that the J92's are ortho to each other at the hexagonal face, whereas in the J92 rhombochoron they were gyro to each other. Over the past few days I've been trying unsuccessfully to figure out how to make a CRF out of gyro J92's with 120° at the hexagon. Looks like the secret is to use ortho orientation when the dichoral angle is 120°. :lol:
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Feb 12, 2014 11:41 pm

student91 wrote:
Marek14 wrote:BTW, I tried to research the augments of square biantiprismatic ring. I'd appreciate if someone checked it, but with dichoral angles from Great Stella, it looks like the cube and both square antiprisms can be independently augmented with pyramids. If both square antiprisms are augmented, two of the pyramids merge into octahedron.

This leads to 2 augmented, 2 biaugmented and 1 triaugmented square biantiprismatic ring.

According to my calculations, the 4-biantipr. ring has dichoral angle 53.5156 between cube and antipr., and 72.96875 between two antipr. Dichoral angle of cube-pyr at square is 53.5156 too, and for antipr.-pyr it is 69.295. This means the cube and the antipr. can be independently augmented. The two antiprisms however, can't be similtaiously augmented while retaining convexity. :) :| :(
I based my conclusions on my calculated list of dichoral angles of the segmentochora, here.
I would really like to know whether you have the same angles, I don't want my list to have more wrong numbers :\


Great Stella shows smaller angle for antiprism/pyramid, 53.5156 which makes the dichoral angle when both antiprisms are augmented exactly 180 degrees, thus allowing the augmentation while fusing two square pyramids into an octahedron. My results also have dichoral angle of cube-pyr at square 45, not 53.5156.

Basically, if I get the polychora in appropriate form, the software can measure the dichoral angles directly.
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Re: Johnsonian Polytopes

Postby student91 » Thu Feb 13, 2014 12:13 am

Marek14 wrote:[..]
Great Stella shows smaller angle for antiprism/pyramid, 53.5156 which makes the dichoral angle when both antiprisms are augmented exactly 180 degrees, thus allowing the augmentation while fusing two square pyramids into an octahedron. My results also have dichoral angle of cube-pyr at square 45, not 53.5156.

Basically, if I get the polychora in appropriate form, the software can measure the dichoral angles directly.

Your calculations are right. for the antipr.-pyr I looked in the wrong coulumn :oops: , and for the cube-pyr I was just plain wrong :cry: . (ofc. it's 90, otherwise the 24-cell wouldn't exist)Thanks for the corrections :)
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 13, 2014 12:57 am

Marek14 wrote:
quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.

Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?


Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.

In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.

EDIT: I basically say the same thing as Keiji, just in a bit different words.

After some further thoughts about this, I have come to agree with Marek. Consider, for example, the pentagon. Its coordinates when laid on the 2D plane involves nested square roots; yet in 3D, when it occurs as a cross-section of the icosahedron, its coordinates involve only the Golden Ratio, which has no nested square roots. Obviously, it makes no sense to say that the pentagon has CVP >2 in 2D, and CVP=2 in 3D. So this is strong proof that it's not about how many square roots you take, but what degree of polynomial you need to solve in order to find the values.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 13, 2014 5:39 am

quickfur wrote:
Marek14 wrote:
quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.

Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?


Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.

In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.

EDIT: I basically say the same thing as Keiji, just in a bit different words.

After some further thoughts about this, I have come to agree with Marek. Consider, for example, the pentagon. Its coordinates when laid on the 2D plane involves nested square roots; yet in 3D, when it occurs as a cross-section of the icosahedron, its coordinates involve only the Golden Ratio, which has no nested square roots. Obviously, it makes no sense to say that the pentagon has CVP >2 in 2D, and CVP=2 in 3D. So this is strong proof that it's not about how many square roots you take, but what degree of polynomial you need to solve in order to find the values.


And consider equilateral triangle: in 2D it will always involve square roots, while in 3D it can be constructed between points (1,0,0),(0,1,0) and (0,0,1).
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Re: Johnsonian Polytopes

Postby quickfur » Thu Feb 13, 2014 5:53 am

Marek14 wrote:
quickfur wrote:
Marek14 wrote:
quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.

Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?


Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.

In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.

EDIT: I basically say the same thing as Keiji, just in a bit different words.

After some further thoughts about this, I have come to agree with Marek. Consider, for example, the pentagon. Its coordinates when laid on the 2D plane involves nested square roots; yet in 3D, when it occurs as a cross-section of the icosahedron, its coordinates involve only the Golden Ratio, which has no nested square roots. Obviously, it makes no sense to say that the pentagon has CVP >2 in 2D, and CVP=2 in 3D. So this is strong proof that it's not about how many square roots you take, but what degree of polynomial you need to solve in order to find the values.


And consider equilateral triangle: in 2D it will always involve square roots, while in 3D it can be constructed between points (1,0,0),(0,1,0) and (0,0,1).

Or, for that matter, the n-simplex, which has integer coordinates in (n+1)-dimensions, but in n-dimensions involves square roots of triangular numbers (so not just roots of the same number, like √5 in things involving the Golden Ratio like the pentagonal polytopes).
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 13, 2014 6:01 am

quickfur wrote:Or, for that matter, the n-simplex, which has integer coordinates in (n+1)-dimensions, but in n-dimensions involves square roots of triangular numbers (so not just roots of the same number, like √5 in things involving the Golden Ratio like the pentagonal polytopes).


Yes, so maybe for now we should simply sort the polytopes into "constructible" and "nonconstructible".
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 13, 2014 6:38 am

OK, I explicitly constructed all augments of square biantiprismatic ring. And guess what? If you augment both antiprismatic cell, the result will be just a cube||octahedron antiprism :) Didn't expect that, though Klitzing probably knows that :D

So we have 6 polychora in this group:

Square biantiprismatic ring (also bidiminished cube||octahedron)
Cube-augmented square biantiprismatic ring (also augmented bidiminished cube||octahedron)
Antiprism-augmented square biantiprismatic ring (also diminished cube||octahedron)
Cube-antiprism-biaugmented square biantiprismatic ring (also augmented diminished cube||octahedron)
Antiprism-biaugmented square biantiprismatic ring (also cube || octahedron)
Triaugmented square biantiprismatic ring (also augmented cube||octahedron)

EDIT: Of course, in hindsight it's quite obvious that cube||gyrated square would have connection to cube||octahedron...
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Re: Johnsonian Polytopes

Postby wendy » Thu Feb 13, 2014 8:48 am

The double-square root thing in the pentagon, comes from using coordinates that are at right angles to each other. The G = sqrt(f sqrt(5)) is a class-4 type number, and is normally found in the span of chords of the decagon.

Because the sitting of the pentagon in the canonical vertices of x3o5o &c, do not involve putting the pentagon in this way, this number does not happen.
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Feb 13, 2014 9:31 am

I checked the tetrahedral ursachoron -- the dichoral angle at its "apex" tetrahedron (the one surrounded by 4 tridiminished icosahedra) is small enough that this tetrahedron can be augmented by pentachoron (resulting dihedral angle is 173.2837).
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Re: Johnsonian Polytopes

Postby Klitzing » Thu Feb 13, 2014 11:39 am

Marek14 wrote:OK, I explicitly constructed all augments of square biantiprismatic ring. And guess what? If you augment both antiprismatic cell, the result will be just a cube||octahedron antiprism :) Didn't expect that, though Klitzing probably knows that :D

So we have 6 polychora in this group:

Square biantiprismatic ring (also bidiminished cube||octahedron)
Cube-augmented square biantiprismatic ring (also augmented bidiminished cube||octahedron)
Antiprism-augmented square biantiprismatic ring (also diminished cube||octahedron)
Cube-antiprism-biaugmented square biantiprismatic ring (also augmented diminished cube||octahedron)
Antiprism-biaugmented square biantiprismatic ring (also cube || octahedron)
Triaugmented square biantiprismatic ring (also augmented cube||octahedron)

EDIT: Of course, in hindsight it's quite obvious that cube||gyrated square would have connection to cube||octahedron...

If you just augment one of the antiprisms, then I wouldn't be surprised if that comes out to be "gyro suippy || cube".
And yes, both antiprism augmented surely then is "oct || cube".
Whereas, whenever the cube will be augmented (besides potentially others), you'd get a bistratic figure (tower):
either "gyro {4} || cube || pt", or "gyro squippy || cube || pt", or "oct || cube || pt".
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