Johnsonian Polytopes

[Klitzing]

OK, here are some renders of the rhombicuboctahedral rotunda:

Centered on great rhombicuboctahedron:


Side-view:


These renders were made with visibility clipping turned off.

The coordinates of this rotunda are the coordinates of the runcitruncated 16-cell where the last coordinate is positive.

And here's the .off file for Marek & other Stella4D users.

Nice, that's a prit-rotunda, or a x3o4x || u3o4x || x3x4x (here u = edge of length 2, x = edge of lngth 1); prit = x3x3o4x. As further all lacings are of length 1 too, you could write it as a lace tower: xux3oox4xxx&#xt.
I like esp. your concentric picture.
--- rk

[Klitzing]

And here are the renders of the octahedral rotunda:

Side view:


View centered on truncated octahedron base:


Coordinates: same as truncated 16-cell, where last coordinate is ≥0.

Stella4D .off file: here.

Enjoy!

That is a thex-rotunda; thex = x3x3o4o. It also could be written as x3o4o || u3o4o || x3x4o, or as xux3oox4ooo&#xt.
--- rk

[Klitzing]

A large number of the Johnson polyhedra are laminate: that is, one can build them by placing successive layers like pyramid+prism+pyramid. We (Richard Klitzing and myself), devised a notion for forming notations, that represent more polytopes by means of "layers of dynkin symbols". For example, the general antiprism, might be considered as xPo || oPx (meaning, xPo on top, and oPx), or xoPox, which represents an array, the n_th layer is represented by layer n.

The tri-diminished dodecahedron, then might be x3o || f3o || o3x, which becomes xfo3oox&#t. (See eg Lace towers in my Polygloss). This occurs as a face of a four-dimensional polytope xf3oox3ooo&#t, a figure bounded by four tri-diminished icosahedra, five tetrahedra, and an octahedron.

Google on Klitzing Segmentotopes , and ye will find a number of interesting things that will help you. Klitzing enumerated the segmentotopes, being polytopes of equal edge, for which all vertices lie in two parallel rings of a sphere. The .pdf file corresponds to a paper listing some 184 segmentatopes or so, many of these are Johnson polychora.

One of the more interesting figures by Klitzing is "cube || icosahedron". The axial symmetry here is 3*2, or pyritohedral, it occurs in the {3,3,5}, where the cube is part of the dodecahedral layer, and the icosahedron is the large icosahedron layer below the mid-ring.

The next step is that ye can stack these, to get various laminatopes. A useful place to start is something like the 500ch {3,3,5}. Its layers give a fruitful source of johnsons, eg oxoo3ooox5ooxo, a polytope which comprises of several rings of the {3,3,5}, the face consist being tetrahedra, 12 diminished icosahedra, and one icosadodecahedron.

This has four bands, which must occur, and variations of putting a fifth band in, to cater for any of 1-12 apiculations (the diminished icosahedra can have a pyramid mounted on it, replacing this face with tetrahedra and a pentagonal pyramid.

You can mount any consecutive number of these bands together, with an prism layer ("elongated"), with a reversal to any earlier point.

Thanx Wendy for bringing in the outcomes of our communications of the "other" list :-)

Has there been meanwhile any attempt to create a bistratic picture, say a concentrical one, for your xfo3oox3ooo&#xt ?

Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

[quickfur]

[...] Yep, 4D Johnsons is quite well a task. I also once went into that direction. But then settled to a more manageable sub-class: the monostratic orbiform ones. That research, Wendy was citing, already has been published:
"Convex Segmentochora", by Dr. R. Klitzing, Symmetry: Culture and Science, vol. 11, 139-181, 2000
[...]

Hi, and welcome to the forum! It's awesome that you can chime in with our search for 4D Johnsons (or as we call them here, CRF polychora, for Convex-Regular-Faced). Your research has been very helpful to us in confirming many of our independently-discovered results.

Have you considered the slightly larger category of monostratic Johnsons with the orbiform requirement relaxed? The extrusions of the Johnson solids are the most obvious ones in this category. I've been trying to find other kinds of non-orbiform monostratic polychora but so far haven't found anything yet.

Another direction of research that I've been doing is to find the maximally-diminished uniform polychora, and so far I've found some interesting derivatives of the 5-cell family, as well as the cantellated tesseract, which was an unexpectedly rich gold mine of CRF polychora. I haven't done the actual research yet, but I'm expecting the cantellated 24-cell to also generate lots of CRFs.

On a slightly higher-level, I wonder if the enumeration of 4D Johnsons can be made somewhat more tractable if we split up the task into more directed categories where more specialized search methods can be employed. Following the way Johnson arranged the 3D Johnson solids in groups, I came up with the tentative categories:

Code: Select all

I. Monostratic polychora
   A. Orbiform polychora -- your list of segmentochora
   B. Non-orbiform polychora -- prisms of the non-orbiform 3D Johnsons, probably with a few others.
   C. Various derivatives of the above, e.g., the laminochora (stacking monostratic polychora on top of each other)

II. Rotundae -- so far, the 5 or 6 that we've found in this thread, plus their derivatives - birotundae, elongated birotundae, augmentations with the monostratics, etc..

III. Modified Uniform Polychora
   A. Augmented duoprisms
      1. Augmentations with prism pyramids: 1633 of them, for 3,n-duoprisms up to 5,n-duoprisms (the largest known is the omniaugmented 5,20-duoprism, which has a ring of 10 augmented square bipyramid cells flanked by square pyramids).
      2. Augmentations with n-gon||n-prism: at least 1633 of them (probably more), for 6,n-duoprisms up to 10,n-duoprisms, the largest of which probably will be the 10,20-duoprism.
   B. Augmented/diminished 5-cell family members
   C. Augmented/diminished tesseract family members
   D. Augmented/diminished 24-cell family members
   E. Augmented/diminished 120-cell family members

IV. Crown Jewels -- only 1 known, if we count cube||icosahedron (though that one can be put in category I). No known search method.


Thanks to your research, category I.A. is completed. Category I.B is probably the easiest next step. Category I.C. can probably be enumerated by computer if we formulate the appropriate requirements for a particular stacking to be CRF. At the very least, the derivatives of I.A. can be run through a computer enumeration since we already have the necessary data.

A few examples in category II are known, as described elsewhere in this thread; some work is needed to find any remaining instances (or prove there are none others). Once that is done, it's just a matter of enumerating derivatives like bicupolae, augmentations with I.C polychora, etc..

Category III is a large category, where I think will probably also be the largest in terms of number. The duoprism augmentations with prism pyramids have been completed by Marek & myself (though our results need to be double-checked). But there remains augmentations with other things, among which the n-gon||n-prism augmentations constitute at least 1600+ members. Augmentations with other CRFs have yet to be studied.

The 600-cell probably deserves its own category, since a very large number of diminishings are possible. I've attempted to find the maximally-diminished 600-cells, but so far have not been able to find a general scheme for generating these things save by brute-force search. Things are very complicated here because adjacent vertices can be deleted while still keeping the result CRF, unlike the 24-cell diminishings where assuming non-adjacency of deleted vertices greatly simplifies enumeration.

As for category IV, I don't know what method can be used to search for them -- even exhaustive brute-force search seems intractible because of non-rigid edges when more than 3 polyhedra surround it. This will probably be the hardest category to crack open.

[quickfur]

[...]Thanx Wendy for bringing in the outcomes of our communications of the "other" list :-)

Has there been meanwhile any attempt to create a bistratic picture, say a concentrical one, for your xfo3oox3ooo&#xt ?

Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

If someone can post the coordinates, I can run it through my polytope viewer and produce various views of it, with/without hidden surface culling, coloring different cells/elements, etc.. Lately, since Ovo introduced me to the technique, I've been doing a lot of cross-eyed stereo pairs, which really helps in conveying the 3D-ness of the 4D-to-3D projections.

[Klitzing]

[...]Thanx Wendy for bringing in the outcomes of our communications of the "other" list :-)

Has there been meanwhile any attempt to create a bistratic picture, say a concentrical one, for your xfo3oox3ooo&#xt ?

Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

If someone can post the coordinates, I can run it through my polytope viewer and produce various views of it, with/without hidden surface culling, coloring different cells/elements, etc.. Lately, since Ovo introduced me to the technique, I've been doing a lot of cross-eyed stereo pairs, which really helps in conveying the 3D-ness of the 4D-to-3D projections.

Well, I do not have them within my hands. But they are more than obvious:
Inner shell: unit-size tetrahedron. Coordinates should be obvious.
Next shell: tetrahedron of size tau = (1 + sqrt5)/2, same orientation.
Outermost shell: unit-size octahedron, with 4 triangles above the vertices and 4 triangles antialigned to the tetrahedral ones.
So the only to be determined thing are the relative heights. But thats a matter of an easy application of the Pythagoras theorem only...
(Further help needed?)
--- rk

[Marek14]

quickfur: Also, I realized that we have barely four years to finish the CRF project -- 2016 will be the 50th anniversary of Norman Johnson's article enumerating the 3D cases, ideal time to have the 4D case solved.

[quickfur]

[...]Thanx Wendy for bringing in the outcomes of our communications of the "other" list :-)

Has there been meanwhile any attempt to create a bistratic picture, say a concentrical one, for your xfo3oox3ooo&#xt ?

Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

If someone can post the coordinates, I can run it through my polytope viewer and produce various views of it, with/without hidden surface culling, coloring different cells/elements, etc.. Lately, since Ovo introduced me to the technique, I've been doing a lot of cross-eyed stereo pairs, which really helps in conveying the 3D-ness of the 4D-to-3D projections.

Well, I do not have them within my hands. But they are more than obvious:
Inner shell: unit-size tetrahedron. Coordinates should be obvious.
Next shell: tetrahedron of size tau = (1 + sqrt5)/2, same orientation.
Outermost shell: unit-size octahedron, with 4 triangles above the vertices and 4 triangles antialigned to the tetrahedral ones.
So the only to be determined thing are the relative heights. But thats a matter of an easy application of the Pythagoras theorem only...
(Further help needed?)
--- rk

I'm having some trouble here. I can't seem to find the relative height between the second tetrahedron and the octahedron: it keeps coming out as the square root of a negative number.

Actually, nevermind that. I just saw the mistake in my calculations. Sorry for the false alarm!

[quickfur]

Or is the result supposed to have non-uniform edge lengths? I've been trying to solve for coordinates for a polychoron with equal edge lengths.

Alright, after fixing a silly mistake in my calculations, here's a projection centered on the octahedron:



The octahedron's edges should be obvious. This viewpoint doesn't give a good look at the tridiminished icosahedra, so here's another view:



One of the tridiminished icosahedron forms most of the projection envelope, with the image of the inner-layer tetrahedron at the top (yellow). I colored the octahedron green so that the way it connects to the 4 other tetrahedra is obvious.

Here are the coordinates I used: (EDIT: I simplified the last coordinate of the first 4 vertices)

Code: Select all

< 1,  1,  1,  phi^2>
< 1, -1, -1,  phi^2>
<-1,  1, -1,  phi^2>
<-1, -1,  1,  phi^2>

< phi,  phi,  phi,  0>
< phi, -phi, -phi,  0>
<-phi,  phi, -phi,  0>
<-phi, -phi,  phi,  0>

< 2,  0,  0,  -phi>
<-2,  0,  0,  -phi>
< 0,  2,  0,  -phi>
< 0, -2,  0,  -phi>
< 0,  0,  2,  -phi>
< 0,  0, -2,  -phi>


where phi is the Golden ratio (sqrt(5)+1)/2 (same as Klitzing's tau -- my program uses phi).

Here's the Stella4D file.

P.S. All edge lengths are confirmed equal by my program.

[quickfur]

quickfur: Also, I realized that we have barely four years to finish the CRF project -- 2016 will be the 50th anniversary of Norman Johnson's article enumerating the 3D cases, ideal time to have the 4D case solved.

Hmm. 4 years seems such a short time for something of this complexity. Unless somebody can devote their full-time to this research.

[quickfur]

[...]

Code: Select all

< 1,  1,  1,  sqrt(8 - 3 / (phi^2))>
< 1, -1, -1,  sqrt(8 - 3 / (phi^2))>
<-1,  1, -1,  sqrt(8 - 3 / (phi^2))>
<-1, -1,  1,  sqrt(8 - 3 / (phi^2))>


Huh, funny, the last complicated quantity there is actually just phi^2. I'll never understand square-root denesting algorithms...

Anyway, Klitzing asked for a concentric projection, so here's one:



This is a sort of Schlegel-diagram like projection. My viewer was designed to do perspective projection at a distance (to simulate a 4D view analogous to how our eyes see 3D objects). So in order to make this image, I took the liberty of scaling the polytope in the W axis and distorting the povray parameters a bit so that the vertices of the second tetrahedron would fall inside the octahedral cell. This has the side-effect of making the antipodal tetrahedron look really far away (very small). But some people may find it easier to locate the cells in this projection. (Note that some of the edges are a bit hard to see unless you do cross-eyed stereo viewing.)

[quickfur]

[...]
Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

Hmm. I do happen to have rendered the various layers of the 600-cell:



I think the image on the left should be a fair representation of Wendy's tristratic polychoron? I had visibility clipping on, so that represents just half of the 600-cell, which would be just the top 4 layers in the right image.

Visually, I can see how between the blue dodecahedron and the yellow icosidodecahedron one could fit 20 tetrahedra and 12 pentagonal antiprisms, and between the blue dodecahedron and the violet icosahedron one could fit 30 tetrahedra and 12 pentagonal pyramids. I'll run the actual coordinates through my viewer tomorrow and do actual renders of Wendy's polychoron (not just a hand-picked subset of the 600-cell's edges, as I have here).

[Klitzing]

Alright, after fixing a silly mistake in my calculations, here's a projection centered on the octahedron:



The octahedron's edges should be obvious. This viewpoint doesn't give a good look at the tridiminished icosahedra, so here's another view:



One of the tridiminished icosahedron forms most of the projection envelope, with the image of the inner-layer tetrahedron at the top (yellow). I colored the octahedron green so that the way it connects to the 4 other tetrahedra is obvious.

P.S. All edge lengths are confirmed equal by my program.

Nicely done. You're right, the 2nd view looks better.
Suppose Wendy would like 'em too, as those polychora were originally hers.
--- rk

[wendy]

At the moment, i am wrangling the metegloss (which is the weights-and-measures side of things).

Still, the lacing-notation is worth discussing, because it really gives the coordinates. In essence, we lace (ie tie) together several polytopes, and pull them out, kind of like a chinese lantern. The actual term of lacing was meant to invoke the resemblence of the zigzag edges of a polygonal antiprism, with the lacing between the top and bottom skins of a drum. so xo5ox is a pentagonal antiprism, the lacing runs between the vertices x5o on top, and the (inverted pentagon) o5x on the bottom.

For something like xfo3oox3ooo3ooo.... the coordinates are relatively easy to extract, i suppose. It's just a number of polytopes laced together, and all of the polytopes are constructed on the same set of mirrors. So if you have means to make these (i have no clue on what you're doing to get these figures, i just look at them and agree etc).

Code: Select all

    xfo 3 oox 3 ooo 3 ooo ...                     v = 0.61803398875 = a(5/2)= $V
     x  3  o  3  o  3  o  ...   =  x3o3o3o3o...   x = 1.0000000000  = a(3)  = $S
     f  3  o  3  o  3  o  ...   =  f3o3o3o3o...   q = 1.41421356238 = a(4)  = $Q = $$
     o  3  x  3  o  3  o  ...   =  o3x3o3o3o...   f = 1.61803398875 = a(5)  = $F
                                                  h = 1.73205080757 = a(6)  = $H = $S$
                                                  u = 2.00000000000 =  2    = $U = $U$


Basically, the xfo3oox etc simply represents three polytopes x3o, f3o, o3x, etc laced together in that order. The little letter represents the edge of the side of the section (which may or may not appear as an edge: the 'f' in the second row is a chord across the pentagon). You can do some sort of arithmetic to get the thing to get the height to get the right edge. It's 'relatively simple' arithmetic, but it stumbles me often.

I don't know if you're into matrices or not, but with matrices, one can generally get the right results. That's what i use.

You make a dynkin matrix, a square matrix of the form a(i,j), where a(i,i) = 2, and a(i,j) = -2 cos(pi/p), ie -v, -x, -q, etc as the branch is 5/2, 3, 4, 5, &c. You then take the inverse of this matrix, to get the 'Stott matrix' S(i,j).

The stott-matrix allows you to find the length of any coordinate, given its vertices. What you do is to take the matrix-dot ie S(i,j) V(i). V(j). This will work regardless of the coordinates of the vector, so if you calculate the matrix-dot of a difference, eg o3x5o = 0,1,0 less o3o5x 0,0,1, gives 0,1,-1, this is the non-height side of the altitude between layers. subtract this from height, to get the distance between layers. It's how i normally do it.

Code: Select all

     o-3-o-5-o         o-3-o-5-o       v  Sv

     2  -1   0        3-f   2   f      0    2-f    0
    -1   2  -f         2    4   2f     1   4-2f   4-2f
     0  -f   2         f   2f   3     -1   2f-3   3-2f
                     * 2/(4-2f)                          -> 2.618033
     Dynkin matrix   Stott matrix                 7-4f   -> 1.381966

    L = lacing, R = value returned by this matrix.

    h =  sqrt( 4L-R ) / 2 =  sqrt(4-1.381966)/2 = 1.61803398875/2.



The dynkin-matrix is a representation of the dynkin symbol. The nodes run down the diagonal, and at ij and ji, one puts a value of -a(N), for a branch marked N. a(N) is the shortchord of N, the chords in the local arithmetic run 1,a,b,c, &c. I think it's call the Gramm matrix or something in real-speak, but it's associated closely with the dynkin-symbol.

The stott-graph is derived by 1/dynkin symbol. I calculated these independently by considering (from a set of coordinates), the dot-product of a.i × a.j, where a.n form the n vectors that make the unit-edge polytopes x3o5o, o3x5o, o3o5x usw. Since stott's method is to expand and contract along (what is here the axies) directions, it is fitting that this matrix has her name. It is simply an oblique coordinate system, so one can calculate the lengths by a 'matrix-dot', or S(ij).V(i).V(j) = V(i)SV(i) = SV². It is a square of length, chosen here for diameter. You can normalise the sections for radius. It is best to work with a known case to get the constants, which -never- change from case to case.

[quickfur]

[...]
For something like xfo3oox3ooo3ooo.... the coordinates are relatively easy to extract, i suppose. It's just a number of polytopes laced together, and all of the polytopes are constructed on the same set of mirrors. So if you have means to make these (i have no clue on what you're doing to get these figures, i just look at them and agree etc).
[...]
Basically, the xfo3oox etc simply represents three polytopes x3o, f3o, o3x, etc laced together in that order. The little letter represents the edge of the side of the section (which may or may not appear as an edge: the 'f' in the second row is a chord across the pentagon). You can do some sort of arithmetic to get the thing to get the height to get the right edge. It's 'relatively simple' arithmetic, but it stumbles me often.
[...]

Looks like I shall have to start learning how to work with Stott matrices. I confess ignorance in that area.

My polytope viewer uses the naïve approach of taking Cartesian coordinates as input and running a convex hull algorithm on them to extract the face lattice, hyperplane normals, etc.. Then there is a scripting/command-prompt interface where one can do basic queries on different polytope elements, e.g., select the set of edges that lie on the cell closest to the currently-set 4D viewpoint, etc.. One can then assign different textures to these selected elements. Once everything is set up, a 'render' command causes the program to generate a povray mesh that can be incorporated into a template povray scene file, which can then be rendered to produce the images.

To simulate 4D vision (as a perspective projection from a distance), the viewer also culls polytope elements that lie on cells whose hyperplane normals are pointing away from the selected 4D viewpoint (visibility clipping). This is especially valuable for complicated objects like the members of the 600-cell family, since otherwise the resulting mesh is a tangle of mutually-intersecting surfaces which is hard to make much sense of from our disadvantaged 3D viewpoint. For simpler shapes, though, like Wendy's bistratic that I rendered above, I turned off visibility clipping so that all cells are visible at once. The self-intersection is minimal in this case, and doesn't detract from the clarity of the image.

I usually work with 4D shapes "visually" (by picturing their perspective projections into 3D from various viewpoints); once I'm reasonably confident that a particular shape can exist, then I try to compute the actual coordinates to run through the viewer to verify that it indeed exists and has the expected properties.

In any case, I do plan on redesigning my viewer at some point -- the current version grew out of an experimental part of a more general program for dealing with arbitrary-dimension polytopes (that unfortunately never materialized), and as such suffers from some bad design decisions. I should probably build in more symmetry-based methods of polytope generation in the new version than just mere Cartesian coordinates. The Dynkin/Stott constructions should prove very useful.

[quickfur]

[...]
Your tristratic oxoo3ooox5ooxo&#xt seems out of scope for visualization by means of pictures, hehe.

--- rk

[...]
I'll run the actual coordinates through my viewer tomorrow and do actual renders of Wendy's polychoron (not just a hand-picked subset of the 600-cell's edges, as I have here).

Alright, here's a render of Wendy's tristratic polychoron:



Due to the huge number of edges inherited from the 600-cell, I decided to omit the edges that do not lie on the vertex/icosahedron/dodecahedron/icosidodecahedron used to construct this polychoron.

To give an idea of where the cells are, though, I decided to highlight a string of tetrahedra that start from the top vertex (shown as the red vertex in the middle of the projection) and going to the icosidodecahedral base. This shows 3 kinds of tetrahedra that lie at different latitudes. There are 20 such triplets of tetrahedra, which one can easily envision by applying the symmetry of the icosahedron.

There is a fourth kind of tetrahedron, one example of which is shown in blue: there are 30 of these, straddling the layer between the icosahedron and dodecahedron.

Finally, the green cell at the bottom shows one of the 12 diminished icosahedra.

To make the shapes of these highlighted cells clearer, I added their edges back in.

And just for the record, here are the coordinates I used:

Code: Select all

<0, 0, 0, ±2>
<0, 0, ±2, 0>
<0, ±2, 0, 0>
<2, 0, 0, 0>

<0, ±1/φ, ±1, ±φ>
<0, ±1, ±φ, ±1/φ>
<0, ±φ, ±1/φ, ±1>
<1, 0, ±1/φ, ±φ>
<1, ±1/φ, ±φ, 0>
<1, ±φ, 0, ±1/φ>
<φ, 0, ±1, ±1/φ>
<φ, ±1/φ, 0, ±1>
<φ, ±1, ±1/φ, 0>

<1, ±1, ±1, ±1>


There are 103 cells in total: 90 tetrahedra, 12 diminished icosahedra, and 1 icosidodecahedron.

And I note, and I believe Wendy has also pointed this out, that one can augment this polychoron with 12 diminished icosahedron pyramids, amounting to adding the vertices of another, slightly larger, icosahedron to the mix. This adds 180 more tetrahedra (15 more for each diminished icosahedron), producing another CRF polychoron with 270 tetrahedra, 12 pentagonal pyramids, and 1 icosidodecahedron. One may think of this as a kind of "bisected 600-cell", as it includes the vertices of exactly half of the 600-cell (though it is not strictly a true bisection, because a number of edges that intersect the cutting hyperplane have been deleted; otherwise those edges would be bisected and the result would be non-CRF). Putting two of these things together makes a 600-cell with 60 missing tetrahedra (every 5 of which corresponds with a deleted edge -- so there are 12 such edges).

[quickfur]

Automatizing the process of generating edges is good idea in any case. As for the nonrigid edges, I wonder if there's a way around that...

An idea just occurred to me. We may be able to generalize the algorithm to degree 4 edges, if we add the restriction that every face must have at least 1 edge that's only degree 3.

Consider again two surface polyhedra A, B joined at face F. By our assumption, at least one of F's edges has only 3 polyhedra surrounding it. So we can use our algorithm to enumerate all possible dichoral angles between A and B, based on the permissible dihedral angles around the degree 3 edges of F. Since degree 4 edges have 1 degree of freedom, whatever solution we find based on just the degree 3 edges also represents the possibilities of the degree 4 edges (they can just be deformed along their degree of freedom to fit the dichoral angle that "works" for the degree 3 edges). So once we fix a dichoral angle between A and B, all the degree 4 edges are also fixed, and there's no need to store vertices as quadratic equations, etc..

Now, taking a step back from the details of handling degree-4 edges, how does this affect our search algo? Since we're trying to find CRF polychora, we don't know in advance which edges are degree 4, etc.. So given two polyhedra A and B joined at face F, any of its edges could be degree 4 or higher. So we need to modify the search algo slightly: instead of simply finding the intersection of the set of dichoral angles that works for every (degree 3) edge, we now need to consider all possible subsets of edges which we assume to be degree 3. The empty subset corresponds with the case where all edges have degree >3; we ignore this case since we don't know how to handle it. Then for each subset of F's edges, we assume they are all degree 3, and compute the possible dichoral angles for them. But then this is just equivalent to taking the union, instead of the intersection, of the sets of possible dichoral angles for each edge.

Then we loop over the possible dichoral angles, and every edge that ends up with a dihedral angle listed in our tables will be known to be degree 3; the remaining edges will be degree 4. Since the dichoral angle is now fixed, we can resolve the cells around the degree 4 edges as well. So the rest of the algo can remain as-is.

[Klitzing]

Two possible CRF ("Johnsonian") polychora:



I came up with these while thinking about how a cell could be attached to a rotated copy of itself in 4D.

Both forms have 16 vertices.

The "ortho" form has 36 edges (16 black, 4 red and 16 yellow).
The "gyro" form has 40 edges (16 black, 8 red and 16 yellow).

The "ortho" form has 31 faces (1 octagon, 14 squares and 16 triangles).
The "gyro" form has 35 faces (1 octagon, 10 squares and 24 triangles).

I like how they turned out somewhat like a duoprism (the projections show an "in-out ring" of 2 cupolae and prism/antiprism, and a "round-the-outside ring" of triangular prisms/tetrahedra and square pyramids)

A few questions...
1. are these two shapes indeed CRF polychora? (i.e. can they be built with all their (two-dimensional) faces regular?)
2. do these appear in wintersolstice's list above? IIRC, his list was mainly composed of prismatoid-like forms (3+1) whereas these are more duoprismatic-like forms (2+2)
3. is the "gyro" form the first explicitly constructed polychoron to include the square antiprism as a cell (other than trivial things like the square antiprism prism/pyramid)? if not, what came before it?
4. any name suggestions for these two?
5. could quickfur render them, pretty-please? I'm aware that my diagrams have the black square faces rotated 22.5 degrees (I think) out of line with where they should be so the square faces do not appear planar even though they are supposed to be

Even so not the fasted one to answer, yes those are true CRF, in fact they even are convex segmentochora! The left one is {8} || cube or equivalently {4} || 4-cupola, the right one is {8} || 4-antiprism or equivalently {4} || gyrated 4-cupola.

You are right in that the {8} is not drawn correctly. Its edges have to be parallel to those of the black squares.

At 5: If you have a VRML-viewer (like Cortona or others), you could view my according files and even turn those polychora around, which are hosted at J. McNeills site: http://www.orchidpalms.com/polyhedra/segmentochora/table.htm.

At 3: In my article (also hosted there in PDF form) there are 4ap || 8g, 4ap || 4g, 4ap || 4ap, 4ap || point.

--- rk

[Klitzing]

[...] Yep, 4D Johnsons is quite well a task. I also once went into that direction. But then settled to a more manageable sub-class: the monostratic orbiform ones. That research, Wendy was citing, already has been published:
"Convex Segmentochora", by Dr. R. Klitzing, Symmetry: Culture and Science, vol. 11, 139-181, 2000
[...]

Hi, and welcome to the forum! It's awesome that you can chime in with our search for 4D Johnsons (or as we call them here, CRF polychora, for Convex-Regular-Faced). Your research has been very helpful to us in confirming many of our independently-discovered results.

Thanx for the welcome. I' overwhelmed by the emphasis of that research, which my article brought up. - In fact that very article itself was partly inspired by the same intend. I was invited by Dinogeorge to the polylist long ago, and so had several discussions there about this topic. Esp. I got to know that in 1979 a german couple already started that 4D research in a somewhat smaller or stricter sense: they extrapolated the Johnson solid conditions in the sense of regular boundaries, i.e. of regular facets. For 4D this would mean regular cells. Those were listed all in those days, except of the various 600-cell diminishings, for sure. But for me that lifting of the settings were artificial, esp. as those changed the wordings. So I clearly asked for what meanwhile has been called CRF.

In fact, those discussions then were the origin for what then became "scaliform" (in those days still: "weakly uniform") or even "orbiform".

The other stem for my research within the millenium change, was to provide a set of polychora which allow for an easy to grasp visualization of 4D content. So think of that research also as kind a didactical one. This is why 2 parallel vertex layers were chosen. Those figures can easily be visualized in 2 concentric settings.

Have you considered the slightly larger category of monostratic Johnsons with the orbiform requirement relaxed? The extrusions of the Johnson solids are the most obvious ones in this category. I've been trying to find other kinds of non-orbiform monostratic polychora but so far haven't found anything yet.

Well, it was kind a direct lifting of what occurs in 3D: convex segmentohedra need to have regular polygons as bases (if not degenerate). So it was easily translated into orbiform bases, as those still bow to an unique circumradius (of the base). And 2 orbiforms arranged atop of each other would allow for an unique circumradius in the next dimension up. At least the lacings would tie them accordingly so that their centers align at the axis. - Only subdimensional bases allow for a shift.

Further these unique circumradii of the bases made it easy to calculate the height of the segmentotopes, i.e. the displacement of the hyperplanes. Additionally this guaranteed that all lacing edges would be of the same length. Thus this choice was quite helpful for the research. - Any weaker setup would get a lot harder.

[...] On a slightly higher-level, I wonder if the enumeration of 4D Johnsons can be made somewhat more tractable if we split up the task into more directed categories where more specialized search methods can be employed. Following the way Johnson arranged the 3D Johnson solids in groups, I came up with the tentative categories:

Code: Select all

I. Monostratic polychora
   A. Orbiform polychora -- your list of segmentochora
   B. Non-orbiform polychora -- prisms of the non-orbiform 3D Johnsons, probably with a few others.
   C. Various derivatives of the above, e.g., the laminochora (stacking monostratic polychora on top of each other)

II. Rotundae -- so far, the 5 or 6 that we've found in this thread, plus their derivatives - birotundae, elongated birotundae, augmentations with the monostratics, etc..

III. Modified Uniform Polychora
   A. Augmented duoprisms
      1. Augmentations with prism pyramids: 1633 of them, for 3,n-duoprisms up to 5,n-duoprisms (the largest known is the omniaugmented 5,20-duoprism, which has a ring of 10 augmented square bipyramid cells flanked by square pyramids).
      2. Augmentations with n-gon||n-prism: at least 1633 of them (probably more), for 6,n-duoprisms up to 10,n-duoprisms, the largest of which probably will be the 10,20-duoprism.
   B. Augmented/diminished 5-cell family members
   C. Augmented/diminished tesseract family members
   D. Augmented/diminished 24-cell family members
   E. Augmented/diminished 120-cell family members

IV. Crown Jewels -- only 1 known, if we count cube||icosahedron (though that one can be put in category I). No known search method.


Could give a try... But sure wont get less hard to do. And you would have the additional task to prove that this partitioning would not omit some cases...

I rather would try to lift the prove of completeness of Normans original research directly, by means of a brute force computer research sticking allowed polyhedra face to face, with lots of freedoms of deformation until those complexes would close finally. - Just could run into problems with the sheer number of to be found polychora, so that such a program wont end within a human lifetime...

Thanks to your research, category I.A. is completed. Category I.B is probably the easiest next step. Category I.C. can probably be enumerated by computer if we formulate the appropriate requirements for a particular stacking to be CRF. At the very least, the derivatives of I.A. can be run through a computer enumeration since we already have the necessary data.

A few examples in category II are known, as described elsewhere in this thread; some work is needed to find any remaining instances (or prove there are none others). Once that is done, it's just a matter of enumerating derivatives like bicupolae, augmentations with I.C polychora, etc..

Category III is a large category, where I think will probably also be the largest in terms of number. The duoprism augmentations with prism pyramids have been completed by Marek & myself (though our results need to be double-checked). But there remains augmentations with other things, among which the n-gon||n-prism augmentations constitute at least 1600+ members. Augmentations with other CRFs have yet to be studied.

The 600-cell probably deserves its own category, since a very large number of diminishings are possible. I've attempted to find the maximally-diminished 600-cells, but so far have not been able to find a general scheme for generating these things save by brute-force search. Things are very complicated here because adjacent vertices can be deleted while still keeping the result CRF, unlike the 24-cell diminishings where assuming non-adjacency of deleted vertices greatly simplifies enumeration.

As for category IV, I don't know what method can be used to search for them -- even exhaustive brute-force search seems intractible because of non-rigid edges when more than 3 polyhedra surround it. This will probably be the hardest category to crack open.

I just gave the bi-stratics some preliminary thoughts. There would be 3 classes: Stacks (i.e. towers) of monostratics with some true dihedral angles between the lacing facets; similar stacks with that angle becoming 180 degrees, so that those facets would be reconnected to towers themselves; and finally the original new ones containing true bistratic lacing facets. - Wendys xfo3oox3ooo&#xt clearly belongs to those last ones.

But then I gave a thought to the diameter of the equatorial vertex layer: Esp. in the last 2 classes that one generally would not match the unique circumsphere condition. So, in view of the usefulness of unique circumradii, as stated above, we probably should restrict orbiformness to the layers only, and not ask for the total figure! But that in turn would open a can of worms for the first class: we have to determine that the dihedral angles of the lacing cells at the equatorial hyperplane still have to bow to convexity. And this asks for additional calculations here...

I.e. the to be chosen setup for a useful extrapolation from mono- to bistratic figures still is not at its end.

--- rk

[quickfur]

[...] Esp. I got to know that in 1979 a german couple already started that 4D research in a somewhat smaller or stricter sense: they extrapolated the Johnson solid conditions in the sense of regular boundaries, i.e. of regular facets. For 4D this would mean regular cells. Those were listed all in those days, except of the various 600-cell diminishings, for sure.

So the regular-celled CRFs are already known? Is this list published or available online somewhere?

[...] The other stem for my research within the millenium change, was to provide a set of polychora which allow for an easy to grasp visualization of 4D content. So think of that research also as kind a didactical one. This is why 2 parallel vertex layers were chosen. Those figures can easily be visualized in 2 concentric settings.

True, that certainly makes it easier to introduce the subject of 4D visualization.

Have you considered the slightly larger category of monostratic Johnsons with the orbiform requirement relaxed? [...]

Well, it was kind a direct lifting of what occurs in 3D: convex segmentohedra need to have regular polygons as bases (if not degenerate). So it was easily translated into orbiform bases, as those still bow to an unique circumradius (of the base). And 2 orbiforms arranged atop of each other would allow for an unique circumradius in the next dimension up. At least the lacings would tie them accordingly so that their centers align at the axis. - Only subdimensional bases allow for a shift.

True, I guess this would keep the numbers of segmentochora manageable in higher dimensions, as well as preserve a useful common property among all of them. Relaxing the orbiform requirement would admit more and more exotic shapes as you go up into higher dimensions, since an exotic shape in one dimension opens up the opportunity for even more exotic shapes constructed from it in the next dimension.

Further these unique circumradii of the bases made it easy to calculate the height of the segmentotopes, i.e. the displacement of the hyperplanes. Additionally this guaranteed that all lacing edges would be of the same length. Thus this choice was quite helpful for the research. - Any weaker setup would get a lot harder.

True. These calculations would get very complicated quickly once the orbiform requirement is relaxed. Still, I'm curious about what kinds of non-orbiform CRF monostratics there are, that aren't just prisms of Johnson solids. I find that the process of discovering how polyhedra can be put together to form polychora is very insightful into the workings of 4D space.

[...] On a slightly higher-level, I wonder if the enumeration of 4D Johnsons can be made somewhat more tractable if we split up the task into more directed categories where more specialized search methods can be employed. Following the way Johnson arranged the 3D Johnson solids in groups, I came up with the tentative categories:
[...]

Could give a try... But sure wont get less hard to do. And you would have the additional task to prove that this partitioning would not omit some cases...

I was thinking of using the crown jewels category as a "catch-all" category for odd shapes that don't belong to any other category. But you're right -- even if we filled out those categories, we still have to prove that we've covered everything. It seems that a computer brute-force search is the only way to conclusively prove the completeness of our list.

I rather would try to lift the prove of completeness of Normans original research directly, by means of a brute force computer research sticking allowed polyhedra face to face, with lots of freedoms of deformation until those complexes would close finally. - Just could run into problems with the sheer number of to be found polychora, so that such a program wont end within a human lifetime...

I've been thinking about that, but the major difficulty is the sheer number of 3D CRFs, which means a giant combinatorial explosion, coupled with the complexity of representing partial CRF complexes in an efficient way that would allow algorithmic detection of when something can be closed up to make a CRF polychoron.

A brute-force algorithm that Marek & I have been considering uses dihedral angles to fix dichoral angles between cells so that viable cell combinations are discovered more quickly. But it turns out that, due to non-rigid edges surrounded by more than 3 polyhedra, this algorithm will not enumerate all possibilities. So it's not suitable for proving CRF completeness, but with some tweaks, it can be used to search for CRFs up to degree-4 edges, with at least one degree-3 edge per face. It's a much smaller class of CRFs, since most of the 600-cell family CRFs would be excluded, but it may help reveal some unusual constructions that can be generalized by hand to more complex shapes.

[...] I just gave the bi-stratics some preliminary thoughts. There would be 3 classes: Stacks (i.e. towers) of monostratics with some true dihedral angles between the lacing facets; similar stacks with that angle becoming 180 degrees, so that those facets would be reconnected to towers themselves; and finally the original new ones containing true bistratic lacing facets. - Wendys xfo3oox3ooo&#xt clearly belongs to those last ones.

I think stacks/towers can be automatically enumerated by computer, since the monostratics are already known. We just compute the dichoral angles for each cell pair in the monostratics, which can also be done by computer, and that will tell us which ones can be stacked on top of each other without becoming non-convex.

The second class needs some special handling, since not all co-planar facets are Johnson. There are a finite number of allowed co-planer facets, like square pyramid + cube = elongated square pyramid, cupola + base prism = elongated cupola, etc.. Things like cube+cube would be illegal, since the convex hull would have cuboidal facets.

The last class I would regard as "true" bistratics, since they cannot be formed by cut-n-paste operations on monostratics. They're also the most interesting class.

But then I gave a thought to the diameter of the equatorial vertex layer: Esp. in the last 2 classes that one generally would not match the unique circumsphere condition. So, in view of the usefulness of unique circumradii, as stated above, we probably should restrict orbiformness to the layers only, and not ask for the total figure! But that in turn would open a can of worms for the first class: we have to determine that the dihedral angles of the lacing cells at the equatorial hyperplane still have to bow to convexity. And this asks for additional calculations here...
[...]

Hmm. While I see the usefulness of the orbiform requirement, I'm more interested in the non-orbiform cases. They would be harder to calculate, but more rewarding to discover IMO.

As for computing angles, it can probably be done by computer, since once vertex coordinates are known, a convex hull algorithm gives the face lattice, and it's easy to automate computation of dihedral/dichoral angles. This should allow us to write programs to quickly eliminate obviously non-convex combinations, then if necessary we can manually check the boundary cases (where the angle is close to 180°) where there may be inaccurate results due to roundoff error in computer arithmetic.

[quickfur]

Today I tried generalizing Wendy's bistratic CRF xfo3oox3ooo by the following observations:

(1) The tridiminished icosahedron may be thought of as a modified truncated tetrahedron, in which 3 alternate edges of one of the hexagonal faces is shrunk to a point. As it does so, the other 3 hexagonal faces become pentagons, and that hexagon gets rectified into a triangle.

(2) Wendy's bistratic polychoron may be thought of as a similar process applied to the truncated 5-cell: the hexagonal faces of one of the truncated tetrahedral cells are "shrunk" so that they become rectified into triangles; this in turn shrinks one hexagonal face of each of the remaining 4 truncated tetrahedra, which causes them to turn into tridiminished icosahedra. The first truncated tetrahedron thus get shrunk into an octahedron.

So I wondered what would happen if the same process is applied to the bisected truncated 16-cell. Topologically speaking, the 8 truncated tetrahedra should turn into tridiminished icosahedra, and the bottom truncated octahedron should turn into a cuboctahedron. This gives a tristratic polychoron with 1 octahedron, 8 tridiminished icosahedra, 8 square pyramids, and 1 cuboctahedron. Unfortunately, calculations show that it cannot be made CRF, because either the cuboctahedron will have unequal edge lengths, or the pentagonal faces will fissure into trapeziums and isosceles triangles.

Then I wondered if edge lengths would line up correctly if we start with an icosahedron and attach 20 tridiminished icosahedra around it. This indeed does work; it produces a tristratic CRF which is a kind of diminished 600-cell. It's basically Wendy's tristratic oxoo3ooox5ooxo&#xt with the top vertex and the dodecahedron deleted, and an icosahedron inserted before the icosidodecahedron (I'm not sure exactly what the scaling factor is -- basically it corresponds with the apexes of 12 truncated icosahedron pyramids which may be placed on top of the truncated icosahedron cells). It consists of 1 icosahedron, 20 tridiminished icosahedra, 12 pentagonal pyramids, and 1 icosidodecahedron. I'm not well-versed with Wendy's notation, but I believe it would be something like xfx3ooo5oox.

Another possible modification to xfo3oox3ooo would be something like xfx3oox3xfo. This should give a CRF with 1 octahedron, 4 tridiminished icosahedra, 6 pentagonal prisms, 4 triangular prisms, 4 triangular cupola, and 1 truncated octahedron. It's sortof like a pseudo-runcinated form of xfo3oox3ooo. I haven't verified this yet, but I believe it should work. I'll try to compute the coordinates tomorrow.

[wendy]

In the end, i decided on the following terminology to be consistent with that laid out in the Polygloss.

One first notes the notion of 'larger sense' and 'smaller sense'. The larger sense is inclusive, the lesser sense excludes those that obey more restrictive rules. So, for example, the outer sense of the johnsons is simply 'any polytope bounded by regular polygons', but the lesser sense excludes the regulars, the archemedians, the prisms, and the antiprisms; collectively the uniforms (in the larger sense).

The list here is that the figure needs to be convex, and constructed of elements that are uniform up to a particular dimension. For UH, we regard this set equal to the regular polygons, since any equalateral polygon.

CUH convex-uniform-hedra (3d regular, uniform, johnson) In all dimensions, it just means that it's made of uniform hedra (regular polygons), with no larger structure. The 3d elements (chora) are from the set of johnson-figures (in the larger sense),

CUC convex -uniform-chora (made of uniform chorids (polyhedra)),

CUT convex uniform-terahedra (made of uniform terids (polychora))

There are CUC's, made of uniform polyhedra, among the CUH's (CRF). These include things like the 335 with one, two or three rings of pentagonal antiprisms, and varying numbers of sundry icosahedra diminished. The icosahedral and octahedral pyramids, and the icosahedral bi-pyramid, are also CUC's.

Quickfur's list is probably a good start, but one needs to realise that things like different arrangements of the same thing can create new polytopes. An example is the quasi-rhomboCO, where one of the caps is rotated.

The vast bulk of johnson's figures can be variously handled by the tower-notation, eg "oxoo4ooxo &tx = bi-apiculated square antiprism", or with some modification, by the use of icosahedrids (dim, bi-dim, tri-dim, and pap) for icosahedra sections. There's a smallish zoo of etwas, but i am not sure what to do there. They just sit there to annoy us that construction <> outcome.

[Marek14]

Got another idea -- how about trying to search for duals of CRF polychora?

Now, there are various ways to define duals for nonorbiform figures etc., but in this particular case we could define duals as "external duals", i.e. CRF polychoron is obtained by joining centroids of the cells of the larger "dual". Or they could be defined in another way, as will be needed.

Why do this? Because the main constraining condition on CRF polychora concerns its faces. Well, on dual, those faces would correspond to edges. This means that a dual has main constraining condition on edges -- and so we could use the edge algorithm (or a suitable variant) and completely avoid the problem with nonrigid edges we have when searching for CRF polychora directly.

I don't know the exact constraining condition on the edges, but I think it could be probably found out.

This just occured to me and I haven't looked into it more deeply, it's just a thought.

EDIT: First obvious hurdle, of course, is that we wouldn't have a database of allowed cell shapes, but maybe that could be bypassed by making a database of possible CRF vertex figures and then dualizing those.

EDIT2: The particular dualizing operation to choose should be easy to reverse, but it doesn't have to be easy to construct from original polychoron (since we will only need to go from dual to CRF). We should find one that will transform the strict condition of CRF faces into equally strict condition on dual's edges. As long as there will be an algorithm to make the CRF dual from that, we're OK. (Specifically, the dual operation doesn't require that you get the original polychoron by applying it twice.)

EDIT3: Another idea: even in the original version, the nonrigid edges problem could be lessened by building a database of vertex figures. In this case, vertex figures wouldn't necessarily have co-hyperplanar vertices, but there's nevertheless a big constrain on them: their vertices must form an abstract polytope whose edge lengths must be selected from a small list of shortchords. Combining with the rule that vertex figures must be convex and there must be a point with distance 1 from every vertex, the list of possible abstract polytopes that can serve as vertex figures shouldn't be unreasonably long since there has to be a maximum amount of vertices for them. (I suspect it's 12 and that you can't have more edges at one vertex than in 600-cell, but I might be wrong.)
In the dualized version, all possible vertex figures would be already known.

In 3D this approach doesn't work too well since vertex figures are polygons which can then be skewed in third dimension, and that has many degrees of freedom. I suspect it might be different in 4D case. For example, adding a vertex to a polygon requires to satisfy the shortchord constraints with 2 other vertices, but in 4D there are at least 3 constraints and there might be more. With 3 constraints, the shape would still be flexible, BUT that would imply a 3-face edge. On the other side, a 4- or more-face edge would lead to 4 or more constraints on the vertex in vertex figure, and that would actually fix it in 4D, wouldn't it?
In other words, the strong points of edge and vertex approach cover each other's weaknesses!

[Marek14]

I compiled all vertex types of regular-faced polyhedra; this should help in enumerating possible vertex figures.

Where the vertex figure is skew, this is noted, furthermore, various types of skews are separated:

Code: Select all

3,3,3: tetrahedron, elongated triangular pyramid, triangular dipyramid, elongated triangular dipyramid, augmented tridiminished icosahedron
3,3,4: square pyramid
3,3,5: pentagonal pyramid
3,4,4: triangular prism, elongated triangular pyramid, gyrobifastigium, augmented triangular prism
3,4,6 (chiral): triangular cupola
3,4,8 (chiral): square cupola
3,4,10 (chiral): pentagonal cupola
3,5,5: metabidiminished icosahedron, tridiminished icosahedron [fits in 2 different ways], augmented tridiminished icosahedron, bilunabirotunda
3,5,10 (chiral): pentagonal rotunda
3,6,6: truncated tetrahedron, augmented truncated tetrahedron [fits in 2 different ways]
3,8,8: truncated cube, augmented truncated cube [fits in 4 different ways, 2 of them chiral], biaugmented truncated cube
3,10,10: truncated dodecahedron, augmented truncated dodecahedron [fits in 10 different ways, 6 of them chiral], parabiaugmented truncated dodecahedron [fits in 4 different ways, 2 of them chiral], metabiaugmented truncated dodecahedron [fits in 20 different ways, 16 of them chiral], triaugmented truncated dodecahedron [fits in 10 different ways, 6 of them chiral]
4,4,4: cube, elongated square pyramid [fits in 3 different ways]
4,4,5: pentagonal prism, elongated pentagonal pyramid, augmented pentagonal prism [fits in 3 different ways, 2 of them chiral], biaugmented pentagonal prism
4,4,6: hexagonal prism, elongated triangular cupola [fits in 2 chiral ways], augmented hexagonal prism [fits in 4 chiral ways], parabiaugmented hexagonal prism, metabiaugmented hexagonal prism [fits in 2 chiral ways]
4,4,7: heptagonal prism
4,4,8: octagonal prism, elongated square cupola [fits in 2 chiral ways]
4,4,9: enneagonal prism
4,4,10: decagonal prism, elongated pentagonal cupola [fits in 2 chiral ways], elongated pentagonal rotunda [fits in 2 chiral ways]
4,5,10 (chiral): diminished rhombicosidodecahedron, paragyrate diminished rhombicosidodecahedron, metagyrate diminished rhombicosidodecahedron [fits in 5 different ways], bigyrate diminished rhombicosidodecahedron [fits in 5 different ways], parabidiminished rhombicosidodecahedron, metabidiminished rhombicosidodecahedron [fits in 5 different ways], gyrate bidiminished rhombicosidodecahedron [fits in 10 different ways], trigyrate rhombicosidodecahedron [fits in 5 different ways]
4,6,6: truncated octahedron
4,6,8 (chiral): truncated cuboctahedron
4,6,10 (chiral): truncated icosidodecahedron
5,5,5: dodecahedron, augmented dodecahedron [fits in 9 different ways], parabiaugmented dodecahedron [fits in 3 different ways], metabiaugmented docecahedron [fits in 15 different ways, 6 of them chiral], triaugmented dodecahedron [fits in 5 different ways]
5,6,6: truncated icosahedron

3,3,3,3: octahedron, square pyramid, elongated square pyramid, gyroelongated square pyramid, elongated square dipyramid, gyroelongated square pyramid, augmented triangular prism [fits in 2 different ways], biaugmented triangular prism [fits in 4 different ways], triaugmented triangular prism [fits in 2 different ways], augmented pentagonal prism [fits in 2 different ways], biaugmented pentagonal prism [fits in 4 different ways], augmented hexagonal prism [fits in 2 different ways], parabiaugmented hexagonal prism [fits in 2 different ways], metabiaugmented hexagonal prism [fits in 4 different ways], triaugmented hexagonal prism [fits in 2 different ways], augmented sphenocorona [fits in 4 different ways]
3,3,3,3 (skew1): triangular dipyramid
3,3,3,3 (skew2): pentagonal dipyramid
3,3,3,3 (skew3): snub disphenoid [fits in 2 different ways]
3,3,3,3 (skew4): sphenomegacorona [fits in 2 different ways]
3,3,3,4: square antiprism, gyroelongated square pyramid
3,3,3,4 (skew1, chiral): augmented triangular prism, biaugmented triangular prism
3,3,3,4 (skew2, chiral): sphenocorona
3,3,3,5: pentagonal antiprism, gyroelongated pentagonal pyramid, metabidiminished icosahedron [fits in 3 different ways, 2 of them chiral], tridiminished icosahedron, augmented tridiminished icosahedron, triangular hebesphenorotunda
3,3,3,6: hexagonal antiprism, gyroelongated triangular cupola [fits in 2 different ways]
3,3,3,7: heptagonal antiprism
3,3,3,8: octagonal antiprism, gyroelongated square cupola [fits in 2 different ways]
3,3,3,9: enneagonal antiprism
3,3,3,10: decagonal antiprism, gyroelongated pentagonal cupola [fits in 2 different ways], gyroelongated pentagonal rotunda [fits in 2 different ways]
3,3,4,4: triangular orthobicupola
3,3,4,4 (skew1): elongated triangular pyramid, elongated triangular dipyramid
3,3,4,4 (skew2): elongated square pyramid, elongated square dipyramid, square orthobicupola
3,3,4,4 (skew3): elongated pentagonal pyramid, elongated pentagonal dipyramid
3,3,4,4 (skew4): pentagonal orthobicupola
3,3,4,4 (skew5): sphenocorona, augmented sphenocorona
3,3,4,4 (skew6): sphenomegacorona
3,3,4,4 (skew7): hebesphenomegacorona [fits in 2 chiral ways]
3,3,4,4 (skew8): disphenocingulum
3,4,3,4: cuboctahedron, triangular cupola [fits in 2 different ways], elongated triangular cupola [fits in 2 different ways], gyroelongated triangular cupola [fits in 2 different ways], triangular orthobicupola [fits in 2 different ways], elongated triangular orthobicupola [fits in 2 different ways], elongated triangular gyrobicupola [fits in 2 different ways], gyroelongated triangular bicupola [fits in 4 chiral ways], augmented truncated tetrahedron [fits in 2 different ways]
3,4,3,4 (skew1, chiral): gyrobifastigium
3,4,3,4 (skew2, chiral): square gyrobicupola
3,4,3,4 (skew3, chiral): pentagonal gyrobicupola
3,3,4,5 (skew1, chiral): pentagonal gyrocupolarotunda
3,3,4,5 (skew2, chiral): augmented pentagonal prism, biaugmented pentagonal prism [fits in 2 different ways]
3,4,3,5 (skew, chiral): pentagonal orthocupolarotunda, bilunabirotunda, triangular hebesphenorotunda
3,3,4,6 (skew1, chiral): augmented hexagonal prism, parabiaugmented hexagonal prism, metabiaugmented hexagonal prism [fits in 2 different ways], triaugmented hexagonal prism
3,3,4,6 (skew2, chiral): triangular hebesphenorotunda
3,4,3,6 (skew, chiral): augmented truncated tetrahedron
3,4,3,8 (skew, chiral): augmented truncated cube, biaugmented truncated cube
3,4,3,10 (skew, chiral): augmented truncated dodecahedron, parabiaugmented truncated dodecahedron, metabiaugmented truncated dodecahedron [fits in 5 different ways], triaugmented truncated dodecahedron [fits in 5 different ways]
3,3,5,5: pentagonal orthobirotunda
3,3,5,5 (skew1): augmented dodecahedron, parabiaugmented dodecahedron, metabiaugmented docecahedron [fits in 5 different ways, 4 of them chiral], triaugmented dodecahedron [fits in 5 different ways, 4 of them chiral]
3,3,5,5 (skew2): augmented tridiminished icosahedron
3,5,3,5: icosidodecahedron, pentagonal rotunda [fits in 4 different ways], elongated pentagonal rotunda [fits in 4 different ways], gyroelongated pentagonal rotunda [fits in 4 different ways], pentagonal orthocupolarotunda [fits in 4 different ways], pentagonal gyrocupolarotunda [fits in 4 different ways], pentagonal orthobirotunda [fits in 4 different ways], elongated pentagonal orthocupolarotunda [fits in 4 different ways], elongated pentagonal gyrocupolarotunda [fits in 4 different ways], elongated pentagonal orthobirotunda [fits in 4 different ways], elongated pentagonal gyrobirotunda [fits in 4 different ways], gyroelongated pentagonal cupolarotunda [fits in 8 chiral ways], gyroelongated pentagonal birotunda [fits in 8 chiral ways], bilunabirotunda, triangular hebesphenorotunda [fits in 2 different ways]
3,4,4,4: rhombicuboctahedron, square cupola, elongated square cupola [fits in 3 different ways, 2 of them chiral], gyroelongated square cupola, square orthobicupola, square gyrobicupola, elongated square gyrobicupola [fits in 3 different ways, 2 of them chiral], gyroelongated square bicupola [fits in 2 chiral ways], augmented truncated cube, biaugmented truncated cube
3,4,4,4 (skew1, chiral): elongated triangular cupola, elongated triangular orthobicupola, elongated triangular gyrobicupola
3,4,4,4 (skew2, chiral): elongated pentagonal cupola, elongated pentagonal orthobicupola, elongated pentagonal gyrobicupola, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda
3,4,4,5 (chiral): gyrate rhombicosidodecahedron, parabigyrate rhombicosidodecahedron, metabigyrate rhombicosidodecahedron [fits in 5 different ways], trigyrate rhombicosidodecahedron [fits in 5 different ways], paragyrate diminished rhombicosidodecahedron, metagyrate diminished rhombicosidodecahedron [fits in 5 different ways], bigyrate diminished rhombicosidodecahedron [fits in 10 different ways], gyrate bidiminished rhombicosidodecahedron [fits in 5 different ways]
3,4,4,5 (skew, chiral): elongated pentagonal rotunda, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, elongated pentagonal orthobirotunda, elongated pentagonal gyrobirotunda
3,4,5,4: rhombicosidodecahedron, pentagonal cupola, elongated pentagonal cupola, gyroelongated pentagonal cupola, pentagonal orthobicupola, pentagonal gyrobicupola, pentagonal orthocupolarotunda, pentagonal gyrocupolarotunda, elongated pentagonal orthobicupola, elongated pentagonal gyrobicupola, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, gyroelongated pentagonal bicupola [fits in 2 chiral ways], gyroelongated pentagonal cupolarotunda [fits in 2 chiral ways], augmented truncated dodecahedron, parabiaugmented truncated dodecahedron, metabiaugmented truncated dodecahedron [fits in 5 ways, 4 of them chiral], triaugmented truncated dodecahedron [fits in 5 ways, 4 of them chiral], gyrate rhombicosidodecahedron [fits in 10 different ways, 6 of them chiral], parabigyrate rhombicosidodecahedron [fits in 6 different ways, 4 of them chiral], metabigyrate rhombicosidodecahedron [fits in 20 different ways, 16 of them chiral], trigyrate rhombicosidodecahedron [fits in 10 different ways, 6 of them chiral], diminished rhombicosidodecahedron [fits in 9 different ways, 6 of them chiral], paragyrate diminished rhombicosidodecahedron [fits in 7 different ways, 4 of them chiral], metagyrate diminished rhombicosidodecahedron [fits in 35 different ways, 32 of them chiral], bigyrate diminished rhombicosidodecahedron [fits in 25 different ways, 22 of them chiral], parabidiminished rhombicosidodecahedron [fits in 3 different ways, 2 of them chiral], metabidiminished rhombicosidodecahedron [fits in 15 different ways, 12 of them chiral], gyrate bidiminished rhombicosidodecahedron [fits in 20 different ways, 16 of them chiral], trigyrate rhombicosidodecahedron [fits in 5 different ways, 2 of them chiral]

3,3,3,3,3: icosahedron, pentagonal pyramid, elongated pentagonal pyramid, gyroelongated pentagonal pyramid [fits in 6 different ways], pentagonal dipyramid, elongated pentagonal dipyramid, augmented dodecahedron, parabiaugmented dodecahedron, metabiaugmented dodecahedron [fits in 5 different ways], triaugmented dodecahedron [fits in 5 different ways], metabidiminished icosahedron [fits in 5 different ways]
3,3,3,3,3 (skew1): gyroelongated square pyramid, gyroelongated square pyramid
3,3,3,3,3 (skew2): biaugmented triangular prism, triaugmented triangular prism
3,3,3,3,3 (skew3): snub disphenoid
3,3,3,3,3 (skew4): snub square antiprism
3,3,3,3,3 (skew5): sphenocorona, augmented sphenocorona [fits in 2 chiral ways]
3,3,3,3,3 (skew6): sphenocorona, augmented sphenocorona [fits in 2 different ways]
3,3,3,3,3 (skew7, chiral): augmented sphenocorona
3,3,3,3,3 (skew8): sphenomegacorona
3,3,3,3,3 (skew9): sphenomegacorona
3,3,3,3,3 (skew10): hebesphenomegacorona [fits in 2 different ways]
3,3,3,3,3 (skew11): hebesphenomegacorona
3,3,3,3,3 (skew12): disphenocingulum
3,3,3,3,4: snub cube [fits in 2 chiral ways]
3,3,3,3,4 (skew1, chiral): gyroelongated triangular cupola, gyroelongated triangular bicupola [fits in 2 chiral ways]
3,3,3,3,4 (skew2, chiral): gyroelongated square cupola, gyroelongated square bicupola [fits in 2 chiral ways]
3,3,3,3,4 (skew3, chiral): gyroelongated pentagonal cupola, gyroelongated pentagonal bicupola [fits in 2 chiral ways], gyroelongated pentagonal cupolarotunda [fits in 2 chiral ways]
3,3,3,3,4 (skew4): snub square antiprism
3,3,3,3,4 (skew5, chiral): augmented sphenocorona
3,3,3,3,4 (skew6, chiral): sphenomegacorona
3,3,3,3,4 (skew7, chiral): hebesphenomegacorona
3,3,3,3,4 (skew8, chiral): disphenocingulum
3,3,3,3,5: snub dodecahedron [fits in 2 chiral ways]
3,3,3,3,5 (skew, chiral): gyroelongated pentagonal rotunda, gyroelongated pentagonal cupolarotunda [fits in 2 chiral ways], gyroelongated pentagonal birotunda [fits in 2 chiral ways]


[quickfur]

Got another idea -- how about trying to search for duals of CRF polychora?

Now, there are various ways to define duals for nonorbiform figures etc., but in this particular case we could define duals as "external duals", i.e. CRF polychoron is obtained by joining centroids of the cells of the larger "dual". Or they could be defined in another way, as will be needed.

Why do this? Because the main constraining condition on CRF polychora concerns its faces. Well, on dual, those faces would correspond to edges. This means that a dual has main constraining condition on edges -- and so we could use the edge algorithm (or a suitable variant) and completely avoid the problem with nonrigid edges we have when searching for CRF polychora directly.

Very good idea, use the regular polygons as constraint on the dual's edges.

However, the edge algorithm rests on the premise that possible dihedral angles are enumerable. Since these duals would have arbitrary-shaped cells, we no longer know what dihedral angles are possible (there can be a continuum of angles for all we know), so the algorithm wouldn't be feasible anymore. Of course, in practice the dihedral angles won't be a continuum, but then trying to find all possible dihedral angles is equivalent to solving non-rigid edges in the original case, so we haven't really eliminated the problem, just rephrased it.

I don't know the exact constraining condition on the edges, but I think it could be probably found out.

If the original face is a regular polygon, then I think it should translate to all cells having equally-distributed dichoral angles around the dual's corresponding edge. This makes resolving edge angles easy, but then what to do with resolving cell shapes?

[...] EDIT: First obvious hurdle, of course, is that we wouldn't have a database of allowed cell shapes, but maybe that could be bypassed by making a database of possible CRF vertex figures and then dualizing those.

But finding CRF vertex figures would be the same as solving flexible edges, wouldn't it?

OTOH, we might be able to get somewhere by finding all 3D vertex figures, which translates to what kinds of polygons are allowed in the duals of 4D CRFs. This may help narrow down what cell shapes are permissible.

[...] EDIT3: Another idea: even in the original version, the nonrigid edges problem could be lessened by building a database of vertex figures. In this case, vertex figures wouldn't necessarily have co-hyperplanar vertices, but there's nevertheless a big constrain on them: their vertices must form an abstract polytope whose edge lengths must be selected from a small list of shortchords. Combining with the rule that vertex figures must be convex and there must be a point with distance 1 from every vertex, the list of possible abstract polytopes that can serve as vertex figures shouldn't be unreasonably long since there has to be a maximum amount of vertices for them. (I suspect it's 12 and that you can't have more edges at one vertex than in 600-cell, but I might be wrong.)
In the dualized version, all possible vertex figures would be already known.

You're right, there might be something about vertex figures that can help constrain our search. As for maximum edges around a vertex, maybe there's some theorem about solid angles that we can use to prove that 12 is the upper limit for CRFs? Because the solid angles of 3D CRFs are constrained by the fact that vertices must always be surrounded by regular polygons, so intuitively there should be an upper limit to how many things will fit. Then in 4D, the limited number of 3D solid angles should also add up to a limited number of 4D 4-angles, I think.

In 3D this approach doesn't work too well since vertex figures are polygons which can then be skewed in third dimension, and that has many degrees of freedom. I suspect it might be different in 4D case. For example, adding a vertex to a polygon requires to satisfy the shortchord constraints with 2 other vertices, but in 4D there are at least 3 constraints and there might be more. With 3 constraints, the shape would still be flexible, BUT that would imply a 3-face edge. On the other side, a 4- or more-face edge would lead to 4 or more constraints on the vertex in vertex figure, and that would actually fix it in 4D, wouldn't it?
In other words, the strong points of edge and vertex approach cover each other's weaknesses!

But wouldn't a polyhedron with degree >3 polygons be flexible too? For example if you remove the faces of a cube and keep the edge skeleton, then the edge skeleton exhibits many degrees of freedom. You can squeeze it into different kinds of parallelopipeds with equal edge lengths, for example. Or am I missing your idea here?

[Klitzing]

[...] Esp. I got to know that in 1979 a german couple already started that 4D research in a somewhat smaller or stricter sense: they extrapolated the Johnson solid conditions in the sense of regular boundaries, i.e. of regular facets. For 4D this would mean regular cells. Those were listed all in those days, except of the various 600-cell diminishings, for sure.

So the regular-celled CRFs are already known? Is this list published or available online somewhere?
[...]

Have a look at my listing at the bottom of http://bendwavy.org/klitzing/explain/johnson.htm (esp. note the asterix marks), which is based on their corresponding papers:

[1] R. BLIND, Konvexe Polytope mit kongruenten regulären (n-1)-Seiten im R^n (n >= 4). Comment. Math. Helv. 54, 304-308 (1979).
[2] R. BLIND, Konvexe Polytope mit regulären Facetten im R^n (n >= 4). In Contributions to Geometry, J. Tölke and J. M. Wills eds. Birkhäuser, Basel 1979.
[3] G. BLIND und R. BLIND, Die konvexen Polytope im R^4, bei denen alle Facetten reguläre Tetraeder sind. Mh. Math. 89, 87-93 (1980).
[4] G. BLIND und R. BLIND, Über die Symmetriegruppen von regulärseitigen Polytopen, Mh. Math. 108, 103-114 (1989).

--- rk

[Klitzing]

Today I tried generalizing Wendy's bistratic CRF xfo3oox3ooo by the following observations:

(1) The tridiminished icosahedron may be thought of as a modified truncated tetrahedron, in which 3 alternate edges of one of the hexagonal faces is shrunk to a point. As it does so, the other 3 hexagonal faces become pentagons, and that hexagon gets rectified into a triangle.

(2) Wendy's bistratic polychoron may be thought of as a similar process applied to the truncated 5-cell: the hexagonal faces of one of the truncated tetrahedral cells are "shrunk" so that they become rectified into triangles; this in turn shrinks one hexagonal face of each of the remaining 4 truncated tetrahedra, which causes them to turn into tridiminished icosahedra. The first truncated tetrahedron thus get shrunk into an octahedron.

So I wondered what would happen if the same process is applied to the bisected truncated 16-cell. Topologically speaking, the 8 truncated tetrahedra should turn into tridiminished icosahedra, and the bottom truncated octahedron should turn into a cuboctahedron. This gives a tristratic polychoron with 1 octahedron, 8 tridiminished icosahedra, 8 square pyramids, and 1 cuboctahedron. Unfortunately, calculations show that it cannot be made CRF, because either the cuboctahedron will have unequal edge lengths, or the pentagonal faces will fissure into trapeziums and isosceles triangles.

What you're trying here is xfo3oox4ooo&#xt = x3o4o || f3o4o || o3x4o. (It should only contain 6 squippy.) But you're calculations seem to show that in that case the teddis cannot run thru without being bended somehow?

Then I wondered if edge lengths would line up correctly if we start with an icosahedron and attach 20 tridiminished icosahedra around it. This indeed does work; it produces a tristratic CRF which is a kind of diminished 600-cell. It's basically Wendy's tristratic oxoo3ooox5ooxo&#xt with the top vertex and the dodecahedron deleted, and an icosahedron inserted before the icosidodecahedron (I'm not sure exactly what the scaling factor is -- basically it corresponds with the apexes of 12 truncated icosahedron pyramids which may be placed on top of the truncated icosahedron cells). It consists of 1 icosahedron, 20 tridiminished icosahedra, 12 pentagonal pyramids, and 1 icosidodecahedron. I'm not well-versed with Wendy's notation, but I believe it would be something like xfx3ooo5oox.

No, wrong. That one would be xfo3oox5ooo&#xt = x3o5o || f3o5o || o3x5o.
Looks crazy to me if xfo3ooxPooo&#x would have true teddis for P=3, 5; but not for P=4.

Another possible modification to xfo3oox3ooo would be something like xfx3oox3xfo. This should give a CRF with 1 octahedron, 4 tridiminished icosahedra, 6 pentagonal prisms, 4 triangular prisms, 4 triangular cupola, and 1 truncated octahedron. It's sortof like a pseudo-runcinated form of xfo3oox3ooo. I haven't verified this yet, but I believe it should work. I'll try to compute the coordinates tomorrow.

Not quite correct: you would have to start with a co in the middle, attaching 4 teddis and 4 trips to the triangles, and 6 pips onto the squares. Other faces of the first level are supposed to match. Then, on top of the top base of the trips you would attach tricus. All should be encompassed by a tut. - The according tower would be: xfo3oox3xxx&#xt = x3o3x || f3o3x || o3x3x.

--- rk

[Marek14]

But wouldn't a polyhedron with degree >3 polygons be flexible too? For example if you remove the faces of a cube and keep the edge skeleton, then the edge skeleton exhibits many degrees of freedom. You can squeeze it into different kinds of parallelopipeds with equal edge lengths, for example. Or am I missing your idea here?

Sure, if you only consider edge skeleton, but there are many other constraints if you want to use a cube-isomorphic abstract polytope as vertex figure:

1. All vertices must lie on a single radius-1 hypersphere.
2. Each face of the cuboid must be vertex figure of an uniform or Johnson polyhedron.

The first condition by itself rules out skewing into parallelopipeds (but it might be possible to shift some vertices into 4th dimension and skew the whole cube into a shape that's no longer limited to one hyperplane. But the second condition uses the list of possible polygons, including skew ones, and only allows cuboid that has these polygons as faces. In this case, my list contains 45 possible quadrangles, and of course sharing edges and geometrical constraints mean that only a fraction of their possible combinations will be possible.

So, step 1: build possible vertex figure faces. Step 2: build vertex figures out of them. Step 3: Use combination of vertices and edges to search for CRF polychora.

[quickfur]

[...]
What you're trying here is xfo3oox4ooo&#xt = x3o4o || f3o4o || o3x4o. (It should only contain 6 squippy.) But you're calculations seem to show that in that case the teddis cannot run thru without being bended somehow?

You're right, I made a mistake in the cell count. There are only 6 square pyramids.

I'll explain how I did my calculation, in case I made an error:
1) Start with an octahedron <apacs<sqrt(2), 0, 0>, H> (I chose sqrt(2) so that edge length=2, I usually standardize on edge length 2 for most of my calculations)
2) Place the second octahedron as <apacs<phi*sqrt(2), 0, 0>, 0> where phi is the golden ratio - because the chord of the pentagon with edge length 2 is phi.
3) Solve for H using the fact that edge length between the two octahedra must be 2.
4) Place the cuboctahedron as <apacs<sqrt(2), sqrt(2), 0>, G>, again, sqrt(2) is chosen so that we have edge length 2 in the bottom cell.
5) Solve for G.
6) Check the resulting coordinates to see if the pentagon spanning 3 layers lies on a plane. In this case, I was lazy and just ran the coordinates through a convex hull algorithm, and it came out with a polychoron where the pentagonal faces are broken up into trapeziums and isosceles triangles, so there are no teddi's, only triangular frustums and irregular octahedra.

Then I wondered if edge lengths would line up correctly if we start with an icosahedron and attach 20 tridiminished icosahedra around it. This indeed does work; it produces a tristratic CRF which is a kind of diminished 600-cell. It's basically Wendy's tristratic oxoo3ooox5ooxo&#xt with the top vertex and the dodecahedron deleted, and an icosahedron inserted before the icosidodecahedron (I'm not sure exactly what the scaling factor is -- basically it corresponds with the apexes of 12 truncated icosahedron pyramids which may be placed on top of the truncated icosahedron cells). It consists of 1 icosahedron, 20 tridiminished icosahedra, 12 pentagonal pyramids, and 1 icosidodecahedron. I'm not well-versed with Wendy's notation, but I believe it would be something like xfx3ooo5oox.

No, wrong. That one would be xfo3oox5ooo&#xt = x3o5o || f3o5o || o3x5o.
Looks crazy to me if xfo3ooxPooo&#x would have true teddis for P=3, 5; but not for P=4.

You're right, I made another mistake. The icosidodecahedron is o3x5o, not x3o5x.

Unless I made a mistake in my calculations for the octahedron/cuboctahedron case, it's pretty crazy. Apparently there are true teddis for P=3,5 but not for P=4. I just calculated the coordinates for the icosahedron case, and the convex hull algorithm did produce true teddis!

Code: Select all

# Top icosahedron
<0, -1, -phi, phi>
<0, -1, phi, phi>
<0, 1, -phi, phi>
<0, 1, phi, phi>
<-1, -phi, 0, phi>
<-1, phi, 0, phi>
<1, -phi, 0, phi>
<1, phi, 0, phi>
<-phi, 0, -1, phi>
<phi, 0, -1, phi>
<-phi, 0, 1, phi>
<phi, 0, 1, phi>

# Middle icosahedron scaled by golden ratio
<0, -phi, -phi^2, 0>
<0, -phi, phi^2, 0>
<0, phi, -phi^2, 0>
<0, phi, phi^2, 0>
<-phi, -phi^2, 0, 0>
<-phi, phi^2, 0, 0>
<phi, -phi^2, 0, 0>
<phi, phi^2, 0, 0>
<-phi^2, 0, -phi, 0>
<phi^2, 0, -phi, 0>
<-phi^2, 0, phi, 0>
<phi^2, 0, phi, 0>

# Bottom icosidodecahedron
<0, 0, -2*phi, -1>
<0, 0, 2*phi, -1>
<0, -2*phi, 0, -1>
<0, 2*phi, 0, -1>
<-1, -phi^2, -phi, -1>
<-1, -phi^2, phi, -1>
<-1, phi^2, -phi, -1>
<-1, phi^2, phi, -1>
<1, -phi^2, -phi, -1>
<1, -phi^2, phi, -1>
<1, phi^2, -phi, -1>
<1, phi^2, phi, -1>
<-2*phi, 0, 0, -1>
<2*phi, 0, 0, -1>
<-phi, -1, -phi^2, -1>
<-phi, -1, phi^2, -1>
<-phi, 1, -phi^2, -1>
<-phi, 1, phi^2, -1>
<phi, -1, -phi^2, -1>
<phi, -1, phi^2, -1>
<phi, 1, -phi^2, -1>
<phi, 1, phi^2, -1>
<-phi^2, -phi, -1, -1>
<-phi^2, -phi, 1, -1>
<-phi^2, phi, -1, -1>
<-phi^2, phi, 1, -1>
<phi^2, -phi, -1, -1>
<phi^2, -phi, 1, -1>
<phi^2, phi, -1, -1>
<phi^2, phi, 1, -1>


I'll post images later, this is another of those 600-cell derivatives that has too many edges, so I have to play with it a little bit to find the best viewpoint.

Another possible modification to xfo3oox3ooo would be something like xfx3oox3xfo. This should give a CRF with 1 octahedron, 4 tridiminished icosahedra, 6 pentagonal prisms, 4 triangular prisms, 4 triangular cupola, and 1 truncated octahedron. It's sortof like a pseudo-runcinated form of xfo3oox3ooo. I haven't verified this yet, but I believe it should work. I'll try to compute the coordinates tomorrow.

Not quite correct: you would have to start with a co in the middle, attaching 4 teddis and 4 trips to the triangles, and 6 pips onto the squares. Other faces of the first level are supposed to match. Then, on top of the top base of the trips you would attach tricus. All should be encompassed by a tut. - The according tower would be: xfo3oox3xxx&#xt = x3o3x || f3o3x || o3x3x.
[...]

You're right, I confused o3x4o with x3o4x. I really have to be more careful to check what I wrote before posting.

[quickfur]

[...] That one would be xfo3oox5ooo&#xt = x3o5o || f3o5o || o3x5o.
Looks crazy to me if xfo3ooxPooo&#x would have true teddis for P=3, 5; but not for P=4.

Here's an image of this CRF:



To make the cells easier to pick out, I highlighted the icosahedron (yellow) and one of the 20 teddi's (brown). The icosidodecahedron is a bit hard to see, so I colored its edges red. Its pentagonal faces are joined to pentagonal pyramids, which appear almost planar in this projection because their angle with the line-of-sight is close to 90°. They overhang the icosidodecahedron a little, so from this view angle (looking up from the bottom at the icosidodecahedron) they appear to protrude a little from the image of the former.

Here's a side-view, to show its stool-like shape:



I turned on visibility clipping for this second image, because otherwise there will be too many lacing edges that obscures the image. I colored the pentagonal pyramids green. The detached green triangles are where the teddi's join with pentagonal pyramids on the far side which have been culled. I also shifted the viewpoint upwards a little, so the top icosahedral cell is visible (appearing as a not-quite-flat icosahedron). The icosidodecahedron is not visible, but half of its outline is seen in the red edges skirting the bottom of the projection.