## D4.10 and D4.11

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

### D4.10 and D4.11

student5 wrote:[...]
I haven't found anything, guess February is over or something, but I did find out how incredibly lucky CJ4.8.2 was
[...]

February is certainly not over! I've found another CRF with tetrahedral symmetry, sporting 8 J92's and 12 J91's. You guys are gonna love this one... It has two identical halves, which are joined in dual orientation to each other. Here is the lace tower:

Code: Select all
`       x3o3o   // tetrahedron       f3o3x   // 4 tetrahedra touching vertices of previous one       o3x3f   // midpoints of J91's       f3x3x   // square faces of J91's       x3o3F   // octahedra above J63's       x3F3o   // opposite edge of J91's       f3x3f   // f3x vertices of 8 J92's (this is the midpoint)       o3F3x       F3o3x       x3x3f       f3x3o       x3o3f       o3o3x`

As you can see, there are two antipodal tetrahedra. EDIT 3: Also, there are no lacing edges between f3x3x and x3o3F, so that part of the tower may be better written as:
Code: Select all
`        ...        o3x3f       /  f3x3x   x3o3F    |         |    x3F3o      | f3x3f |      o3F3x   |      |    F3o3x     x3x3f   /        f3x3o       ...`

Well let's look at a render to see what's going on here:

Here's a projection centered on one of the antipodal tetrahedra. Around this central tetrahedron (yellow; you can faintly see it buried deep in the middle where the J91's meet) are 6 J91's touching its edges (brown), with 4 J63's (tridiminished icosahedra) and 4 other tetrahedra (green -- but not the ones on the outside) filling in the gaps in between. The 4 triangular faces of each J63 opposite the antipodal tetrahedra are joined to a formation of 4 octahedra, 3 surrounding a central one (central one shown in cyan). The central octahedron also has 3 pentagonal antiprisms surrounding it, lying on top of the J91's. The opposite faces of the 4 tetrahedra surrounding the antipodal tetrahedra are joined to triangular cupola, which in turn are joined to 4 J92's. The central octahedra with the neighbouring pentagonal antiprisms are joined to another 4 J92's on the far side (which are a mirror image, in dual orientation, of the near side 4 J92's).

Another interesting thing: besides the tetrahedra surrounding these antipodal tetrahedra, there are 6 other tetrahedra that connect the J91's on the near side to the J91's on the far side, and together, these 8 tetrahedra lie at the vertices of a 16-cell!

Now, the J92 cells may be a bit hard to see, so here's another render highlighting them:

Now you can see that square pyramids link the J92's to the cyan octahedra. Notice that the projection envelope has cubic symmetry? (I'm using parallel projection.) So the cyan octahedra are joined to another 4 J92's on the far side, in the dual positions of where they are on the near side. Similarly, the J92's on this side are joined to 4 octahedra on the far side -- the far side counterparts of the cyan octahedra.

Now, let's look at the side-view of this CRF, just to show you how closely it approximates 16-cell symmetry.

The top and bottom squares are the images of the aforementioned antipodal tetrahedra (the x3o3o and o3o3x in the lace tower). The brown cells are J91's, which, as you can see, link these tetrahedra to the 8 equatorial tetrahedra. But the equatorial tetrahedra aren't linked by J91's, they are linked by pairs of J92's (sorry, the blue coloring is a bit ambiguous, each blue patch is actually a pair of J92's, not a single one). If you look carefully at the nearest part of the projection (in 3D, not in 4D) to the upper right of the green square (image of an equatorial tetrahedron), between the brown J91 on top and the blue J92 to the right, you can see the outline of a pentagonal antiprism in an oblique orientation.

So there you have it. A CRF with tetrahedral symmetry, having J91 and J92 cells, that almost has 16-cell symmetry.

February is far from over, folks. CRFebruary, that is.

EDIT: uploaded the .def and .off files to D4.10.
quickfur
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### Re: Johnsonian Polytopes

Cells:
16 tetrahedra (2 + 6 + 8)
24 square pyramids
32 octahedra (24 + 8)
8 tridiminished icosahedra
8 triangular cupolas
24 pentagonal antiprisms
12 J91's
8 J92's

Faces:
464 triangles, 48 squares, 72 pentagons, 8 hexagons

636 edges
176 vertices

This is the first bilbrothawroid that includes pentagonal antiprisms, right?
Marek14
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### Re: Johnsonian Polytopes

quickfur wrote:I'm not against renumbering, though, but if we're going to do that, we'll need to fix the references to those numbers in previous posts in this thread, otherwise we will lose referential consistency for when we come back to read this thread in the future.

Let's leave 4.9.x alone then, but for future reference, .0 is weird and it really doesn't matter that it "doesn't count", since it's just an arbitrary index for reference (hence why we made them D numbers instead of CJ numbers).

student91 wrote:I fully agree on D4.4 not being a bilbirothawroid. I think we should make it CJ4.4 again. If it weren't a crown jewel, we wouldn't be consistent:
the wiki wrote:Crown jewels are a catch-all term for unusual CRF polytopes with unique structures that cannot be obtained from the uniform polytopes or other simpler CRFs by simple "cut-and-paste" operations.
D4.4 clearly can't be obtained by simple cut-n-paste operations. For the same reason, I think the D4.3, and the D4.8's may be reconsidered as well.

Whether we classify D4.4 as a bilbirothawroid or not is up to you guys, who understand how these CRFs come about far more than I do I just came up with the name so we could refer to this class of similar figures easily. However I would not go renaming it back to CJ4.4, as there is nothing wrong with any figure having a D number, it is just an arbitrary index for reference. Furthermore, we may redefine "crown jewels" to be something more strict - the 3D crown jewels have very little in common between them, so I feel we should not have any "families" of shapes count as crown jewels as if they did, it would not highlight the truly unique ones, which is what the term "crown jewel" was intended for. Exactly how we go about redefining it though is up in the air, and may well have to wait until we have come along much further with the discovery project so that we can come up with the most useful criteria.

Also remember we have not yet declared bilbirothawroids as non-crown-jewels, indeed on the wiki they are still categorised as such. So no need to worry about your favorite BT polychoron not being a crown jewel

Keiji

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Location: Torquay, England

### Re: Johnsonian Polytopes

quickfur wrote:... February is certainly not over! I've found another CRF with tetrahedral symmetry, ...
... February is far from over, folks. CRFebruary, that is. ...

Hey guys, you're way too fast!

I just came around to finish the incidence matrices of those other 4 axially tetrahedral ones. (Thus stay tuned for my next webpage update. )

And still I have to consider my own ideas: "Idea 3" was disproven, "idea 2" was elaborated. One of those 2 was definitely non-convex, as including stips for cells. So far I suppose that the other one so would be convex. - Any support for that assumption? - And "idea 1" still remains to be evaluated by my side. (quickfur here already gave some Input with respect to some dihedral angle restrictions. But so far I had no sparetime to look more closely into these...)

@quickfur: Congratulations - both for finding & for that appealing rendering!

--- rk
Klitzing
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### Re: Johnsonian Polytopes

quickfur wrote:[...]
Code: Select all
`       x3o3o   // tetrahedron       f3o3x   // 4 tetrahedra touching vertices of previous one       o3x3f   // midpoints of J91's       f3x3x   // square faces of J91's       x3o3F   // octahedra above J63's       x3F3o   // opposite edge of J91's       f3x3f   // f3x vertices of 8 J92's (this is the midpoint)       o3F3x       F3o3x       x3x3f       f3x3o       x3o3f       o3o3x`

As you can see, there are two antipodal tetrahedra. EDIT 3: Also, there are no lacing edges between f3x3x and x3o3F, so that part of the tower may be better written as:
Code: Select all
`        ...        o3x3f       /  f3x3x   x3o3F    |         |    x3F3o      | f3x3f |      o3F3x   |      |    F3o3x     x3x3f   /        f3x3o       ...`

To be honest, I don't really understand that middle part, I do understand the diagram (I use similar things as well), but the structure itself is just so awesome I can't comprehend it. I mean, Pentagonal antiprisms?? although I can spot them, I don't understand how and why these occur from that lace-tower. It's just wow
Well let's look at a render to see what's going on here:

Here's a projection centered on one of the antipodal tetrahedra. Around this central tetrahedron (yellow; you can faintly see it buried deep in the middle where the J91's meet) are 6 J91's touching its edges (brown), with 4 J63's (tridiminished icosahedra) and 4 other tetrahedra (green -- but not the ones on the outside) filling in the gaps in between. The 4 triangular faces of each J63 opposite the antipodal tetrahedra are joined to a formation of 4 octahedra, 3 surrounding a central one (central one shown in cyan). The central octahedron also has 3 pentagonal antiprisms surrounding it, lying on top of the J91's. The opposite faces of the 4 tetrahedra surrounding the antipodal tetrahedra are joined to triangular cupola, which in turn are joined to 4 J92's. The central octahedra with the neighbouring pentagonal antiprisms are joined to another 4 J92's on the far side (which are a mirror image, in dual orientation, of the near side 4 J92's).

Another interesting thing: besides the tetrahedra surrounding these antipodal tetrahedra, there are 6 other tetrahedra that connect the J91's on the near side to the J91's on the far side, and together, these 8 tetrahedra lie at the vertices of a 16-cell!

Now, the J92 cells may be a bit hard to see, so here's another render highlighting them:

Now you can see that square pyramids link the J92's to the cyan octahedra. Notice that the projection envelope has cubic symmetry? (I'm using parallel projection.) So the cyan octahedra are joined to another 4 J92's on the far side, in the dual positions of where they are on the near side. Similarly, the J92's on this side are joined to 4 octahedra on the far side -- the far side counterparts of the cyan octahedra.

Now, let's look at the side-view of this CRF, just to show you how closely it approximates 16-cell symmetry.

The top and bottom squares are the images of the aforementioned antipodal tetrahedra (the x3o3o and o3o3x in the lace tower). The brown cells are J91's, which, as you can see, link these tetrahedra to the 8 equatorial tetrahedra. But the equatorial tetrahedra aren't linked by J91's, they are linked by pairs of J92's (sorry, the blue coloring is a bit ambiguous, each blue patch is actually a pair of J92's, not a single one). If you look carefully at the nearest part of the projection (in 3D, not in 4D) to the upper right of the green square (image of an equatorial tetrahedron), between the brown J91 on top and the blue J92 to the right, you can see the outline of a pentagonal antiprism in an oblique orientation.

So there you have it. A CRF with tetrahedral symmetry, having J91 and J92 cells, that almost has 16-cell symmetry.

February is far from over, folks. CRFebruary, that is.

EDIT: uploaded the .def and .off files to D4.10.

because the bilbiro's are oriented x2o||f2x||o2f||f2x||x2o, I think we can adjust the bilbiro's into x2o||f2x||o2f||x2o, i.e. a metabidiminished ike. The lace-tower would then become: x3x3o||(awesomeness)|| o3x3x
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Johnsonian Polytopes

Marek14 wrote:Cells:
[...]
Faces:
[...]
636 edges
176 vertices

Please put this information on the wiki page.

This is the first bilbrothawroid that includes pentagonal antiprisms, right?

Possibly. Though I should point on that in the course of constructing this CRF, I initially explored in a different direction that didn't have pentagonal antiprisms where you see them now. Instead, I had metabidiminished icosahedra. Basically, that region of the polychoron has some kind of subset of an icosahedron's vertices, and there are multiple CRF solutions that give you a different diminishing of it. The pentagonal antiprism is just a parabidiminished icosahedron, in this case.

Keiji wrote:
quickfur wrote:I'm not against renumbering, though, but if we're going to do that, we'll need to fix the references to those numbers in previous posts in this thread, otherwise we will lose referential consistency for when we come back to read this thread in the future.

Let's leave 4.9.x alone then, but for future reference, .0 is weird and it really doesn't matter that it "doesn't count", since it's just an arbitrary index for reference (hence why we made them D numbers instead of CJ numbers).

Agreed. We chose the .0 suffix before we switched over to the D numbers, back when we were still trying to (prematurely) classify them as soon as they're discovered, so it's kinda inconsistent with what we decided to do with the D numbers later.

[...]
Whether we classify D4.4 as a bilbirothawroid or not is up to you guys, who understand how these CRFs come about far more than I do I just came up with the name so we could refer to this class of similar figures easily. However I would not go renaming it back to CJ4.4, as there is nothing wrong with any figure having a D number, it is just an arbitrary index for reference.

Yeah, rather than waste time fighting over how to assign numbers to these things, let's just accept the D numbers as an arbitrary unique identifier that we can use to unambiguously refer to these polytopes. The final classification may have nothing to do with the assigned D numbers at all -- we still need much more data before we can spot more trends and identify what are the correct categories!

Furthermore, we may redefine "crown jewels" to be something more strict - the 3D crown jewels have very little in common between them, so I feel we should not have any "families" of shapes count as crown jewels as if they did, it would not highlight the truly unique ones, which is what the term "crown jewel" was intended for. Exactly how we go about redefining it though is up in the air, and may well have to wait until we have come along much further with the discovery project so that we can come up with the most useful criteria.

Even among the 3D crown jewels, there are some things in common: sphenocorona and augmented sphenocorona, for example, or, for that matter, J91 and J92 themselves, which are both related to the icosidodecahedron, and also among themselves (you can make a J92 by taking the convex hull of 3 mutually intersecting J91's). So I wouldn't dismiss this too quickly.

But I agree that in 4D, it's turning out that there are many more "crown jewels", or crown-jewel-like CRFs, than we initially realized, so it seems that where we had just individual 3D CRFs in the crown jewel category, in 4D we may very well have a crown jewel super-category that contains families of crown jewels. In which case we may question the label "crown jewel", since they are no longer so rare, even if their construction is by no means obvious! (You may have noticed my remarks in the wiki's update history that crown jewels are depreciating in value due to their unexpected commonness. )

Also remember we have not yet declared bilbirothawroids as non-crown-jewels, indeed on the wiki they are still categorised as such. So no need to worry about your favorite BT polychoron not being a crown jewel

Yes, I noticed that.

Perhaps the "soft" crown jewel vs. "hard" crown jewel distinction will turn out to be very useful, since we're finding so many "soft" crown jewels (CVP=2)! In 3D, J91 and J92 are soft crown jewels that are related to the icosahedral CRFs, including the diminished icosahedra (which includes the pentagonal antiprism) and the icosahedral uniforms. It just so happens that besides J91 and J92, all of the modified icosahedral polyhedra are simple diminishings. But in 4D, not only are there far vaster numbers of 120-cell family diminishings; there are also correspondingly vaster numbers of modified diminishings ala J91 and J92, yielding all sorts of bilbirothawroids, and also some of these modification appear to also apply to things that are only tenuously related to the 120-cell family polytopes -- there may be some connection with the ursachora here, where you have the icosahedral J63 appearing in octahedral symmetry, for example, which has no CRF analogue in 3D (although, having said that, the direct 3D analogue is simply the square ursahedron, which has isosceles triangles having a √2-edge: I've recently been thinking that if we relaxed the 3D definition of a Johnsonian polyhedron to allow triangles with a √2-edge and phi-edge, then we'd find many more 3D analogues of some of the 4D CRFs that we have been finding -- of course, they would no longer be CRF per definition, but they will at least allow us to trace more common patterns from 3D where they would otherwise appear to have popped up out of nowhere in 4D).
quickfur
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### Re: Johnsonian Polytopes

Klitzing wrote:
quickfur wrote:[...]
... February is far from over, folks. CRFebruary, that is. ...

Hey guys, you're way too fast!

I just came around to finish the incidence matrices of those other 4 axially tetrahedral ones. (Thus stay tuned for my next webpage update. )

Yes, CRFebruary isn't over yet. They're still turning up in large numbers at our door.

And still I have to consider my own ideas: "Idea 3" was disproven, "idea 2" was elaborated. One of those 2 was definitely non-convex, as including stips for cells. So far I suppose that the other one so would be convex. - Any support for that assumption? - And "idea 1" still remains to be evaluated by my side. (quickfur here already gave some Input with respect to some dihedral angle restrictions. But so far I had no sparetime to look more closely into these...)

Which one would be convex again? I only remember two convex findings from your ideas, one was the same as an n,12-duoprism, and the other was the same as x4o3o3x. Did I miss another possible convex figure in there?

@quickfur: Congratulations - both for finding & for that appealing rendering!

Thanks!! But just in case you were holding your breath: there's more coming! I'm currently in the process of constructing something that I'm reasonably certain will be CRF, and it will have something to do with the 24-cell. (Oh yes, CRFebruary has only just begun! )

Not to mention, that this CRF is only one of two possibilities that I've found just yesterday -- there is another way to close it up around the pentagonal antiprisms, that would have produced something not quite so symmetric (that's why I didn't fully explore that option -- yet), but nonetheless CRF. I decided to go with the more symmetric solution, because it's becoming quite obvious to me that there is something special going on in 4D with the golden ratio, that allows a huge number of polytopes to close up in a CRF way, and at almost every step you can find many possible choices -- so the more symmetric solutions are preferable as more recognizable "landmarks" in what is turning out to be a vast landscapes of phi-related CRFs.

This has led me to wonder, is there some kind of 4D tiling involving the golden ratio that I'm not aware of? Some kind of grid-like structure with large numbers of unit edges? Because while working with the coordinates of this latest CRF, that's the feeling that I'm getting: at almost every step, the numbers just keep working out "magically", and I can't help feeling that this is just the expression of some underlying phi-based grid of some sort, and that once your vertices get on this grid, you just have to keep following the grid and all your edges will "magically" turn out to be unit length. It feels as though these CRFs are just convex sections cut out from the "cloth" of this underlying grid. If indeed such a grid/tiling exists, it would allow us to classify these CRFs in a far more generic way, and perhaps even allow us to calculate an upper bound to their number -- because I'm almost certain now that there will be a very large number of them, as they allow so many variations at every step! Getting at their underlying phi-based grid, if such exists, would allow us to analyze them collectively, rather than just individually, and discover global properties that apply to all of them.
quickfur
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### Temp1

Alright, so my guess was correct: there is a CRF featuring 24 J91's with demitesseractic symmetry. It's actually just a slight modification from the previous CRF (D4.10). The lace tower is almost the same:
Code: Select all
`x3o3of3o3xo3x3ff3x3xx3F3ox3o3FF3x3oo3x3FF3o3xo3F3xx3x3ff3x3ox3o3fo3o3x`

Basically, the only difference is that the middle f3x3f got deleted, and replaced with F3x3o || o3x3F. This deletes the J92 cells and introduces 12 equatorial J91's. With this change, the asymmetry caused by the J92's is removed, and we now have full-blown demitesseractic symmetry. Let's take a look at this beauty:

The green patches are the 8 axial tetrahedra, lying along the 4 coordinate axes of 4D space. The red, orange, and yellow show 6 of the J91's around the nearest tetrahedron to the 4D viewpoint. The cyan cells are octahedra surrounded by 4 other octahedra: these fill the space where previously the J92's were. These 8 octahedra lie on the vertices of an alternated tesseract. On the vertices of the other alternated tesseract, are 8 cuboctahedra, here outlined in blue edges. These fill the space of where previously there were octahedron + 3 octahedra combos interfacing with J92 cells from the far side -- since the J92 cells are no more, cuboctahedra now fill that space. So here you can see demitesseractic symmetry very clearly.

The far side of the polytope is the mirror image of the above arrangement. The two sides are connected via the 12 J91's on the equator, which are shown next:

I omitted the non-equatorial cells for clarity's sake. As you can see, this is very clearly a demitesseractic symmetry.

So what we have here is almost a uniform arrangement, with each axial tetrahedron surrounded by 4 tridiminished icosahedra (at the faces), 4 J91's (at the edges), and 4 non-axial tetrahedra (at the vertices). Above the tridiminished icosahedra you have an arrangement of a central octahedron surrounded by 4 other octahedra, and above the non-axial tetrahedra are cuboctahedra. This pattern is symmetric across all of the axial tetrahedra, thus completing the polytope in a CRF manner. Beautiful, ain't it?

Edit: assigned D4.11 to this CRF (I decided to just go with a flat numbering, since we don't fully know yet how these things relate to each other, so it makes little sense to attempt to classify them with suffixes at this point in time). The Stella4D and polyview models can be found there.
quickfur
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### D4.10 temp

In this post quickfur wrote:
... February is certainly not over! I've found another CRF with tetrahedral symmetry, sporting 8 J92's and 12 J91's. You guys are gonna love this one... It has two identical halves, which are joined in dual orientation to each other. Here is the lace tower:

Code: Select all
`       x3o3o   // tetrahedron       f3o3x   // 4 tetrahedra touching vertices of previous one       o3x3f   // midpoints of J91's       f3x3x   // square faces of J91's       x3o3F   // octahedra above J63's       x3F3o   // opposite edge of J91's       f3x3f   // f3x vertices of 8 J92's (this is the midpoint)       o3F3x       F3o3x       x3x3f       f3x3o       x3o3f       o3o3x`

As you can see, there are two antipodal tetrahedra. EDIT 3: Also, there are no lacing edges between f3x3x and x3o3F, so that part of the tower may be better written as:
Code: Select all
`        ...        o3x3f       /  f3x3x   x3o3F    |         |    x3F3o      | f3x3f |      o3F3x   |      |    F3o3x     x3x3f   /        f3x3o       ...`

Well let's look at a render to see what's going on here:

...

Then, in that post he wrote:
... Alright, so my guess was correct: there is a CRF featuring 24 J91's with demitesseractic symmetry. It's actually just a slight modification from the previous CRF (D4.10). The lace tower is almost the same:
Code: Select all
`x3o3of3o3xo3x3ff3x3xx3F3ox3o3FF3x3oo3x3FF3o3xo3F3xx3x3ff3x3ox3o3fo3o3x`

Basically, the only difference is that the middle f3x3f got deleted, and replaced with F3x3o || o3x3F. This deletes the J92 cells and introduces 12 equatorial J91's. With this change, the asymmetry caused by the J92's is removed, and we now have full-blown demitesseractic symmetry. Let's take a look at this beauty:

...

Now, comparing those in a bit more detail, it occurs to me that there is a further switch of 2 layers - at least in your provided tower descriptions. - Which then is the correct one?
Code: Select all
`Older: Newer:-------------x3o3o  x3o3of3o3x  f3o3xo3x3f  o3x3ff3x3x  f3x3xx3o3F  x3F3o  <-- additional Switch, by purpose?x3F3o  x3o3F  <--   "f3x3f  F3x3o  <-- equatorial change, by purpose!o3F3x  o3x3F  <--   "F3o3x  F3o3xx3x3f  o3F3xf3x3o  x3x3fx3o3f  f3x3oo3o3x  x3o3f       o3o3x`

--- rk
Klitzing
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
Now, comparing those in a bit more detail, it occurs to me that there is a further switch of 2 layers - at least in your provided tower descriptions. - Which then is the correct one?
Code: Select all
`Older: Newer:-------------x3o3o  x3o3of3o3x  f3o3xo3x3f  o3x3ff3x3x  f3x3xx3o3F  x3F3o  <-- additional Switch, by purpose?x3F3o  x3o3F  <--   "f3x3f  F3x3o  <-- equatorial change, by purpose!o3F3x  o3x3F  <--   "F3o3x  F3o3xx3x3f  o3F3xf3x3o  x3x3fx3o3f  f3x3oo3o3x  x3o3f       o3o3x`

--- rk

Oops!! I think I have a typo in the D4.10 (the first column): the x3o3F should not be there! It should be just x3o3o || f3o3x || o3x3f || f3x3x || x3F3o || f3x3f (equator) || ... . Sorry for the confusion!!

OK, looks like I made multiple errors in my original post, but I have to run now, so I'll fix them later. Sorry!! The lace tower for D4.11 (second column) seems to be correct, though. (I hope! )

EDIT: Actually, I just checked again, and it seems that I confused myself because in my algebraic coordinates source file some layers are out-of-order w.r.t. their height from the top (because some of the layers fill up unrelated regions of the polytope, so they don't lace to other vertex layers with nearby heights).

So, for clarity, here's the lace tower for D4.10 (with height values inserted for clarity):
Code: Select all
`# D4.10x3o3o (h = -(2*phi+3)/√2 = -4.409565...)f3o3x (h = -(phi+3)/√2 = -3.265443...)o3x3f (h = -(phi+2)/√2 = -2.558336...)f3x3x (h = -phi^2/√2 = -1.851229...)x3o3F (h = -phi/√2 = -1.144122...)x3F3o (h = -1/√2 = -0.707106...)f3x3f (h = 0 -- equator)o3F3x (h = 1/√2 = 0.707106...)F3o3x (h = phi/√2 = 1.144122...)x3x3f (h = phi^2/√2 = 1.851229...)f3x3o (h = (phi+2)/√2 = 2.558336...)x3o3f (h = (phi+3)/√2 = 3.265443...)o3o3x (h = (2*phi+3)/√2 = 4.409565...)`

For D4.11, here is the lace tower with height values:
Code: Select all
`# D4.11x3o3o (h = -(2*phi+3)/√2)f3o3x (h = -(phi+3)/√2)o3x3f (h = -(phi+2)/√2)f3x3x (h = -phi^2/√2)x3o3F (h = -phi/√2)x3F3o (h = -1/√2)F3x3o (h = -1/(phi*√2) = -0.437016...)# (equator has no vertices)o3x3F (h = 1/(phi*√2) = 0.437016...)o3F3x (h = 1/√2)F3o3x (h = phi/√2)x3x3f (h = phi^2/√2)f3x3o (h = (phi+2)/√2)x3o3f (h = (phi+3)/√2)o3o3x (h = (2*phi+3)/√2)`

Now it should be clear that the only difference is in the deletion of the h=0 layer from D4.10 and the addition of the h=±1/(phi*√2) layers in D4.11. I apologize for the confusion earlier, caused by my inaccurate description.

P.S. by "height" I mean displacement from the equator, IOW the value of the 4th coordinate. Just in case it's not clear.
quickfur
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### D4.10 temp

quickfur wrote:

Did you already make up a good name for that one?
Or should we ask HedronDude to set up one?

Btw., here is the incmats of that figure (in its full hexadecachoral symmetry):
Code: Select all
`32  *  * |  3  3  0  0   0  0 |  3  6  3  0  0  0  0  0  0  0 | 1  3  3  1 0  0 0 * 96  * |  0  1  2  2   2  0 |  0  2  2  1  1  2  1  2  0  0 | 0  1  2  1 1  1 0 *  * 48 |  0  0  0  0   4  4 |  0  2  0  0  0  0  1  2  2  2 | 0  2  1  0 0  2 1---------+--------------------+-------------------------------+------------------ 2  0  0 | 48  *  *  *   *  * |  2  2  0  0  0  0  0  0  0  0 | 1  2  1  0 0  0 0 1  1  0 |  * 96  *  *   *  * |  0  2  2  0  0  0  0  0  0  0 | 0  1  2  1 0  0 0 0  2  0 |  *  * 96  *   *  * |  0  0  1  1  0  1  0  0  0  0 | 0  0  1  1 1  0 0 0  2  0 |  *  *  * 96   *  * |  0  0  0  0  1  1  0  1  0  0 | 0  0  1  0 1  1 0 0  1  1 |  *  *  *  * 192  * |  0  1  0  0  0  0  1  1  0  0 | 0  1  1  0 0  1 0 0  0  2 |  *  *  *  *   * 96 |  0  0  0  0  0  0  1  0  1  1 | 0  1  0  0 0  1 1---------+--------------------+-------------------------------+------------------ 3  0  0 |  3  0  0  0   0  0 | 32  *  *  *  *  *  *  *  *  * | 1  1  0  0 0  0 0 2  2  1 |  1  2  0  0   2  0 |  * 96  *  *  *  *  *  *  *  * | 0  1  1  0 0  0 0 1  2  0 |  0  2  1  0   0  0 |  *  * 96  *  *  *  *  *  *  * | 0  0  1  1 0  0 0 0  3  0 |  0  0  3  0   0  0 |  *  *  * 32  *  *  *  *  *  * | 0  0  0  1 1  0 0 0  3  0 |  0  0  0  3   0  0 |  *  *  *  * 32  *  *  *  *  * | 0  0  0  0 1  1 0 0  4  0 |  0  0  2  2   0  0 |  *  *  *  *  * 48  *  *  *  * | 0  0  1  0 1  0 0 0  1  2 |  0  0  0  0   2  1 |  *  *  *  *  *  * 96  *  *  * | 0  1  0  0 0  1 0 0  2  1 |  0  0  0  1   2  0 |  *  *  *  *  *  *  * 96  *  * | 0  0  1  0 0  1 0 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  * 32  * | 0  1  0  0 0  0 1 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  *  * 32 | 0  0  0  0 0  1 1---------+--------------------+-------------------------------+------------------ 4  0  0 |  6  0  0  0   0  0 |  4  0  0  0  0  0  0  0  0  0 | 8  *  *  * *  * *  tet 3  3  3 |  3  3  0  0   6  3 |  1  3  0  0  0  0  3  0  1  0 | * 32  *  * *  * *  teddi 4  8  2 |  2  8  4  4   8  0 |  0  4  4  0  0  2  0  4  0  0 | *  * 24  * *  * *  bilbiro 1  3  0 |  0  3  3  0   0  0 |  0  0  3  1  0  0  0  0  0  0 | *  *  * 32 *  * *  tet 0 12  0 |  0  0 12 12   0  0 |  0  0  0  4  4  6  0  0  0  0 | *  *  *  * 8  * *  co 0  3  3 |  0  0  0  3   6  3 |  0  0  0  0  1  0  3  3  0  1 | *  *  *  * * 32 *  oct 0  0  6 |  0  0  0  0   0 12 |  0  0  0  0  0  0  0  0  4  4 | *  *  *  * *  * 8  oct`

--- rk
Klitzing
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### Re: Johnsonian Polytopes

Klitzing wrote:
quickfur wrote:

Did you already make up a good name for that one?
Or should we ask HedronDude to set up one?

No, we don't have a name for it yet. It's currently just called D4.11 (you can see the current assignment of D-numbers on the Discovery index page).

Btw., here is the incmats of that figure (in its full hexadecachoral symmetry):
Code: Select all
`32  *  * |  3  3  0  0   0  0 |  3  6  3  0  0  0  0  0  0  0 | 1  3  3  1 0  0 0 * 96  * |  0  1  2  2   2  0 |  0  2  2  1  1  2  1  2  0  0 | 0  1  2  1 1  1 0 *  * 48 |  0  0  0  0   4  4 |  0  2  0  0  0  0  1  2  2  2 | 0  2  1  0 0  2 1---------+--------------------+-------------------------------+------------------ 2  0  0 | 48  *  *  *   *  * |  2  2  0  0  0  0  0  0  0  0 | 1  2  1  0 0  0 0 1  1  0 |  * 96  *  *   *  * |  0  2  2  0  0  0  0  0  0  0 | 0  1  2  1 0  0 0 0  2  0 |  *  * 96  *   *  * |  0  0  1  1  0  1  0  0  0  0 | 0  0  1  1 1  0 0 0  2  0 |  *  *  * 96   *  * |  0  0  0  0  1  1  0  1  0  0 | 0  0  1  0 1  1 0 0  1  1 |  *  *  *  * 192  * |  0  1  0  0  0  0  1  1  0  0 | 0  1  1  0 0  1 0 0  0  2 |  *  *  *  *   * 96 |  0  0  0  0  0  0  1  0  1  1 | 0  1  0  0 0  1 1---------+--------------------+-------------------------------+------------------ 3  0  0 |  3  0  0  0   0  0 | 32  *  *  *  *  *  *  *  *  * | 1  1  0  0 0  0 0 2  2  1 |  1  2  0  0   2  0 |  * 96  *  *  *  *  *  *  *  * | 0  1  1  0 0  0 0 1  2  0 |  0  2  1  0   0  0 |  *  * 96  *  *  *  *  *  *  * | 0  0  1  1 0  0 0 0  3  0 |  0  0  3  0   0  0 |  *  *  * 32  *  *  *  *  *  * | 0  0  0  1 1  0 0 0  3  0 |  0  0  0  3   0  0 |  *  *  *  * 32  *  *  *  *  * | 0  0  0  0 1  1 0 0  4  0 |  0  0  2  2   0  0 |  *  *  *  *  * 48  *  *  *  * | 0  0  1  0 1  0 0 0  1  2 |  0  0  0  0   2  1 |  *  *  *  *  *  * 96  *  *  * | 0  1  0  0 0  1 0 0  2  1 |  0  0  0  1   2  0 |  *  *  *  *  *  *  * 96  *  * | 0  0  1  0 0  1 0 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  * 32  * | 0  1  0  0 0  0 1 0  0  3 |  0  0  0  0   0  3 |  *  *  *  *  *  *  *  *  * 32 | 0  0  0  0 0  1 1---------+--------------------+-------------------------------+------------------ 4  0  0 |  6  0  0  0   0  0 |  4  0  0  0  0  0  0  0  0  0 | 8  *  *  * *  * *  tet 3  3  3 |  3  3  0  0   6  3 |  1  3  0  0  0  0  3  0  1  0 | * 32  *  * *  * *  teddi 4  8  2 |  2  8  4  4   8  0 |  0  4  4  0  0  2  0  4  0  0 | *  * 24  * *  * *  bilbiro 1  3  0 |  0  3  3  0   0  0 |  0  0  3  1  0  0  0  0  0  0 | *  *  * 32 *  * *  tet 0 12  0 |  0  0 12 12   0  0 |  0  0  0  4  4  6  0  0  0  0 | *  *  *  * 8  * *  co 0  3  3 |  0  0  0  3   6  3 |  0  0  0  0  1  0  3  3  0  1 | *  *  *  * * 32 *  oct 0  0  6 |  0  0  0  0   0 12 |  0  0  0  0  0  0  0  0  4  4 | *  *  *  * *  * 8  oct`

--- rk

Thanks!

@Keiji: add this to the D4.11 page?
quickfur
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### temp

I knew about the mere number. Quickfur in his first post also suggested something along "distended snub 24-cell", but soon didn't support that name any longer, because it is no true Stott expansion, it rather asks for some rebuild before.

Thus, meanwhile I contacted privately PolyhedronDude for a potential nameing.
PolyhedronDude wrote:I've been glancing through the thread, it moves so fast, I've barely been able to keep up. I can see why that polychoron is hard to name - I notice that it has demitessic symmetry. The bilbiroes look like some sort of pentagonal rhombus, then there are the three types of 32 cells - how about something like "pentagonorhombic trisnub trisoctachoron" - a short name could be "pretasto". The conjugate could be called "pentagramorhombic trisnub trisoctachoron - "partesto".

Hehe, PolyhedronDude always thinks about non-convex conjugates as well... - But I cannot wrap my mind around that one so far. It then should include the conjugates of bilbiro and teddi for cells. Several cells potentially might be used retrograde then too. - But instead of a try to build that one from scratch, just by "translation" of the final polychoron, one well could start from the conjugate of sadi, and then apply the same rebuild and expansion onto that. Perhaps this might be easier to follow...

--- rk
Klitzing
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### Re: Johnsonian Polytopes

Klitzing wrote:I knew about the mere number. Quickfur in his first post also suggested something along "distended snub 24-cell", but soon didn't support that name any longer, because it is no true Stott expansion, it rather asks for some rebuild before.

Thus, meanwhile I contacted privately PolyhedronDude for a potential nameing.
PolyhedronDude wrote:I've been glancing through the thread, it moves so fast, I've barely been able to keep up. I can see why that polychoron is hard to name - I notice that it has demitessic symmetry. The bilbiroes look like some sort of pentagonal rhombus, then there are the three types of 32 cells - how about something like "pentagonorhombic trisnub trisoctachoron" - a short name could be "pretasto". The conjugate could be called "pentagramorhombic trisnub trisoctachoron - "partesto".

Sounds like a tasteful name, though I'm not sure I fully understand why it's "trisnub"?

I like "pentagonorhombic", it follows the same pattern as the J92 rhombochoron in the sense that you have 4 pentagons (resp. J92's) in a rhombus-like arrangement with side faces (resp. cells) filling up the gaps.

Hehe, PolyhedronDude always thinks about non-convex conjugates as well... - But I cannot wrap my mind around that one so far. It then should include the conjugates of bilbiro and teddi for cells. Several cells potentially might be used retrograde then too. - But instead of a try to build that one from scratch, just by "translation" of the final polychoron, one well could start from the conjugate of sadi, and then apply the same rebuild and expansion onto that. Perhaps this might be easier to follow...
[...]

Interesting. What's the general procedure for constructing a conjugate? Does it apply to any polytope? So far I only know of the conversion of icosahedron -> great icosahedron, but I'm not sure I understand the general procedure yet.
quickfur
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### Re: Johnsonian Polytopes

quickfur wrote:Sounds like a tasteful name, though I'm not sure I fully understand why it's "trisnub"?

I got the trisnub from the three groups of 32 cells which were tets, octs, and teddis which have trigon pyramid symmetric arrangements instead of some full symmetric arrangements - so they act more like snub cells.
Whale Kumtu Dedge Ungol.
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### Re: D4.10 and D4.11

I just realized that D4.11 can be represented it it's full demitesseractic symmetry. first of all, I'll express the bilbiro in it's 2-2-2 symmetry. Then it is the union of f2o2o, x2x2f and o2F2x. This can be written as fxo2oxF2ofx&U, where U means union.
Now D4.11 can be seen as fxo3xoo3oxF2ofx3*b&U. The union can also be seen as a lace simplex, where all heights are 0. Therefore, we might also write fxo3xoo3oxF2ofx3*b&#x. what do you think is favorable?

Why this is interesting, is because student5 has recently been posting partiall stott-expansions, all the time with respect to a lower-dimensional symmetry. These post showed to me how stott-expansion works for lower symmetries. Now because the D4.11 has demicube symmetry, it should be derived by partiall stott-expansion with respect to demitesseractic symmetry, and with this representation, we can apply the partiall stott-expansions/contractions.
Because we suspect D4.11 to be derived from the snub demitesseract or the 600-cell, we should have a representation of one of these in demitesseractic symmetry. As far as I know, such a representation doesn't exist, but I hope I'm wrong here.

According to this:
Code: Select all
`Bilbiro -> ikef2o2o -> f2(-x)2o "=" f2x2ox2x2f -> x2o2fo2F2x -> o2f2xD4.11     Parent?f3x3o2o3*b <- f3x3(-x)2o3*b "=" f3o3x2o3*bx3o3x2f3*b <- x3o3o2f3*bo3o3F2x3*b <- o3o3f2x3*b`
the parent should be the union of f3o3x2o3*b, x3o3o2f3*b and o3o3f2x3*b. I have no idea what this union might look like, I'll try to find that out soon
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
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student91
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### Re: D4.10 and D4.11

student91 wrote:[...] the parent should be the union of f3o3x2o3*b, x3o3o2f3*b and o3o3f2x3*b. I have no idea what this union might look like, I'll try to find that out soon

All my ugly drawings seem to be telling me that the described union is the snub demicube. Yet wikipedia (and my intuition) is telling me that the snub demicube doesn't have full demicubic symmetry, and thus can't be expressed as such. I would really like it if someone would help me to find out what fxo3ooo3xof2ofx3*b&U is, or at least can tell me if the snub 24-cell has a demicube subsymmetry yes or no

EDIT: it turns out that it is the snub demicube after all
all
Last edited by student91 on Mon Jul 28, 2014 11:48 pm, edited 1 time in total.
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student91
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### Re: D4.10 and D4.11

student91 wrote:I just realized that D4.11 can be represented it it's full demitesseractic symmetry. first of all, I'll express the bilbiro in it's 2-2-2 symmetry. Then it is the union of f2o2o, x2x2f and o2F2x. This can be written as fxo2oxF2ofx&U, where U means union.
Now D4.11 can be seen as fxo3xoo3oxF2ofx3*b&U. The union can also be seen as a lace simplex, where all heights are 0. Therefore, we might also write fxo3xoo3oxF2ofx3*b&#x. what do you think is favorable?

This is a quite interesting idea! So not fully thought thru. So, if you allow, I would take it up.

Wendy's lace terms include the followings:
• lace prisms ("&#x") = axial stack of 2 base layers, both based on the same symmetry, laced by some further edges
• lace tower ("&#xt") = axial stack of more than 2 layers (bases + sections), all based on the same symmetry, consecutive ones being laced by some further edges
• lace rings ("&#xr") = circular stack of layers (border margins), all based on the same symmetry, consecutive ones being laced by some further edges
• lace simplices ("&#x") = simplicial stack of more than 2 base layers, all based on the same symmetry, pairwise laced by some further edges
• compounds (" ") = coincident "stack" of 2 or more "layers" (components), all based on the same symmetry, no lacings
• lace cities ( - ) = 2-dimensional plan of several perp-space "layers" (subfacet + subsections), all based on the same perp-space symmetry, lacings usually not shown

Your considered figures now definitely are laced. E.g. the height between f2o2o and x2f2x, as well as the height between x2f2x and o2x2F surely is zero. In fact, you describe subsets of vertices of a 3D body (bilbiro) by a 3D Dynkin symbol. Thus stacking ought to be co-realmic. But f2o2o and o2x2F definitely are not laced in bilbiro. In fact, calculating the squared height results in a negative value (-tau). - Therefore you definitely have a lace tower here. Sure, one with zero heights between consecutive layers.

Thus bilbiro could indeed be written as degenerate lace tower   fxo2ofx2oxF&#xt

Now to your description of that 4D CRF D4.11 (= pretasto). Again you describe subsets of vertices of that 4D figure by means of 4D Dynkin symbols. Thus again these "layers" ought to be co-hyperrealmic. And indeed, the height between f3x3o *b3o and x3o3f *b3x, as well as the height between x3o3f *b3x and o3o3x *b3F comes out to be zero (as it should). But the squared height between f3x3o *b3o and o3o3x *b3F again evaluates to a negative value (-tau). - This again shows, that the first and third layer cannot be laced. So in consequence we again cannot have a lace simplex here, rather again a pure lace tower.

That is, pretasto likewise can be given as degenerate lace tower   fxo3xoo3ofx *b3oxF&#xt

Btw., layer f3x3o *b3o describes an edge-scaled variation of thex. Accordingly it has 48 vertices. Layer x3o3f *b3x describes an edge-scaled variation of rico. Accordingly it has 96 vertices. And layer o3o3x *b3F describes an edge-scaled variation of rit. Accordingly it has 32 vertices. - And by no surprize, the vertices of pretasto indeed decompose into 48+96+32, each class thereby having a different vertex figure type!

--- rk
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### Re: D4.10 and D4.11

I've split the notation to another topic. In this topic I think I can prove that the "parent" is indeed the snub 24-cell.

the union can be taken apart in lace-towe notation as follows:
Code: Select all
`f3o3x*b3o:   x3o3o*b3f:   o3o3f*b3x:  f3o3x        x3o3o        o3o3f  F3o3o        x3f3o        o3x3f  o3o3F        o3f3x        f3x3o  x3o3f        o3o3x        f3o3o`

when sorted to circumradius you get:
x3o3o o3o3f f3o3x o3x3f x3f3o F3o3o o3o3F o3f3x f3x3o x3o3f f3o3o o3o3x, which is indeed the lace tower of the snub 24-cell. This means that the snub 24-cell has full demiteseractic symmetry!!

This also means the above "parent" is indeed the snub 24-cell, and it means D4.11 is a true complex partiall stott-expansion of the snub 24-cell.
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### Re: D4.10 and D4.11

student91 wrote:This means that the snub 24-cell has full demiteseractic symmetry!!

Well, on the one hand sadi (s3s4o3o) is just the diminishing of ex (x3o3o5o) at the positions of a vertex inscribed ico (x3o4o3o). In fact there is a 25 ico compound, the hull of which is just ex.

On the other hand hex (x3o3o4o) can be vertex inscribed into ico (x3o4o3o). In fact there is a 3 hex compound, the hull of which is just ico.

Thus sadi clearly could be described in hexic symmetry. But it has not only full hexic symmetry, it even has full icoic symmetry!

Consider the incidence matrix of sadi:
Code: Select all
`s3s4o3odemi( . . . . ) | 96 |   3   6 |  3   9  3 |  3  1  4----------------+----+---------+-----------+---------      . s4o .   |  2 | 144   * |  0   2  2 |  1  1  2sefa( s3s . . ) |  2 |   * 288 |  1   2  0 |  2  0  1----------------+----+---------+-----------+---------      s3s . .   |  3 |   0   3 | 96   *  * |  2  0  0sefa( s3s4o . ) |  3 |   1   2 |  * 288  * |  1  0  1sefa( . s4o3o ) |  3 |   3   0 |  *   * 96 |  0  1  1----------------+----+---------+-----------+---------      s3s4o .   | 12 |   6  24 |  8  12  0 | 24  *  *      . s4o3o   |  4 |   6   0 |  0   0  4 |  * 24  *sefa( s3s4o3o ) |  4 |   3   3 |  0   3  1 |  *  * 96`

That one shows up "24" (or multiples therefrom) everywhere!

--- rk
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### Re: D4.10 and D4.11

Klitzing wrote:[...]
Thus sadi clearly could be described in hexic symmetry. But it has not only full hexic symmetry, it even has full icoic symmetry!
[...]
I'll believe you there, but then why doesn't the wikipedia article of the snub 24-cell mention all these symmetries, while it does mention the chiral demitesseractic resp. icosaic symmetries? Is wikipedia wrong here .(Also, how does a snub not give chiral symmetry, is that something awesome of 4D?)
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student91
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### Re: D4.10 and D4.11

student91 wrote:
Klitzing wrote:[...]
Thus sadi clearly could be described in hexic symmetry. But it has not only full hexic symmetry, it even has full icoic symmetry!
[...]
I'll believe you there, but then why doesn't the wikipedia article of the snub 24-cell mention all these symmetries, while it does mention the chiral demitesseractic resp. icosaic symmetries?[...]

This is only a "half" 24-cell symmetry, because it's missing some of the reflective symmetries of the 24-cell. So you have to divide the number of symmetries by 2. The full 24-cell symmetry has twice the number of symmetries.
quickfur
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### Re: D4.10 and D4.11

Quickfur is right. Only 1/2 F4.
And by means of that 3 hex compound within ico, this halving runs thru: likewise 1/2 BC4, resp. 1/2 D4 only.

--- rk
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### Re: D4.10 and D4.11

I guess that means the snub 24-cell has only chiral demitesseractic symmetry, yet I represented it as a union of things with full demitesseractic symmetry. Doesn't this representation show that it has full demitesseractic symmetry? (I mean, all fundamental domains are filled with the same arrangement of vertices) I'm really puzzled right now.

What I'm having struggles with:

Wikipedia says the snub demitesseract has [31,1,1]+ as a subsymmetry (and you seem to agree on this)
I found a union that seems to have [31,1,1] symmetry, and that seems to coincide with the snub demicube.

My compound clearly has a higher symmetry than wikipedia mentions. Therefore I at first thought that my union wouldn't be the snub demicbe, but I found something that seems to be a proof that it is. so I guess my "proof" is wrong, or wikipedia is. (I think that means I'm wrong)
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: D4.10 and D4.11

Please don't confuse alternation with chirality. AFAIK, the snub 24-cell is not chiral, even though it does have reduced symmetry due to alternation from the full 24-cell symmetry. Chirality sometimes happens with alternated polytopes, but not always. For example, the alternated tesseract is a 16-cell: non-chiral. The alternated cube is the tetrahedron: non-chiral. But the alternated omnitruncated cube (great rhombicuboctahedron) is chiral. So chirality and alternation are related, but they are not equivalent.
quickfur
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### Re: D4.10 and D4.11

I've discovered a CRF with 32 icosahedra in demitesseractic symmetry.

It's basically D4.11 augmented with bilbiro pseudopyramids. The pentagonal pyramids in the pseudopyramid augments merge with the J63's to form icosahedra. I've to run now, will post renders later. I'm pretty sure somebody must've thought of doing this before??
quickfur
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### Re: D4.10 and D4.11

quickfur wrote:I've discovered a CRF with 32 icosahedra in demitesseractic symmetry.

It's basically D4.11 augmented with bilbiro pseudopyramids. The pentagonal pyramids in the pseudopyramid augments merge with the J63's to form icosahedra. I've to run now, will post renders later. I'm pretty sure somebody must've thought of doing this before??

Great!
That then ought to be   ooxf3foox3oxfo *b3xFxo&#zxr.

Cell count should be: 116 136 tet + 32 ike + 8 co + 40 oct + 48 trip + 96 squippy.
Trips work still as digonal cupolae. Their base square attaches to co, the lacing squares attach to squippies.

--- rk
Last edited by Klitzing on Wed Apr 02, 2014 8:57 pm, edited 1 time in total.
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### Re: D4.10 and D4.11

How many CRFs is this anyway? If each bilbro is augmentable individually...

I looked at D4.10. It seems that bilbros are also augmentable (and once again, pentagonal faces merge).

The scary thing that dichoral angles in Stella indicate that the thawros are also augmentable with pseudopyramids! This time both pentagonal AND hexagonal faces merge.

And each of these is augmentable separately.

The omniaugment would therefore have 136 tetrahedra, 96 square pyramids, 40 octahedra, 32 icosahedra, 8 cuboctahedra and 48 triangular prisms, if I count correctly.

Partial augments with all bilbiros/all thawros augmented might be also interesting...

Is it possible that the existence of these pseudopyramid augments is in fact a general rule?

EDIT: The numbers of cells are suspiciously similar to Klitzing's calculation in previous post, so I thought it might be in fact the same figure -- except for the tetrahedra. D4.10 has 16 tetrahedra, 12 bilbiros and 8 thawros. Each bilbiro augment adds another 4 tetrahedra and each thawro argument adds 9, so 16 + 4*12 + 8*9 = 16 + 48 + 72 = 136. So I guess the omniaugmented D4.10 is a different figure, though all the other cell counts are the same?
Last edited by Marek14 on Wed Apr 02, 2014 8:50 pm, edited 1 time in total.
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### Re: D4.10 and D4.11

... and here finally comes the incmat of "icau pretasto" (icositetra-augmented pentagonorhombic-trisnub-trisoctachoron = 24-augm. D4.11)

Code: Select all
`ooxf3foox3oxfo *b3xFxo&#zxr   → all heights = 0 – except that of not existing lacing(2,4)o...3o...3o... *b3o...      | 48  *  *  * |  1  2   4  2  0  0  0  0   0  0 |  1  2  4  2   4   4  0  0  0  0  0  0  0  0  0 |  2  2  2  4 0  0 0  0 0.o..3.o..3.o.. *b3.o..      |  * 32  *  * |  0  3   0  0  3  3  0  0   0  0 |  3  0  0  0   6   0  3  3  0  0  0  0  0  0  0 |  3  3  0  0 1  1 0  0 0..o.3..o.3..o. *b3..o.      |  *  * 96  * |  0  0   2  0  0  1  2  2   2  0 |  0  2  2  0   2   2  0  2  1  1  2  1  2  0  0 |  1  2  2  2 0  1 1  1 0...o3...o3...o *b3...o      |  *  *  * 48 |  0  0   0  2  0  0  0  0   4  4 |  0  0  0  1   0   4  0  0  0  0  0  4  2  2  2 |  2  0  0  2 0  0 0  2 1----------------------------+-------------+---------------------------------+------------------------------------------------+------------------------.... .... ....    x...      |  2  0  0  0 | 24  *   *  *  *  *  *  *   *  * |  0  0  4  2   0   0  0  0  0  0  0  0  0  0  0 |  0  0  2  4 0  0 0  0 0oo..3oo..3oo.. *b3oo..&#x   |  1  1  0  0 |  * 96   *  *  *  *  *  *   *  * |  1  0  0  0   2   0  0  0  0  0  0  0  0  0  0 |  2  1  0  0 0  0 0  0 0o.o.3o.o.3o.o. *b3o.o.&#x   |  1  0  1  0 |  *  * 192  *  *  *  *  *   *  * |  0  1  1  0   1   1  0  0  0  0  0  0  0  0  0 |  1  1  1  1 0  0 0  0 0o..o3o..o3o..o *b3o..o&#x   |  1  0  0  1 |  *  *   * 96  *  *  *  *   *  * |  0  0  0  1   0   2  0  0  0  0  0  0  0  0  0 |  1  0  0  2 0  0 0  0 0.... .... .x..    ....      |  0  2  0  0 |  *  *   *  * 48  *  *  *   *  * |  1  0  0  0   0   0  2  0  0  0  0  0  0  0  0 |  2  0  0  0 1  0 0  0 0.oo.3.oo.3.oo. *b3.oo.&#x   |  0  1  1  0 |  *  *   *  *  * 96  *  *   *  * |  0  0  0  0   2   0  0  2  0  0  0  0  0  0  0 |  1  2  0  0 0  1 0  0 0..x. .... ....    ....      |  0  0  2  0 |  *  *   *  *  *  * 96  *   *  * |  0  1  0  0   0   0  0  1  1  0  1  0  0  0  0 |  0  1  1  0 0  1 1  0 0.... .... ....    ..x.      |  0  0  2  0 |  *  *   *  *  *  *  * 96   *  * |  0  0  1  0   0   0  0  0  0  1  1  0  1  0  0 |  0  0  1  1 0  0 1  1 0..oo3..oo3..oo *b3..oo&#x   |  0  0  1  1 |  *  *   *  *  *  *  *  * 192  * |  0  0  0  0   0   1  0  0  0  0  0  1  1  0  0 |  1  0  0  1 0  0 0  1 0.... ...x ....    ....      |  0  0  0  2 |  *  *   *  *  *  *  *  *   * 96 |  0  0  0  0   0   0  0  0  0  0  0  1  0  1  1 |  1  0  0  0 0  0 0  1 1----------------------------+-------------+---------------------------------+------------------------------------------------+------------------------.... .... ox..    ....&#x   |  1  2  0  0 |  0  2   0  0  1  0  0  0   0  0 | 48  *  *  *   *   *  *  *  *  *  *  *  *  *  * |  2  0  0  0 0  0 0  0 0o.x. .... ....    ....&#x   |  1  0  2  0 |  0  0   2  0  0  0  1  0   0  0 |  * 96  *  *   *   *  *  *  *  *  *  *  *  *  * |  0  1  1  0 0  0 0  0 0.... .... ....    x.x.&#x   |  2  0  2  0 |  1  0   2  0  0  0  0  1   0  0 |  *  * 96  *   *   *  *  *  *  *  *  *  *  *  * |  0  0  1  1 0  0 0  0 0.... .... ....    x..o&#x   |  2  0  0  1 |  1  0   0  2  0  0  0  0   0  0 |  *  *  * 48   *   *  *  *  *  *  *  *  *  *  * |  0  0  0  2 0  0 0  0 0ooo.3ooo.3ooo. *b3ooo.&#x   |  1  1  1  0 |  0  1   1  0  0  1  0  0   0  0 |  *  *  *  * 192   *  *  *  *  *  *  *  *  *  * |  1  1  0  0 0  0 0  0 0o.oo3o.oo3o.oo *b3o.oo&#x   |  1  0  1  1 |  0  0   1  1  0  0  0  0   1  0 |  *  *  *  *   * 192  *  *  *  *  *  *  *  *  * |  1  0  0  1 0  0 0  0 0.... .o..3.x..    ....      |  0  3  0  0 |  0  0   0  0  3  0  0  0   0  0 |  *  *  *  *   *   * 32  *  *  *  *  *  *  *  * |  1  0  0  0 1  0 0  0 0.ox. .... ....    ....&#x   |  0  1  2  0 |  0  0   0  0  0  2  1  0   0  0 |  *  *  *  *   *   *  * 96  *  *  *  *  *  *  * |  0  1  0  0 0  1 0  0 0..x.3..o. ....    ....      |  0  0  3  0 |  0  0   0  0  0  0  3  0   0  0 |  *  *  *  *   *   *  *  * 32  *  *  *  *  *  * |  0  0  0  0 0  1 1  0 0.... ..o. .... *b3..x.      |  0  0  3  0 |  0  0   0  0  0  0  0  3   0  0 |  *  *  *  *   *   *  *  *  * 32  *  *  *  *  * |  0  0  0  0 0  0 1  1 0..x. .... ....    ..x.      |  0  0  4  0 |  0  0   0  0  0  0  2  2   0  0 |  *  *  *  *   *   *  *  *  *  * 48  *  *  *  * |  0  0  1  0 0  0 1  0 0.... ..ox ....    ....&#x   |  0  0  1  2 |  0  0   0  0  0  0  0  0   2  1 |  *  *  *  *   *   *  *  *  *  *  * 96  *  *  * |  1  0  0  0 0  0 0  1 0.... .... ....    ..xo&#x   |  0  0  2  1 |  0  0   0  0  0  0  0  1   2  0 |  *  *  *  *   *   *  *  *  *  *  *  * 96  *  * |  0  0  0  1 0  0 0  1 0.... ...x3...o    ....      |  0  0  0  3 |  0  0   0  0  0  0  0  0   0  3 |  *  *  *  *   *   *  *  *  *  *  *  *  * 32  * |  1  0  0  0 0  0 0  0 1.... ...x3.... *b3...o      |  0  0  0  3 |  0  0   0  0  0  0  0  0   0  3 |  *  *  *  *   *   *  *  *  *  *  *  *  *  * 32 |  0  0  0  0 0  0 0  1 1----------------------------+-------------+---------------------------------+------------------------------------------------+------------------------.... foox3oxfo    ....&#xr  |  3  3  3  3 |  0  6   6  3  3  3  0  0   6  3 |  3  0  0  0   6   6  1  0  0  0  0  3  0  1  0 | 32  *  *  * *  * *  * *  ikeoox. .... ....    ....&#xr  |  1  1  2  0 |  0  1   2  0  0  2  1  0   0  0 |  0  1  0  0   2   0  0  1  0  0  0  0  0  0  0 |  * 96  *  * *  * *  * *  teto.x. .... ....    x.x.&#x   |  2  0  4  0 |  1  0   4  0  0  0  2  2   0  0 |  0  2  2  0   0   0  0  0  0  0  1  0  0  0  0 |  *  * 48  * *  * *  * *  trip.... .... ....    x.xo&#x   |  2  0  2  1 |  1  0   2  2  0  0  0  1   2  0 |  0  0  1  1   0   2  0  0  0  0  0  0  1  0  0 |  *  *  * 96 *  * *  * *  squippy.o..3.o..3.x..    ....      |  0  4  0  0 |  0  0   0  0  6  0  0  0   0  0 |  0  0  0  0   0   0  4  0  0  0  0  0  0  0  0 |  *  *  *  * 8  * *  * *  tet.ox.3.oo. ....    ....&#x   |  0  1  3  0 |  0  0   0  0  0  3  3  0   0  0 |  0  0  0  0   0   0  0  3  1  0  0  0  0  0  0 |  *  *  *  * * 32 *  * *  tet..x.3..o. .... *b3..x.      |  0  0 12  0 |  0  0   0  0  0  0 12 12   0  0 |  0  0  0  0   0   0  0  0  4  4  6  0  0  0  0 |  *  *  *  * *  * 8  * *  co.... ..ox .... *b3..xo&#x   |  0  0  3  3 |  0  0   0  0  0  0  0  3   6  3 |  0  0  0  0   0   0  0  0  0  1  0  3  3  0  1 |  *  *  *  * *  * * 32 *  oct.... ...x3...o *b3...o      |  0  0  0  6 |  0  0   0  0  0  0  0  0   0 12 |  0  0  0  0   0   0  0  0  0  0  0  0  0  4  4 |  *  *  *  * *  * *  * 8  oct`

(Yep, Marek, 96+8+32=136, you're right. - So the omni augmented D4.10 might be the same after all.)

--- rk
Klitzing
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### Re: D4.10 and D4.11

OK, so D4.10 and D4.11 can be transformed into each other just by gluing and removing caps? Then what are the "partially augmented" D4.10 (with one or the other type of cells augmented)? Maybe the thawro-augmented one is D4.11? (It's too late for me to compute it and be sure.) But then the bilbiro-augmented one would be tetrahedrally symmetrical with thawros and no bilbiros... did we already have one like that?
Marek14
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