## Partial Stott expansion of nonconvex figures

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

### Re: Johnsonian Polytopes

I didn't get very far with the 4-dimensional partial stott-expansions (I am a bit stuck), so I'll just post what I've got so far:
considering the expansions x3o5o"="(-x)3x5o=>bilbiro. One part of this expansion looks like (-x)3x5o => o3x5o, and the other part looks like (-x)3x5o => x3o5x. When I tried to understand why it is exactly the expansion (-x)3x5o=>x3o5x, I got stuck, so I went on, just accepting that this happens. Then I got to consider why there is a part of x5o3x3x in the bilbiro'd o5x3o3x.
When I considered the expansion o5o3x3o"="o5x3(-x)3x => bilbiro'd o5x3o3x, I saw one part looked like the expansion o5x3(-x)3x => o5x3o3x, and the other part looked like o5x3(-x)3x => x5o3x3x. You can see the similarity with the bilbiro-expansion, yet without any explanation why that happens. (the .5.3.-part is practically identical to the bilbiro'd ike)
The orthocupolarotunda'd ike also has a part going (-x)3x5o=>o3x5o and one part going (-x)3x5o => x3o5x. this also happens in the bilbiroing of o5x3o3x, giving the orthocupolarotunda's.

Then I wanted to see how far this analogy holds, so I went on considering the expansion o5o3o3x "=" o5o3x3(-x) "=" o5x3(-x)3o => bilbiro'd o5x3o3o. This should, if my analogy is correct, have partg going like o5x3(-x)3o=>o5x3o3o, and parts going like o5x3(-x)3o => x5o3x3o. So I went on searching for parts of x5o3x3o in the bilbiro'd o5x3o3o. This isn't directly present, but in analogy of inserting a patch of x5o3x3x in o5x3o3x, I tried to insert a patch of x5o3x3o in o5x3o3o, and it worked (you even have the "missing" parts of the bilbiro's in there). This "patch" is shown when you insert a pentagonal magnabicupolic ring in the decagonal prisms of the bilbiro'd o5x3o3o, such that the squippy's become oct's, and the pentagonal cupolae become orthocupolarotundae. The resulting figure should be considered the "real" bilbiro'd o5x3o3o, and the former figure a diminishing of that.

To make the analogy complete, I tried to find the orthocupolarotundae in the bilbiro'd o5o3x3o. these are present as well (I think), and they can be shown by removing the following vertices:you should look at the nearest bilbiro. the penatgon next to the edge of a triangle of the bilbiro (so the pentagon touching the triangle of the bilbiro, but not part of the bilbiro itself)should be deleted, together with the vertex that makes a pentagonal pyramid out of this.

I just made up the stott-deriviation of this bilbiro'd o5o3x3o. o5o3x3o"="o5x3(-x)3x "=" o5x3o3(-x). this should be the product of the o5x3(-x)3A=>o5x3o3A procedure, so A=-x, and the "parent" figure is o5x3(-x)3(-x). this is not "equal" to a uniform CRF, so the stott-deriviation is a bit weird. nevertheless, via analogy, it should also have parts of x5o3x3(-x) "=" x5o3o3x. this is true, it does have a pentagonal orthobicupolic ring, borrowed from x5o3o3x.

so in summary all bilbiro-ings are weird stott-expansions, although "we" (I) don't know why. Also orthocupolarotunda-ing is the same procedure as bilbiro-ing. More complex weird stott-expansions such as D4.11 and thawro-ing are still unexplained as well. Furthermore I think this kind of expansion shouldn't be called "partial", because it's a full, combined expansion of both ike=>id and ike=>x5o3x. I guess I'll stick to "weird" expansions as I don't know any better name. lastly, I don't know what ike=>thawro should be seen as, all analogy's seem to fail. please help me out, I think we've got something important here. If you don't understand something of this post, please tell me, I will explain it as best as I can, in order to work this out
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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wendy wrote:Here is what i know of ursulatopes.

Ursulatopes are a kind of partially truncated fostrums, literally the prototype is xu&#ft. If the truncation happens at 1/F of the height from the large base, and the large base is rectified, the style of the figure is xfo&xt, a pentagon. The base figure is rectified.

The ursulation of a polygon gives xfoPoop&#t, where p is the shortchord of P (eg x=3, q=4, f=5, h=6, u=infinity. The largest value of P which leads to mostly unit-edge ursulate is P=10. A polytope with this size is x3o3o5o, so xfo3oox3ooo5ooo&#xt is in fact flat.

So you're saying that the 5D 600-cell ursateron is actually flat? So we should remove it from the list of CRF ursatopes?

On the flip side, this means that the icosahedral ursachoron can (locally) tile 4-space with 600-cells? If so, this is very remarkable, since the icosahedral ursachoron itself is a diminished hemi-600-cell, so this would imply that there exists a tiling of 4-space with some manner of diminished 600-cells!

Or did I misunderstand what you said?
quickfur
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### Re: Johnsonian Polytopes

student91 wrote:[...]
so in summary all bilbiro-ings are weird stott-expansions, although "we" (I) don't know why. Also orthocupolarotunda-ing is the same procedure as bilbiro-ing. More complex weird stott-expansions such as D4.11 and thawro-ing are still unexplained as well. Furthermore I think this kind of expansion shouldn't be called "partial", because it's a full, combined expansion of both ike=>id and ike=>x5o3x. I guess I'll stick to "weird" expansions as I don't know any better name. lastly, I don't know what ike=>thawro should be seen as, all analogy's seem to fail. please help me out, I think we've got something important here. If you don't understand something of this post, please tell me, I will explain it as best as I can, in order to work this out

the cool thing here, altough I think it has been already seen, is that the thawro contains part of x3x5o and o3x5o and the orthocup parts of o3x5o and x3o5x. I guess this is interesting, and might be caused by the fact that the lace tower of ike is the same upside-down, in other words, if you add x to the left column, you get past both the right and left column, (because it's right when upside-down) e.g. xofo3ofox&#xt becomes oxFo3xfox&#xt, which is the same as xofx3oFxo&#xt and therefore it has o.x3x (x3o + x on the right) and o.o3x (x3o"="(-x)3x +x on the left) which makes x3x5o and o3x5o
in the same way, the bilb is a x.o expansion and therefore has o3x.x (x.o +x on the left) and x3o.o (o3x.o"="x3(-x).o + x on the right) which makes x3o5x and o5x3o
the orthocup is a o5o expansion and is more difficult, x3o5o according to the above analogy expanded, would have x3o5x (o5o+x on the right) and x3x5o (o5o + x on the left), while it has x3o5x and o3x5o this seems random to me, but I believe it has got something to do with the fact that the change of x only appears in the second layer of the lace pyramid, I just have no idea why maybe the rule should be expanded to mention the place aswell?
student5
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### Re: Johnsonian Polytopes

Skimmed over your post, will come back and read it in more detail later, but wanted to just comment on your last paragraph:

student91 wrote:[...]
so in summary all bilbiro-ings are weird stott-expansions, although "we" (I) don't know why. Also orthocupolarotunda-ing is the same procedure as bilbiro-ing. More complex weird stott-expansions such as D4.11 and thawro-ing are still unexplained as well. Furthermore I think this kind of expansion shouldn't be called "partial", because it's a full, combined expansion of both ike=>id and ike=>x5o3x. I guess I'll stick to "weird" expansions as I don't know any better name.

In my mind, it's a "partial" Stott expansion because we're expanding along some symmetry axes but not others (since a full Stott expansion will correspond to adding an x to an o-node in the CD diagram, and that just produces a uniform polychoron). Obviously, there are several layers of expansions going on with these bilbiro'd and thawro'd polychora, since we have some 3D elements that are being (partially or fully) expanded, but only to a selected subset of them, rather than everywhere (which yields a uniform polychoron).

lastly, I don't know what ike=>thawro should be seen as, all analogy's seem to fail. please help me out, I think we've got something important here. If you don't understand something of this post, please tell me, I will explain it as best as I can, in order to work this out

This is just my gut feeling, since I haven't studied this rigorously yet, but it seems that what we're seeing here is a kind of modified, or "complex" Stott expansion / contraction (depending on which way you look at it), where some elements are identified with a non-convex element with the same vertices, and the Stott expansion is applied to (parts of) the non-convex structure rather than the usual convex structure. Technically, there is no Stott expansion (in the traditional sense) that can transform an ike into a bilbiro; the way we did it was by substituting parts of ike with parts of the great dodecahedron and applying Stott expansion to that instead, while simultaneously keeping other parts of ike as-is. So in a sense, it's really a pseudo-ike, not ike itself, that's transforming into a bilbiro; it's a kind of "frankenstein monster" double of ike, which is not really ike but a hybrid of ike and great dodecahedron. It just so happens that great dodecahedron has the same vertices as ike, so this hybridization "works", and seemingly it's ike itself that's transforming into bilbiro. But actually, it's the frankensteinian pseudo-ike that is undergoing the transformation.

Similarly, things like D4.11 are produced technically not from the snub 24-cell itself, but from a "frankenstein double" of snub 24-cell where the icosahedra are hybrids of ike and great dodecahedron. The coincidence of vertices means that these "pseudo-ike"'s can be interpreted either way.

Now the interesting part about some of the bilbiro'd and thawro'd polychora that we found, is that most of our constructions were by deleting some vertices or substituting some other vertices. So it's more like a partial (pseudo) Stott contraction than a Stott expansion. Also, this happens to a hybridized "frankensteinian" version of the polychoron instead of the "pure" convex version. For example, we produced thawro's in o5x3o3o by shrinking some vertex layer (sorry I can't remember the exact symbol off-hand, it was something like x5f3f) to x5x3x. This could be understood as the substitution of o5x3o3o with parts of some non-convex uniform with the same vertices, where the Stott contraction of the non-convex unform produces a x5x3x cross-section. Since the vertices of the non-convex and o5x3o3o are the same, the vertices can be interpreted either way, which gives us the flexibility to apply the Stott expansion/contraction to them, partly as if they are part of o5x3o3o and partly as if they are part of the non-convex polychoron.

This somewhat ties in with my previous observation about something fishy going on with 4D and the golden ratio, in the sense that the numbers somehow keep working out "coincidentally", sometimes in unexpected ways. This phenomenon can be seen to be less strange if we understand that given, say, some 600-cell family convex uniform, there are entire regiments (to use Bowers' term) of non-convex uniforms with the same vertices, each of which is related to other non-convex uniforms of various relative sizes with different vertices via Stott expansion/contraction/faceting/stellating. I suspect that many of the CRFs we have been finding are actually CRF pieces of some of the non-convex uniforms, or some combination of multiple non-convex uniforms. The fact that there are so many non-convex uniforms, and they are related to each other in many ways via various Stott contractions/expansions and stellations/facetings, means that there are many combinations of vertices that will have unit edge lengths (e.g., the relation between a uniform and its Stott-expanded version is a unit edge length difference), so they form some kind of pentagonal grid in 4D space that we can cut many CRFs out of. The coincidence of vertices of many non-convex uniforms with convex uniforms means that many of these pieces may have an underlying non-convex derivation, but they can be reinterpreted as convex, and therefore many of them become CRF.
quickfur
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### Re: Johnsonian Polytopes

Stott expansion in that sense should be possible. But then you should do that consistently, i.e. in the same sense to all left sides of the perp space symbols:
Code: Select all
                x3o                                                           u3o   x3f F3o   x3x                                                                               x3o x3f   F3x       u3f   F3o x3o                                       F3o       x3F   X3o       x3f                                                                         x3x   u3f X3o   F3f   x3F F3x   u3o                                                                         F3o       x3F   X3o       x3f                                       x3o x3f   F3x       u3f   F3o x3o                                                                               u3o   x3f F3o   x3x                                                           x3o

(here I use f = 1.618 x, u = 2x, F = ff = f+x, X = f+u = F+x.)

As you can see quite clearly, your intended instances of thawro here no longer appear...

In fact it happens to be nothing but sectioning ex at its equatorial hyperplane, and inserting prisms at this cut. - Whether in that orientation again some tetrahedra get dissected - as in the pentagonal orientation - I can not see at the moment. (And esp. that those then could be replaced by something CRF...)

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

Klitzing wrote:Stott expansion in that sense should be possible. But then you should do that consistently, i.e. in the same sense to all left sides of the perp space symbols:
Code: Select all
                x3o                                                           u3o   x3f F3o   x3x                                                                               x3o x3f   F3x       u3f   F3o x3o                                       F3o       x3F   X3o       x3f                                                                         x3x   u3f X3o   F3f   x3F F3x   u3o                                                                         F3o       x3F   X3o       x3f                                       x3o x3f   F3x       u3f   F3o x3o                                                                               u3o   x3f F3o   x3x                                                           x3o

but would it be possible to "invert" x3o to (-x)3x before expanding? you'd then get o3x when expanded and the thawros, thus the following city.
Code: Select all
                x3o                                                           o3x   x3f F3o   x3x                                                                               x3o x3f   F3x       u3f   F3o x3o                                       F3o       x3F   X3o       x3f                                                                         x3x   u3f X3o   F3f   x3F F3x   o3x                                                                         F3o       x3F   X3o       x3f                                       x3o x3f   F3x       u3f   F3o x3o                                                                               o3x   x3f F3o   x3x                                                           x3o

In fact it happens to be nothing but sectioning ex at its equatorial hyperplane, and inserting prisms at this cut. - Whether in that orientation again some tetrahedra get dissected - as in the pentagonal orientation - I can not see at the moment. (And esp. that those then could be replaced by something CRF...)

--- rk

in this picture, I guess it is visible that quite some tetrahedra are dissected, if I understand it correctly, all visible tetrahedra do not lie flat, the have a point sticking out to either side, but they'd become, depending on dichoral angle, an elongated pentagonal bipyramid, or 2 pentagonal pyramids with a prism in between
student5
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### Re: Johnsonian Polytopes

student5 wrote:
student91 wrote:... orthocupolarotunda ...
... orthocup ...

Just want to throw in the OBSA of that thingy for a future neater reference: it's called pocuro (for Pentagonal OrthoCUpolaROtunda).
--- rk
Klitzing
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### Re: Johnsonian Polytopes

Klitzing wrote:Just want to throw in the OBSA of that thingy for a future neater reference: it's called pocuro (for Pentagonal OrthoCUpolaROtunda).
--- rk

Added to the list on the wiki

Keiji

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### Re: Johnsonian Polytopes

student5 wrote:
Klitzing wrote:Stott expansion in that sense should be possible. But then you should do that consistently, i.e. in the same sense to all left sides of the perp space symbols:
Code: Select all
                x3o                                                           u3o   x3f F3o   x3x                                                                               x3o x3f   F3x       u3f   F3o x3o                                       F3o       x3F   X3o       x3f                                                                         x3x   u3f X3o   F3f   x3F F3x   u3o                                                                         F3o       x3F   X3o       x3f                                       x3o x3f   F3x       u3f   F3o x3o                                                                               u3o   x3f F3o   x3x                                                           x3o

but would it be possible to "invert" x3o to (-x)3x before expanding? you'd then get o3x when expanded and the thawros, thus the following city.
Code: Select all
                x3o                                                           o3x   x3f F3o   x3x                                                                               x3o x3f   F3x       u3f   F3o x3o                                       F3o       x3F   X3o       x3f                                                                         x3x   u3f X3o   F3f   x3F F3x   o3x                                                                         F3o       x3F   X3o       x3f                                       x3o x3f   F3x       u3f   F3o x3o                                                                               o3x   x3f F3o   x3x                                                           x3o

Ahhh, now I see how you'd get there. That part then might work after all.
In fact it happens to be nothing but sectioning ex at its equatorial hyperplane, and inserting prisms at this cut. - Whether in that orientation again some tetrahedra get dissected - as in the pentagonal orientation - I can not see at the moment. (And esp. that those then could be replaced by something CRF...)

--- rk

in this picture, I guess it is visible that quite some tetrahedra are dissected, if I understand it correctly, all visible tetrahedra do not lie flat, the have a point sticking out to either side, but they'd become, depending on dichoral angle, an elongated pentagonal bipyramid, or 2 pentagonal pyramids with a prism in between

Sorry, this is the wrong pic. Here quickfur was running through ex with icosahedral symmetry. But your lace city was showing up threefold one only. Thus, in order to remain comparable, you should run through ex with some trigonal symmetry, i.e. not starting vertex first, but rather either cell first or triangle first!

In fact, this I was already refering to by "as in pentagonal orientation". There indeed you can replace the chunks by peppies. But I do not see in the moment what would be the equatorial crossection for triangle first sectioning. - The vertices in that equatorial layer of either possible axial directions so are provided in my webpage already, if that would help.

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

student91 wrote:... I found another way to partially stott-expand the ike, this time into a orthocupolarotunda!! ugly drawing:

the pentagons are black, red, yellow, blue and green. those should be moved according to the pink arrows, every pentagon has five arows at its vertices that are somewhat parralel. the central vertex should be pulled apart into a pentagon, the 1-further-vertices should be placed atop the ones next to them, and the next vertices should be pulled apart int an edge. the last vertex (not drawn) should be pulled into a pentagon as well,completing the cupola.

maybe the occurence of the orthocupolarotunda's in the bilbiro'd o5x3o3x isn't that random.
I think this is the last such partial stott-contraction of ike

Finally got around that one as well. Again, you are not starting with ike itself, rather with some faceting thereof in order to apply a mere partial Stott expansion. In fact, there are 2 possible ones, one resulting in pero (pentagonal rotunda) and one resulting in pocuro (pentagonal orthocupolarotunda).

Below I show 2 close relatives of those starting facetings:

infact you would have to blend those at the top view with a further peppy each. This would reduce the (topmost) axis orthogonal pentagon (blending it out) and introduce 5 further triangles, all incident to the top vertex. Then, you can imagine, the 5 edges incident to the top vertex all get doubled! One being incident to 2 pentagons, one to 2 triangles.

Applying Stott expansion (in fact wrt. their axial pentagonal symmetry, which clearly is just a subsymmetry of the full icosahedral one) to those 2 figures, would result in pocuro for the left one, resp. in pero for the right one.

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

quickfur wrote:Skimmed over your post, will come back and read it in more detail later, but wanted to just comment on your last paragraph:

student91 wrote:[...]
so in summary all bilbiro-ings are weird stott-expansions, although "we" (I) don't know why. Also orthocupolarotunda-ing is the same procedure as bilbiro-ing. More complex weird stott-expansions such as D4.11 and thawro-ing are still unexplained as well. Furthermore I think this kind of expansion shouldn't be called "partial", because it's a full, combined expansion of both ike=>id and ike=>x5o3x. I guess I'll stick to "weird" expansions as I don't know any better name.

In my mind, it's a "partial" Stott expansion because we're expanding along some symmetry axes but not others (since a full Stott expansion will correspond to adding an x to an o-node in the CD diagram, and that just produces a uniform polychoron). Obviously, there are several layers of expansions going on with these bilbiro'd and thawro'd polychora, since we have some 3D elements that are being (partially or fully) expanded, but only to a selected subset of them, rather than everywhere (which yields a uniform polychoron).

lastly, I don't know what ike=>thawro should be seen as, all analogy's seem to fail. please help me out, I think we've got something important here. If you don't understand something of this post, please tell me, I will explain it as best as I can, in order to work this out

This is just my gut feeling, since I haven't studied this rigorously yet, but it seems that what we're seeing here is a kind of modified, or "complex" Stott expansion / contraction (depending on which way you look at it), where some elements are identified with a non-convex element with the same vertices, and the Stott expansion is applied to (parts of) the non-convex structure rather than the usual convex structure. Technically, there is no Stott expansion (in the traditional sense) that can transform an ike into a bilbiro; the way we did it was by substituting parts of ike with parts of the great dodecahedron and applying Stott expansion to that instead, while simultaneously keeping other parts of ike as-is. So in a sense, it's really a pseudo-ike, not ike itself, that's transforming into a bilbiro; it's a kind of "frankenstein monster" double of ike, which is not really ike but a hybrid of ike and great dodecahedron. It just so happens that great dodecahedron has the same vertices as ike, so this hybridization "works", and seemingly it's ike itself that's transforming into bilbiro. But actually, it's the frankensteinian pseudo-ike that is undergoing the transformation.

I always assumed, and my analogies only work with this assumption for some reason, that the "frankenstein monster" you're talking about should be described as o5x3(-x). Now what is o5x3(-x)? First of all, it has pentagons (o5x). those pentagons are connected to weird triangles (x3(-x) ) with their x-edges. As Klitzing pointed out, those are actually hexagons, but I always say x3(-x) "=" o3x, because actually its just a weird o3x. Now those "triangles" are connected to o2(-x)'s, which is nothing, and then the (-x) is connected to another x3(-x). this thus gives a ike made of x3(-x)-triangles, with the loose edges (the x's) connected to pentagons. those pentagons are layed out in a great dodecahedron-like fashion. This means, o5x3(-x) is an ike with a great dodecahedron at it's interior, connected to the triangles by double edges.
Normally I don't bother about the faces of those negative node-things, and thus just say o5x3(-x) "=" o5o3x.
Similarly, things like D4.11 are produced technically not from the snub 24-cell itself, but from a "frankenstein double" of snub 24-cell where the icosahedra are hybrids of ike and great dodecahedron. The coincidence of vertices means that these "pseudo-ike"'s can be interpreted either way.
About this D4.11, I was thinking, could it be derived from the 600-cell instead of the snub 24-cell? This would mean, that if you insert teddi-pyramids in the teddi's, bilbiro-pseudopyramids in the bilbiro's, and maybe a tetrahedron in the cuboctahedra, does it become a convex shape? this shape would certainly be non-convex if you do it the described way, but my hope is that when you do it this way, the peppi's of the bilbiro-pseudopyramid and the teddi-pyramid would produce a connecting edge, just as what happens when you place two hemi-600-cells together.
Now the interesting part about some of the bilbiro'd and thawro'd polychora that we found, is that most of our constructions were by deleting some vertices or substituting some other vertices. So it's more like a partial (pseudo) Stott contraction than a Stott expansion. Also, this happens to a hybridized "frankensteinian" version of the polychoron instead of the "pure" convex version. For example, we produced thawro's in o5x3o3o by shrinking some vertex layer (sorry I can't remember the exact symbol off-hand, it was something like x5f3f) to x5x3x. This could be understood as the substitution of o5x3o3o with parts of some non-convex uniform with the same vertices, where the Stott contraction of the non-convex unform produces a x5x3x cross-section. Since the vertices of the non-convex and o5x3o3o are the same, the vertices can be interpreted either way, which gives us the flexibility to apply the Stott expansion/contraction to them, partly as if they are part of o5x3o3o and partly as if they are part of the non-convex polychoron.

This somewhat ties in with my previous observation about something fishy going on with 4D and the golden ratio, in the sense that the numbers somehow keep working out "coincidentally", sometimes in unexpected ways. This phenomenon can be seen to be less strange if we understand that given, say, some 600-cell family convex uniform, there are entire regiments (to use Bowers' term) of non-convex uniforms with the same vertices, each of which is related to other non-convex uniforms of various relative sizes with different vertices via Stott expansion/contraction/faceting/stellating. I suspect that many of the CRFs we have been finding are actually CRF pieces of some of the non-convex uniforms, or some combination of multiple non-convex uniforms. The fact that there are so many non-convex uniforms, and they are related to each other in many ways via various Stott contractions/expansions and stellations/facetings, means that there are many combinations of vertices that will have unit edge lengths (e.g., the relation between a uniform and its Stott-expanded version is a unit edge length difference), so they form some kind of pentagonal grid in 4D space that we can cut many CRFs out of. The coincidence of vertices of many non-convex uniforms with convex uniforms means that many of these pieces may have an underlying non-convex derivation, but they can be reinterpreted as convex, and therefore many of them become CRF.
I hope we don't have to go consider all the non-convex uniforms . Instead, I hope we only have to consider those that have a negative-node representation. e.g. the great dodecahedron can be written as o5x3(-x) "=" o5o3x, and the stellated dodecahedron as f5(-x)3o "=" o5o3x. I'm not able to do such a thing for the great stellated dodecahedron and the great icosahedron, probably because they are based on cuts that are deeper than one edge-length of the convex hull. Anyway, I thought as well this weird stott-expansion/contraction was the grid you were looking for, and that's why I said this might be something important
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### Re: Johnsonian Polytopes

Both for the bilbiro-2-ike(faceting) and the thawro-2-ike(faceting) transitions I've designed already some pics. They will be part of my next incmats webpage update. But I want to show them here already before.

bilbiro-2-ike(faceting) is clearly a true Stott expansion (not in the sense of Dynkin symbols but in the sense of Stott, incorporating the full symmetry of the object). It was only the false additional application of the convex hull, which resulted in the full symmetrical ike. And with respect to that one then it would be just a subsymmetry. But not so for the to be used here ike faceting.
ike2bilbiro.png

thawro-2-ike(faceting) in the same sense is a true Stott expansion. Again it is just a matter of selecting the correct ike faceting, to which it shall apply.
ike2thawro.png

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:Both for the bilbiro-2-ike(faceting) and the thawro-2-ike(faceting) transitions I've designed already some pics. They will be part of my next incmats webpage update. But I want to show them here already before.

bilbiro-2-ike(faceting) is clearly a true Stott expansion (not in the sense of Dynkin symbols but in the sense of Stott, incorporating the full symmetry of the object). It was only the false additional application of the convex hull, which resulted in the full symmetrical ike. And with respect to that one then it would be just a subsymmetry. But not so for the to be used here ike faceting.
ike2bilbiro.png

thawro-2-ike(faceting) in the same sense is a true Stott expansion. Again it is just a matter of selecting the correct ike faceting, to which it shall apply.
ike2thawro.png

--- rk

Nice pics!

And yes, this is what I was trying to say in my previous post, that the ike->bilbiro transformation is not really applied to ike itself, but to a (non-fully-symmetric) faceting of it. Ditto for the ike->thawro transformation.

What is interesting, is that because of the equivalence of vertices, we can substitute ikes with faceted_ike in a polychoron, and then apply the transformation to it to get a CRF result. So we have the apparent transformation of snub 24-cell (snub demitesseract) -> D4.11, whereas it's actually a kind of modified snub 24-cell in which the ikes are substituted with faceted_ike, to which Stott expansion is then applied.

In fact, now that I think of it more carefully, can the modified snub 24-cell with faceted_ike even be regarded as a valid polychoron? Because after the substitution, the cells no longer connect to ike in the way that a polytope should, since the pentagons in the faceting would be unconnected, and the tetrahedra touching the triangles of the full symmetric ike would also be unconnected, since those triangles no longer exist in faceted_ike! So I'm not sure how to analyse this situation.

In any case, the possibility of this kind of substitution (cell -> faceted_cell, for some arbitrary cell shape) means that there should be a very large number of CRFs that can be derived in this way from the uniform polychora. So it's a kind of substitution + Stott-expansion operation, which differs from the usual expansion processes in the initial substitution step, even though the expansion itself isn't fundamentally different.
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### Re: Johnsonian Polytopes

Well, substituting faceted ikes for real ikes clearly would not lead to a true polychoron. It surely has holes and unconnected faces. But you could substitute real ikes by an overlapping corealmic complex of 1 faceted ike plus 4 peppies (attached to the pentagons of the faceted ikes). This then closes again to full dyadicity. (Just that some triangles will be coplanar, but in an abstract sense to be distinguished, either one being incident to exactly 2 cells.)

It then is that this substitution itself already would break the full symmetry of sadi. In fact all formerly full symmetrical ikes become oriented. And then you could apply the normal Stott expansion to that thingy. But note, that intermedial thingy clearly can not be described by a Dynkin symbol anymore. Thus this Stott expansion as well cannot be meant to substitute some o nodes by x nodes in that Dynkin symbol. It rather is meant in the original sense, stretching the figure in its full symmetry (which here already is reduced!), blowing up "additional zero sized edges" to unity.

This then would result in local complexes of bilbiroes and 4 attached peppies each. - But you might want to do a local rearangement of final cells then. (Kind of in the sense of diminishings or augmentations.)

And, for sure, you could consider all that too in the other way round: Starting with your final shape with the bilbiroes (and no adjacent peppies). Then apply the invers transformation of the former, i.e. a normal Stott contraction thereof. This then would result in something similar to sadi, but ikes being replaced by my mentioned faceted ikes. And the former cells, adjacent to the bilbiroes pentagons then too would become transformed into something. And that "something" then assures that this contracted version still is a valide polychoron. I.e. it still will be dyadic. But it might come out to use retrograde cells for example, or other crude stuff. (I have not applied this thought to your example in detail so far, just speaking abstractly.)

quickfur wrote:So it's a kind of substitution + Stott-expansion operation, which differs from the usual expansion processes in the initial substitution step, even though the expansion itself isn't fundamentally different.

Yep. And it often is that this substitution process itself is what breaks symmetry. - Partial Stott expansion / contraction then is different as well: that one would transform only along a subsymmetry of the to be transformed figure. And the symmetry of that substituted figure clearly differs in general from the symmetry of the non-substituted figure! So even if the total aggregated transformation seems to act with respect to a subsymmetry only, it still might contain normal (non-partial) Stott transformations only.

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
quickfur wrote:So it's a kind of substitution + Stott-expansion operation, which differs from the usual expansion processes in the initial substitution step, even though the expansion itself isn't fundamentally different.

Yep. And it often is that this substitution process itself is what breaks symmetry. - Partial Stott expansion / contraction then is different as well: that one would transform only along a subsymmetry of the to be transformed figure. And the symmetry of that substituted figure clearly differs in general from the symmetry of the non-substituted figure! So even if the total aggregated transformation seems to act with respect to a subsymmetry only, it still might contain normal (non-partial) Stott transformations only.
[...]

This makes me wonder... are there CRFs which can be derived by full-symmetric Stott expansion? Obviously, the uniform polychora themselves are examples of this, but I'm thinking of something that's no longer uniform, e.g., is it possible to Stott-expand the bitruncated 24-cell o3x4x3o to get a CRF with augmented 24-cell symmetry? By this I mean, push out the truncated cubes radially and fill in the gaps with other CRF polyhedra. The result will not be uniform (since otherwise we'd have known about it before ), but can it still be CRF (or made CRF with some suitable modifications)?
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### Re: Johnsonian Polytopes

I've been thinking about the J92 rhombochoron. In another post, I suggested that it may be a kind of siamese-twin version of a 4D "fat teddy" made with J92 side cells around a triangular ridge.

However, I think perhaps there's a more mundane derivation: as a Stott expansion of the 600-cell. I haven't thought through the details yet, but note that the 600-cell has icosidodecahedral cross-sections (well, if you subtract the florets of 5 tetrahedra from it). I'm not sure if there's a non-convex uniform that has these cross-sections as cells, but suppose we take 4 such cross-sections, such that they make dichoral angles of 60° and 120° with each other. Consider these as part of a (hypothetical?) non-convex faceting of the 600-cell. Group them in two pairs, where within each pair the dichoral angle is 60°. Note that the icosidodecahedra within each pair will bisect each other. Now apply Stott expansion by pulling the pairs apart, until the top triangles of the two pairs coincide. Within each pair, the icosidodecahedra still bisect each other, but now we truncate them at the plane of bisection, and then shrink the phi-scaled hexagons to unit-edged hexagons to turn them into pairs of J92's. I believe the result should be the J92 rhombochoron.

This process is somewhat analogous to the (faceted) ike -> bilbiro transformation, in which 2 pairs of pentagons are pulled outwards until their top vertices coincide. Here we have 4 icosidodecahedra that are pulled apart until their tip triangles coincide. Well, we also truncate some stuff to make J92's, so I'm not 100% sure how that comes about, but I'm reasonably confident that at least the hyperplanes of the 4 J92 cells should correspond with 4 icosidodecahedral cross-sections of the 600-cell.
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### Re: Johnsonian Polytopes

quickfur wrote:
Klitzing wrote:[...]
quickfur wrote:So it's a kind of substitution + Stott-expansion operation, which differs from the usual expansion processes in the initial substitution step, even though the expansion itself isn't fundamentally different.

Yep. And it often is that this substitution process itself is what breaks symmetry. - Partial Stott expansion / contraction then is different as well: that one would transform only along a subsymmetry of the to be transformed figure. And the symmetry of that substituted figure clearly differs in general from the symmetry of the non-substituted figure! So even if the total aggregated transformation seems to act with respect to a subsymmetry only, it still might contain normal (non-partial) Stott transformations only.
[...]

This makes me wonder... are there CRFs which can be derived by full-symmetric Stott expansion? Obviously, the uniform polychora themselves are examples of this, but I'm thinking of something that's no longer uniform, e.g., is it possible to Stott-expand the bitruncated 24-cell o3x4x3o to get a CRF with augmented 24-cell symmetry? By this I mean, push out the truncated cubes radially and fill in the gaps with other CRF polyhedra. The result will not be uniform (since otherwise we'd have known about it before ), but can it still be CRF (or made CRF with some suitable modifications)?

For sure can true CRFs be derived by full Stott expansions.
E.g. bilbiro was shown to be derivable by true Stott expansion from some ike faceting.
And thawro too was shown to be derivable by true Stott expansion from some other ike faceting.

Note, the refered to facetings then show up themselves a full symmetry which is reducible (brique, resp. 3-fold axial) only! Accordingly "full" refers to any arbitrary irreducible component independently only, i.e. the mere 1D expansion required for bilbiro and the 3fold expansion required for thawro then are examples for full symmetric expansions for the respective facetings. - But surely not for the full symmetrical icosahedral hull thereof.

Just for comparision: the octahedron has as full symmetry clearly the octahedral one. But it could be Stott expanded wrt. any combination of individual axial directions as well, i.e. wrt. mere brique symmetry. This then derives either J15 (esquidpy - by 1D expansion), J28 (squobcu - by 2D expansion), or sirco (by 3d expansion). Only the last of those then is full (by coincidence, as a symmetrical brique is just a cube again), but the other ones correspond to a true subgroup. Thus those Stott expansions then are just partial ones (wrt. octahedral symmetries - but surely full valide ones wrt. mere brique subsymmetry). - This is what was intended by that attribute "partial".

--- rk
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### Re: Johnsonian Polytopes

The {3,3,5} ursulate xfo3oox5ooo&#xt, as well as xfo3oox5xxx&#xt are like the decagon surrounded by petagons, indde flat.
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### Re: Johnsonian Polytopes

wendy wrote:The {3,3,5} ursulate xfo3oox5ooo&#xt, as well as xfo3oox5xxx&#xt are like the decagon surrounded by petagons, indde flat.

Aha! So this must be it. This is the phi-based grid that I was looking for. If a 600-cell with the icosidodecahedral ursachoron tiles 4D space, then it should be possible to find a minimal set of 4D tiles, based on 600-cell sections (said ursachoron among them), that also does. This, together with the 5-cell + 16-cell complex that exhibits a concave phi-scale surface that most likely can be filled with some manner of pentagonal polychora, produces a 4D tessellation that allows a large number of phi-based CRFs to be "cut from the cloth", so to speak.
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### Re: Johnsonian Polytopes

There are pentagonal tilings in 2, 3, and 4 dimensions, these are infinitely dense, but just involve numbers of the form a+b\phi.

The 3d tiling centres around the group o5o5/2oAo. In 4D, there are many dynkin symbols for it, but o5/2o3o3z3o3o and o5o3o3o5/2o are examplers of it.

Of the 15 polytopes of the {5,3,3} type, all but the smallest and largest can be laced together in a flat tower. The ursulated {3,3,5} connects the x3o3o5o and o3x3o5o.
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### Re: Johnsonian Polytopes

The 4D locally dense groups, I'm aware of, are
Code: Select all
o5o3o3o5/2oo5/2o5o3o5/2oo5o5/2o5o5/2oo5o3o5/2o5oo3o5o5/2o3oo3o3o3o3o5/2*co3o3o3o3/2o5*c...

Some of the therefrom derived flat segmentotera are
Code: Select all
pt || gaghi        = ox5oo5/2oo3oo&#xragaghi || tigaghi = ox5xx5/2oo3oo&#xrigfix || sirgaghi = ox5oo5/2xx3oo&#xgofix || quipdohi  = ox5oo5/2oo3xx&#xpt || sishi        = ox5/2oo5oo3oo&#xrofix || sirsashi  = ox5/2oo5xx3oo&#xfix || padohi      = ox5/2oo5oo3xx&#xpt || sidtixhi     = ox3oo3oo3oo5/2*b&#xpt || gidtixhi     = ox3oo3oo3/2oo5*b&#x...

--- rk
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### Re: Johnsonian Polytopes

You forgot the tiling of octogonny, which is a different, non-wythoff group, whose dual is a tiling of octagrammy.

I generally count there to be five quarterion integers, numbered from Q2 to Q5 (Q1 is the regular integers). The eutactic star is the set of points on the integer glome, which is closed to multiplication, and the span of them form the set of integers. The size of the sets look suspiciously like the orders of the 3d groups.

Q2 has 8 members, the group forms {4,3,3,4}

Q3 has 24 members, the group form {3,3,4,3}

Q4 has 48 members, forms the non-wythoffian tiling of octagonny, the dual being the isomorph (like 8,8/3 and 12,12/5). The tiling is mirror-edge and mirror-margin, but it needs six mirrors to create the group.

Q5 has 120 members, forming the {5/2,3,3,5}. There are ten regular starry groups, and a mitful of others. One notes, for example, that the pentagon-pentagram tegum (as well as the prism), is equalateral, and would be a RH figure, but sadly, not a CRH figure. Lots of interesting things are to be found among Q5. Note that there are figures that exist in 5v (ie {5/2.3.3.5} etc, that can not be directly constructed in unit-edge Q5 figures (the pentachoron is an example: the vertices of larger examples exist, but a unit-edge pentachoron can not be made.
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### Re: Johnsonian Polytopes

wendy wrote:There are pentagonal tilings in 2, 3, and 4 dimensions, these are infinitely dense, but just involve numbers of the form a+b\phi.

The 3d tiling centres around the group o5o5/2oAo. In 4D, there are many dynkin symbols for it, but o5/2o3o3z3o3o and o5o3o3o5/2o are examplers of it.

Of the 15 polytopes of the {5,3,3} type, all but the smallest and largest can be laced together in a flat tower. The ursulated {3,3,5} connects the x3o3o5o and o3x3o5o.

Interesting. So this means there should be lots of CRFs made from convex fragments of these tilings / flat lace towers.

Also, do you have a page listing the various node symbols (x, f, F, q, z, etc.) and their corresponding values? I know a few, but the rest elude me. I've also been thinking if it may not be amiss to use numerical notation for them instead, since they are after all an oblique coordinate system!
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### Re: Johnsonian Polytopes

quickfur wrote:...
Also, do you have a page listing the various node symbols (x, f, F, q, z, etc.) and their corresponding values? I know a few, but the rest elude me. I've also been thinking if it may not be amiss to use numerical notation for them instead, since they are after all an oblique coordinate system!

You might be looking for http://bendwavy.org/klitzing/explain/dynkin-notation.htm?

--- rk
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### Re: Johnsonian Polytopes

Removed full last post quote. ~Keiji

Thanks, that's exactly what I was looking for.
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### Re: Johnsonian Polytopes

The list of used nodes as follows.

• a atom node. Like x, but does not add. Used to show compounds
• b bevel-node. It works like m, but no addition. A bevelled cube would be o4b3b.
• d divisor-node Used with p or w, to show a star form, eg p5d2 for the shortchord of {5/2}
• f A(5) = 1.61803398875 also /f
• g dual of 's' node
• h A(6) = 1.73205080757 also /h
• i Creates a supplement, or negative edge, eg fi = -1.61803398875, 5i = 5/4
• o zero node = A(2) = 0.0000000 also /r
• p polygon node, followed by edge-count or girthed polytope, eg p10, p(3,3,5) are the same thing
• q A(4) = 1.41421356238 also /q
• r A(2), used for creating a zero-sized polytope, eg r4o3o = 0-sized cube also /r
• s snub node (alternated x in that position)
• u 2.00000000 = W4 also /u
• v A(5/2) = 0.61803398875 also /v
• w Used to introduce a numeric shortchord-square, eg w4 = horogon. Mainly hyperbolic.
• x A(3) also /
• z first loop-node, also ':' (*a) zz represents the second loop-node (*b)
• * Klitzing's re-node mark. Governs a letter counted from the beginning
• . Context suppress node. Used to show the context of a sub-symmetry in a group, eg .3o3x vs x3o3. are both tetrahedra, but different positions in the runcinated pentachoron.

Note that the re-node mark is itself a node, different to the node it abutts. That is, there is an implied comma between the two nodes.

These symbols are reserved with specific meanings

• ",,," quotes may be used to separate a symbol from the surrounds, eg "x5o5x"
• (...) what is in brackets resolve to a shortchord or polygon of the same size. Also (W=q+1) is acceptable.
• [...] A segment that may be freely repeated, eg x[3x] is any of x, x3x, x3x3x, x3..x3x

These letters are in various use, to mean specific non-three branches. It is particularly used when a number represents a chain of 3's. eg replaces coxeter's 2_21 style. A branch connects a 'first subject' to a 'first object'. If the branch is redirected to an earlier node, then a higher-order object or subject node is used.

X_Y1 becomes (X+Y) + Y-subject-node, eg 4_21 becomes 6B. Slashes may be inserted to show positions of x-nodes, eg 1/5B is the rectified 4_21. This is used, eg in preference to o3x3o3o3o3o3oBo.

• A A second-subject 3-branch
• B A third-subject 3-branch
• C A fourth-subject 3-branch
[*}D a divisor-branch, eg P8D3 is a branch marked 8/3
• E A second-object branch
• F A branch marked '5'
• G A third-object branch
• H A branch marked 8
• I An inversion-marker, eg PI/D = P/(P-D)
• O The sphere-node. A sphere is rated as a regular polytope, its truncates are increasing diameters of ellipsoids.
• P A polygon by edge count
• Q A branch marked 4
• R A branch marked 2, when it governs a number, it reduces the denominator and numerator, eg P8D1D2R4 is (8,4)
• S A branch marked 3
• U branch marked \infty, but also refers to the horotopic infinity only.
• V branch marked 5/2
• W branch representing
• X a branch marked '2'
• & A discontinuous branch. The node-counts are reset after this branch, for example

Note that Richard Klitzing uses a different order in the CRH (ie CRF) project. I have not traced all of his forms as yet, but this list is instrumentive.

• w = q+1
• F = f+1
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