## Bilbirothawroids (D4.3 to D4.9)

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

### Re: Johnsonian Polytopes

quickfur wrote:Before we get too deep into the details, don't forget that the triangles in the triangular tiling of a grand antiprism's rings are not coplanar, and they are not all equivalent. Consider pairs of triangles that share an edge perpendicular to the direction of the ring. Some pairs form an obtuse angle where 2 tetrahedra may fit, whereas the other pairs form a reflex angle where three tetrahedra may meet. So, the triangles are not all equivalent, and this may change the possible CRF topologies in a profound way.

Edit: hmm, it seems that I was wrong, the pairs are all equivalent. I take that back.

No problem
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### Re: Johnsonian Polytopes

gap (the great antiprism) has 3 types of triangles:

Even the edges of gap fall into 3 classes:
- those sides of the pentagons,
- the lacings of the paps (pentagonal antiprisms), and
- the joining ones of those 2 "rings".

In terms of the latter triple the former one can be described as having side class counts:
- (1,2,0),
- (1,0,2),
- (0,1,2).

--- rk
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### Re: Johnsonian Polytopes

Building GAP-like polytopes is something i toyed around with ever since i wrote it as j5j2j5j.

It's most unlikely to work if the numbers are of different integer systems, that is, the only cases i would consider is jMj2jNj, where M is odd, and N=2M, or where M=N. Otherwise, the thing isn't going to close. j7j2j14j or j9j2j18j might be the most viable cases, in this case.

That is, you start with m 2M antiprisms, and 2M m antiprisms, but that won't work if M is odd. So you might look at 2m m antiprisms on both arms.
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### Re: Johnsonian Polytopes

wendy wrote:Building GAP-like polytopes is something i toyed around with ever since i wrote it as j5j2j5j.

It's most unlikely to work if the numbers are of different integer systems, that is, the only cases i would consider is jMj2jNj, where M is odd, and N=2M, or where M=N. Otherwise, the thing isn't going to close. j7j2j14j or j9j2j18j might be the most viable cases, in this case.

That is, you start with m 2M antiprisms, and 2M m antiprisms, but that won't work if M is odd. So you might look at 2m m antiprisms on both arms.

Have you ever explored the option of middle layer of vertices? It looks like the extra freedom gained from dihedral angles on this layer could help close the shapes...
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### Re: Johnsonian Polytopes

Kind of you are investigating the sequence   "gudap -> gap -> ???"

gudap = s5s2s5/3s, consisting of 10 pentagonal antiprisms (paps), 10 pentagrammal retro(-grade anti-)prisms (starps), and 50 joining tetrahedra (tets). Here the vertices of the 2 conjugate rings of antiprisms do coincide. Thus you will need for rather few glueing tets.

(A close relative here is s2s2s2s = hex. Those then are already the only uniform members of the sNs2sMs series.)

gap = j5j2j5j, consisting of 2x10 paps, 2x100 adjoining tets (used as trigonal pyramids), and 1x100 isolated tets (used as digonal antiprisms). Here the vertices of either ring is separated from those of the other one, but there are supporting edges (of the same size!), which bind those 2 rings together, as well as ensuring the right distance between them.

(A close relative to that one is padiap = j5/3j2j5/3j, using 2x10 starps instead of those paps. - In fact, just as gap was a faceting of ex = x3o3o5o, padiap can be obtained as a similar faceting of gax = x3o3o5/2o. - Again those 2 are already the only uniform members of the series jNj2jMj.)

And now you are considering a further set of vertices in kind an "equatorial layer" between those 2 rings of antiprisms...

--- rk
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### Re: Johnsonian Polytopes

Klitzing: Yes, based on the sudden revelation that the midpoints of trans-layer edges in grand antiprism form a layer that contains a snub square tiling. I suspect that this tiling could work as a middle layer of vertices because it's exactly what you need for transition between two rings of antiprisms.
But I'm lost in the particulars of the math, can't compute the coordinates properly

It's a bit similar logic as taking a tetrahedron as point||triangle, finding that midpoints of lateral edges form another triangle and deducing existence of tower point||triangle||triangle (which would be elongated triangular pyramid). It's not exact operation that would guarantee existence, it's more of a signpost showing that something interesting might lie that way.

For example, if you try the same thing with tetrahedron as line || perpendicular line, breaking it in half and scaling the square to have unit edge, you'll end up with gyrobifastigium...

In this case, I'm trying to "break in half" the layer of tetrahedra between the antiprismatic rings. Some tetrahedra will break into triangular prism + tetrahedron, while other will break into triangular prism + gyrated triangular prism. And while there's of course no space to do it with the real rings in grand antiprism, the structure of the middle layer is repeating and can accomodate antiprismatic rings of any size, so it's possible that there will be a size that fits (personally, I suspect two rings of 20 decagonal antiprisms could work, but it's just a hunch without any evidence but intuition).
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### Re: Johnsonian Polytopes

Yesterday, I got inspired by Marek's ideas to write a program that applies m,n-duoprismic symmetry to a set of input seed points. It's a kind of generalization of the Wythoff construction where given a seed point, it rotates it through all combinations of WX and YZ rotations according to m,n-duoprism symmetry. The idea is that instead of grappling with individual coordinates, we just have to find the seed point(s) of the shape we want to construct (assuming it has m,n-duoprism symmetry), and it will generate the coordinates of the full polychoron for you.

I haven't gotten very far yet, but I might work a little more on it today and test it on the grand antiprism -- I think it's possible to generate the grand antiprism from 4 seed points and 5,5-duoprism symmetry: two points per ring to specify the antiprisms, to get the 100 vertices. It is, of course, not limited to the grand antiprism; you can insert intermediate layers of vertices by adding the appropriate seed points, so hopefully this will turn out to be a useful tool for investigating this class of shapes.

One idea I have, related to this, is CRF expansions of the duoprisms: by expanding the rings of a duoprism outwards (resp. shrink the cells in each ring and insert more cells per ring to fill the gaps), then fill up the intermediate gaps with CRF cells. A kind of "runcinated" duoprism, if you will. I don't know what intermediate cells would "work" yet; I'm thinking probably square pyramids on the square faces of the prisms, so that the orientation change between the rings can be effected in a CRF way.
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### Re: Johnsonian Polytopes

quickfur wrote:Yesterday, I got inspired by Marek's ideas to write a program that applies m,n-duoprismic symmetry to a set of input seed points. It's a kind of generalization of the Wythoff construction where given a seed point, it rotates it through all combinations of WX and YZ rotations according to m,n-duoprism symmetry. The idea is that instead of grappling with individual coordinates, we just have to find the seed point(s) of the shape we want to construct (assuming it has m,n-duoprism symmetry), and it will generate the coordinates of the full polychoron for you.

I haven't gotten very far yet, but I might work a little more on it today and test it on the grand antiprism -- I think it's possible to generate the grand antiprism from 4 seed points and 5,5-duoprism symmetry: two points per ring to specify the antiprisms, to get the 100 vertices. It is, of course, not limited to the grand antiprism; you can insert intermediate layers of vertices by adding the appropriate seed points, so hopefully this will turn out to be a useful tool for investigating this class of shapes.

One idea I have, related to this, is CRF expansions of the duoprisms: by expanding the rings of a duoprism outwards (resp. shrink the cells in each ring and insert more cells per ring to fill the gaps), then fill up the intermediate gaps with CRF cells. A kind of "runcinated" duoprism, if you will. I don't know what intermediate cells would "work" yet; I'm thinking probably square pyramids on the square faces of the prisms, so that the orientation change between the rings can be effected in a CRF way.

That's great! Should prove interesting.

One of the things I've mentioned here was that the runcinated tesseract is actually a convex envelope of (4,8)-duoprism (sign changes of (1,1,1,1+sqrt(2)) and (1,1,1+sqrt(2),1) and (8,4)-duoprism (sign changes of (1,1+sqrt(2),1,1) and (1+sqrt(2),1,1,1) -- the cubic cells of both prisms will stay while the octagonal prisms disappear, replaced by tetrahedra, cubes and triangular prisms formed by joinings of vertices of both duoprisms. This illustrates something in this vein.

Next interesting type of layer would probably be prism/antiprism alternating rings. But I'm not sure whether there's a way to smoothly change this layer into another.

Another idea might be to expand a duoprismatic (square) layer into a truncated square layer of squares and octagons, but not sure yet how the lateral cells would look...
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### Re: Johnsonian Polytopes

Hi Marek14,

your idea was to insert a medial vertex layer into generalized great antiprisms (gap), i.e. having then some n-ap and m-ap rings (perhaps n and m to be spezified a posteriori) and some portion of medial snasquat (as it would occur from medially dissecting the glueing tets of gap: the ap-adjacent ones would have trigonal sections, while the isolated ones would have square sections).

You further plan to connect that scrambled snasquat portion to the ap-triangles (i.e. a scrambled trat portions) by means of trip = {3}||{3} whenever an n-gon side is used in the trat portion, tet = {3}||pt for the other snasquat triangles, and trip = {4}||line.

This results for the antiprism base sides to having 2 antiprisms (with angle between n-gon and triangle) + 2 trips (with angle between square and triangle) incident. For n > 2 this results already in an angle sum > 360 degrees, which is impossible for non-hyperbolic geometries.

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:Hi Marek14,

your idea was to insert a medial vertex layer into generalized great antiprisms (gap), i.e. having then some n-ap and m-ap rings (perhaps n and m to be spezified a posteriori) and some portion of medial snasquat (as it would occur from medially dissecting the glueing tets of gap: the ap-adjacent ones would have trigonal sections, while the isolated ones would have square sections).

You further plan to connect that scrambled snasquat portion to the ap-triangles (i.e. a scrambled trat portions) by means of trip = {3}||{3} whenever an n-gon side is used in the trat portion, tet = {3}||pt for the other snasquat triangles, and trip = {4}||line.

This results for the antiprism base sides to having 2 antiprisms (with angle between n-gon and triangle) + 2 trips (with angle between square and triangle) incident. For n > 2 this results already in an angle sum > 360 degrees, which is impossible for non-hyperbolic geometries.

--- rk

Darn, looks like it...
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### Re: Johnsonian Polytopes

I'm not quite giving up hope yet, though. I'm almost certain that some manner of lacing cells, probably involving intermediate vertex layers, will be able to interface two rings of antiprisms in a CRF way. I've just been busy at work so I haven't been able to do much in this direction yet.
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### Re: Johnsonian Polytopes

My main problem with linking up a set of antiprisms, is that one is going to produce coordinates in different integer systems, like a+rb.5 vs a+b.r3. (eg pentagon-dodecagon schemes). But there are not many easy polytopes that make this jump. You might get something close, but i ought be supprised that it would close up.
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### Re: Johnsonian Polytopes

wendy wrote:My main problem with linking up a set of antiprisms, is that one is going to produce coordinates in different integer systems, like a+rb.5 vs a+b.r3. (eg pentagon-dodecagon schemes). But there are not many easy polytopes that make this jump. You might get something close, but i ought be supprised that it would close up.

What about if the two rings of antiprisms belong to the same integer system? Is it possible that they might close up given enough intermediate layers between them?
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### Re: Johnsonian Polytopes

They evidently do, since the GAP is an example of that. But it pretty much limits one to have p q-AP's in each circle. You could have p q-AP against q p-AP would work, eg 10 dodecagons vs 12 decagons could work, because the integer system here is Z5Z6 vs Z6Z5, ie the same thing. But to expect something like a heptagon to hook up to a decagon is beyond the pale.
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### Re: Johnsonian Polytopes

wendy wrote:They evidently do, since the GAP is an example of that. But it pretty much limits one to have p q-AP's in each circle. You could have p q-AP against q p-AP would work, eg 10 dodecagons vs 12 decagons could work, because the integer system here is Z5Z6 vs Z6Z5, ie the same thing. But to expect something like a heptagon to hook up to a decagon is beyond the pale.

Well, I don't think we were insisting on hooking up heptagons to decagons. We were mainly looking for CRFs with duoprismic symmetry that aren't themselves duoprisms or some augmentations thereof.

One possible direction I've been considering is how square pyramid interfacing cells would work in a duoprismic CRF. Suppose we have a ring of n m-prisms. The first thing to note is that we cannot put square pyramids on adjacent squares between two members of the ring, because that would require a tetrahedron to close the shape up, but square_pyramid + tetrahedron + square_pyramid has a total angle of 180°, so you won't be able to bend the m-prisms around in 4D to form a ring, you'll end up with an infinite sequence of m-prisms, square pyramids and tetrahedra lying flat on a 3D hyperplane. The only way to not have to insert a tetrahedron between the square pyramids is if the ring's dichoral angle is small enough that the apices of the square pyramids meet, in which case the shape becomes the equivalent of an m,n-duoprism with prism pyramids erected on the m rings of n-prisms (and the CRFability depends on the specific combination of m and n, which have been studied before). Similarly, we cannot insert anything larger than a tetrahedron between the square pyramids, because then it becomes hyperbolic (angle defect < 0). Nor can we insert anything else between them, because the smallest triangle-triangle dihedral angle is that which is present in the tetrahedron.

So this means that if we're going to use square pyramids as linking cells for the ring of n m-prisms, then they must be non-adjacent (i.e., cannot share an edge on the m-prisms). This leads to the question of what else we can attach to the square faces on the m-prisms adjacent to the square pyramids, in order to form a linking layer of cells. One possibility is to use triangular prisms, so that every other m-prism will have a triangular prism on its square face, and the neighbouring m-prisms have square pyramids adjacent to it. This leads to a pattern of alternating square pyramids and triangular prisms around the ring, which immediately imposes the constraint that n must be even. Furthermore, since the ring of n m-prisms is assumed to project to a regular n-gon, this additionally limits n to be ≤10, because otherwise, it would imply the existence of a CRF (n/2)-gonal cupola for n/2 > 5, which we know is impossible. For the case n=8, this produces the runcinated tesseract or any of its gyrated variants. I'm not sure if n=10 or n=6 are possible yet (n=4 would require the orthogonal ring to become degenerate; probably not what we were looking for but it could yield a CRF, possibly -- but for now I'm ignoring that case). This is one possible direction to look.

Other directions to look are having additional vertex layers between the two rings, which means 2 or more layers of intermediate cells. I'm not sure whether this is possible with pure duoprismic symmetry yet. What I've been seeing in my experiments with my new duoprism vertices generator is that CRFs with duoprismic symmetry are much more constrained than might seem at first glance, so there may not be that many CRFs in this direction after all.
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### Re: Johnsonian Polytopes

Just dreamed up 2 such duoprismatic symmetrical figures. But so far I do not know whether those can truely exist, nor for which n and m. - Someone likes to come in?

First one:
Take a stack of 2m n-prisms and form a ring out of it within 4D. Then think of those n-prisms of being alternating ones. Attach to the squares of the even ones triangular prisms, the square-opposite edge then being parallel to the lacing edges of the n-prism. Additionally insert into the gap (i.e. incident to the n-prism lacing edge) a further triangular prism. Thus both being used as digonal cupolae, the first ones pointing up, the second ones pointing down. - Then to the other n-prisms in odd positions. Attach to their squares square pyramids each, and incident to the lacing edges of the n-prism tetrahedra into those gaps. - Then the remainder should be fillable by a second ring of prisms. Those prisms then should have m squares each (attaching to the second type of triangular prisms). And there should be n of those m-prisms, because there are n such trips of the second kind around each former n-prism.
That is, in conclusion, we need 2m n-prisms, 2nm trips, nm squippies, nm tets, and n m-prisms.

Second kind:
Take a stack of 2m 2n-prisms and form a ring out of it within 4D. Again think of those 2n-prisms of being alternating ones. Attach to the squares of the even ones alternatingly triangular prisms and cubes. Again the opposite edge of the triangular prism (seen as digonal cupola) will be parallel to the lacing edges of the 2n-prism. Nothing else would fill the gap here. - Then the other 2n-prisms in odd positions. Here attach alternatingly square pyramids and triangular prisms onto the squares. Obviously the triangular prisms in here would serve as digonal cupolae again, but here the opposite edge would be parallel to the base edges of the 2n-prism. - Then the remainder should be fillable by a second stack of prisms. Those prisms again would have m squares again. And we will need exactly n of those, because we have exactly n cubes being attached to each of the former 2n-prisms.
That is, in conclusion, here we need 2m 2n-prisms, 2nm trips, nm cubes, nm squippies, and n m-prisms.

What do you think about those?

--- rk
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### Re: Johnsonian Polytopes

Thought up a third one too:

Take a stack of 2m n-prisms. Again alternating ones. Even ones having cubes attached to all squares, nothing in between. Odd ones having triangular prisms attached, all being used as digonal cupolae, with opposite edge being parallel to base sides of the n-prism. - Then the remainder should be fillable by a further orthogonal ring of prisms again: those should then be n m-prisms.
Thus in conclusion we use 2m n-prisms, nm cubes, nm trips, and n m-prisms.

Possible at all? And then: for which n, m?

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### Re: Johnsonian Polytopes

Hm, this is something quickfur's program should be able to determine
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### Re: Johnsonian Polytopes

Marek14 wrote:Hm, this is something quickfur's program should be able to determine

Unfortunately it's not that simple... all the program currently does is to apply WX and YZ rotations to the input points; it doesn't attempt to deduce what those points should be in order to be CRF, etc.. Given one or more seed points (w,x,y,z), it computes (w cos T - w sin T, x sin T + x cos T, y cos U - y sin U, z sin U + z cos U) for all combinations of T and U pertaining to m,n-duoprism symmetry, for some specified m and n. If someone can compute (w,x,y,z) for the seed points needed to generate these shapes, then I can run it through to see if it comes out CRF, but otherwise this does still require some manual calculation.
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### Re: Johnsonian Polytopes

Klitzing wrote:Just dreamed up 2 such duoprismatic symmetrical figures. But so far I do not know whether those can truely exist, nor for which n and m. - Someone likes to come in?

First one:
Take a stack of 2m n-prisms and form a ring out of it within 4D. Then think of those n-prisms of being alternating ones. Attach to the squares of the even ones triangular prisms, the square-opposite edge then being parallel to the lacing edges of the n-prism. Additionally insert into the gap (i.e. incident to the n-prism lacing edge) a further triangular prism.

Hmm. So the cells around the n-prism lacing edge would be 2 n-prisms + 3 triangular prisms? Wouldn't that have an angle defect of 0? EDIT Sorry, my mistake, I misread your description. There is only a single n-prism here, so there should be positive angle defect.

[...]
Second kind:
Take a stack of 2m 2n-prisms and form a ring out of it within 4D. Again think of those 2n-prisms of being alternating ones. Attach to the squares of the even ones alternatingly triangular prisms and cubes. Again the opposite edge of the triangular prism (seen as digonal cupola) will be parallel to the lacing edges of the 2n-prism. Nothing else would fill the gap here. - Then the other 2n-prisms in odd positions. Here attach alternatingly square pyramids and triangular prisms onto the squares. Obviously the triangular prisms in here would serve as digonal cupolae again, but here the opposite edge would be parallel to the base edges of the 2n-prism. - Then the remainder should be fillable by a second stack of prisms. Those prisms again would have m squares again. And we will need exactly n of those, because we have exactly n cubes being attached to each of the former 2n-prisms.
That is, in conclusion, here we need 2m 2n-prisms, 2nm trips, nm cubes, nm squippies, and n m-prisms.

This one looks promising. If I read it correctly, I think it should produce the runcinated tesseract x4o3o3x for the case m=n=4. Not sure if the other cases are possible, but I was thinking along similar lines, so I might try to investigate this sometime.
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### Re: Johnsonian Polytopes

Some more thoughts about Klitzing's first idea:

So the odd n-prisms in the first ring would be surrounded by alternating square pyramids and tetrahedra, right? So the one edge of the tetrahedra joins to the lacing edge of the n-prism, and the opposite edge connect the apices of the square pyramids. Assuming CRF cells, this means these opposite edges around the n-prism must make a regular n-gon in dual orientation to the prism. Therefore the edges joining this n-gon to either of the prism's n-gon faces will trace out an n-antiprism. So if we were to take a single n-prism and the square pyramids and tetrahedra around it in isolation, the convex hull should be a n-prism||gyro n-gon wedge, or n-antiprism||n-gon.

Now the even n-prisms in the first ring are surrounded by triangular prisms: the triangular prism that touches only a single edge of the n-prism has triangular faces with edges that link the vertices (of the other triangular prisms) that don't lie on the n-prism into a regular n-gon, so if we were to take a single n-prism with the triangular prisms around it in isolation, the convex hull should be an n-antiprism prism.

So what you're doing here essentially amounts to stacking a bunch of alternating n-antiprism||n-gon and n-antiprism prism segmentochora together at their n-antiprism cells (which become internal to the resulting polytope), to form a 4D ring that wraps around an orthogonal ring of m-prisms. If this ring closes up, then the result should be CRF (barring any unexpected overhanging bits that may introduce concave cell patches), by construction. So, what determines whether this ring closes up, is whether the dichoral angle of the two n-antiprisms in the n-antiprism||n-gon segmentochora evenly divides the circle (the n-antiprism prisms have a dichoral angle of 90° at the joining n-gons, so they don't contribute to the overall turning angle of the m-gon). Since these segmentochora are in your list, you probably can answer that question yourself.
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### Re: Johnsonian Polytopes

And now, for your second idea:

Klitzing wrote:[...]
Second kind:
Take a stack of 2m 2n-prisms and form a ring out of it within 4D. Again think of those 2n-prisms of being alternating ones. Attach to the squares of the even ones alternatingly triangular prisms and cubes. Again the opposite edge of the triangular prism (seen as digonal cupola) will be parallel to the lacing edges of the 2n-prism. Nothing else would fill the gap here.

OK, so if we take a single 2n-prism and its surrounding triangular prisms and cubes in isolation, the convex hull should be an n-cupola prism.

- Then the other 2n-prisms in odd positions. Here attach alternatingly square pyramids and triangular prisms onto the squares. Obviously the triangular prisms in here would serve as digonal cupolae again, but here the opposite edge would be parallel to the base edges of the 2n-prism.

OK, so if we take a single 2n-prism here with its surrounding square pyramids and triangular prisms, then the convex hull is an n-gonal magnabicupolic ring, or n-cupola||2n-gon.

So again, whether this ring of 2n-prisms will close up, will be determined by whether the dichoral angle between the cupola cells in n-cupola||2n-gon evenly divides the circle, since the n-cupola prisms don't contribute to the overall turning angle of the m-gon.

Here, we know at least one CRF solution, n=4, since 4-cupola||8-gon has a dichoral angle of exactly 90°, so we can stack exactly 4 pairs of 4-cupola||8-gon + 4-cupola prism around a ring of 8 cubes, and the result is x4o3o3x.

What are the dichoral angles between the n-cupola in the other two n-cupola||2n-gon segmentochora? Do they evenly divide the circle?
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
Take a stack of 2m n-prisms. Again alternating ones. Even ones having cubes attached to all squares, nothing in between. Odd ones having triangular prisms attached, all being used as digonal cupolae, with opposite edge being parallel to base sides of the n-prism. - Then the remainder should be fillable by a further orthogonal ring of prisms again: those should then be n m-prisms.
Thus in conclusion we use 2m n-prisms, nm cubes, nm trips, and n m-prisms.
[...]

The even n-prisms with their surrounding cubes, taken in isolation, form n-prism prisms (aka n,4-duoprisms). The odd n-prisms, with their surrounding square pyramids, form n,3-duoprisms. So this arrangement should be CRF if the dichoral angle between the n-prisms in the n,3-duoprism evenly divides the circle.

Hmm. Seems we are in luck here! If I didn't make a careless mistake, the dichoral angle should always be pi/3, independently of n, so it should always be possible to close things up with a 12-membered ring, basically the gluing together of 6 n,3-duoprisms with 6 n,4-duoprisms into a ring, and closing up the remaining gaps with an orthogonal ring of m-prisms.

Oh, hold on a sec. I think i've studied this case before... The conclusion I got was that the cubes and triangular prisms will be coplanar with the m-prisms, so they will merge into 2m-prisms, and therefore the result is nothing other than an n,12-duoprism. So this doesn't give us anything new.
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### Re: Johnsonian Polytopes

quickfur wrote:And now, for your second idea:

Klitzing wrote:[...]
Second kind:
Take a stack of 2m 2n-prisms and form a ring out of it within 4D. Again think of those 2n-prisms of being alternating ones. Attach to the squares of the even ones alternatingly triangular prisms and cubes. Again the opposite edge of the triangular prism (seen as digonal cupola) will be parallel to the lacing edges of the 2n-prism. Nothing else would fill the gap here.

OK, so if we take a single 2n-prism and its surrounding triangular prisms and cubes in isolation, the convex hull should be an n-cupola prism.

- Then the other 2n-prisms in odd positions. Here attach alternatingly square pyramids and triangular prisms onto the squares. Obviously the triangular prisms in here would serve as digonal cupolae again, but here the opposite edge would be parallel to the base edges of the 2n-prism.

OK, so if we take a single 2n-prism here with its surrounding square pyramids and triangular prisms, then the convex hull is an n-gonal magnabicupolic ring, or n-cupola||2n-gon.

So again, whether this ring of 2n-prisms will close up, will be determined by whether the dichoral angle between the cupola cells in n-cupola||2n-gon evenly divides the circle, since the n-cupola prisms don't contribute to the overall turning angle of the m-gon.

Here, we know at least one CRF solution, n=4, since 4-cupola||8-gon has a dichoral angle of exactly 90°, so we can stack exactly 4 pairs of 4-cupola||8-gon + 4-cupola prism around a ring of 8 cubes, and the result is x4o3o3x.

What are the dichoral angles between the n-cupola in the other two n-cupola||2n-gon segmentochora? Do they evenly divide the circle?

Well, the height of the n-gon atop the 2n-p in that segmentochoron is h = sqrt[3 - 1/sin^2(pi/n)]/2. And the latteral displacement of the 2n-g clearly is 1/2 (half an edge length). This readily provides the tangens of half of the desired angle right from those 2 cathetes. You even could rewrite that as Sinus (provided that tangens), and then the cosinus of the doubled angle, i.e. of the desired one. This results in

cos(dichoral angle between the n-cupola) = (2 sin^2(pi/n) - 1)/(4 sin^2(pi/n) - 1)

esp.
n = 3 -> arccos(1/4) = 75.522488 degrees
n = 4 -> 90 degrees
n = 5 -> 144 degrees

That is, n = 4 provides the only convex solution, which then has a cell consist of 8 ops + 32 trips + 16+4 cubes, and 16 squippies. - Here the 8 ops form the one ring. The latteral squares of those form a 8x8 chessboard, whare alternate black fields get squippies, the remaining black ones get cubes. The white fields get digonal cupolae (trips) in the obvious orientations. The remainder here is a second orthogonal ring from 4 cubes. And indeed, the dihedral angle of a cube is 360 - 90 - 2*90 = 90 degrees.

But n = 5 would also provide a solution here, even so it isn't convex any more. There we use 10 dips + 50 trips + 25 cubes + 25 squippies + 5 stips. - Here the 10 dips form a first ring. The latteral squares thereof here form a larger 10x10 chessboard, again likewise been filled. The remainder here will be a ring of 5 stips (pentagrammal prisms). Indeed, the dihedral angle at their lacing edges is 360 - 144 - 2*90 = 36 degrees.

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:[...]
Take a stack of 2m n-prisms. Again alternating ones. Even ones having cubes attached to all squares, nothing in between. Odd ones having triangular prisms attached, all being used as digonal cupolae, with opposite edge being parallel to base sides of the n-prism. - Then the remainder should be fillable by a further orthogonal ring of prisms again: those should then be n m-prisms.
Thus in conclusion we use 2m n-prisms, nm cubes, nm trips, and n m-prisms.
[...]

The even n-prisms with their surrounding cubes, taken in isolation, form n-prism prisms (aka n,4-duoprisms). The odd n-prisms, with their surrounding square pyramids, form n,3-duoprisms. So this arrangement should be CRF if the dichoral angle between the n-prisms in the n,3-duoprism evenly divides the circle.

Hmm. Seems we are in luck here! If I didn't make a careless mistake, the dichoral angle should always be pi/3, independently of n, so it should always be possible to close things up with a 12-membered ring, basically the gluing together of 6 n,3-duoprisms with 6 n,4-duoprisms into a ring, and closing up the remaining gaps with an orthogonal ring of m-prisms.

Oh, hold on a sec. I think i've studied this case before... The conclusion I got was that the cubes and triangular prisms will be coplanar with the m-prisms, so they will merge into 2m-prisms, and therefore the result is nothing other than an n,12-duoprism. So this doesn't give us anything new.

Don't be afraid. It just has been a try.

And it is indeed amasing that it becomes Independent of n, but m has always to be 6.
And then that some cells become corealmic, is also not a usual behavior.
It is just that those resulting n,12-duoprisms are nothing new...

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:
quickfur wrote:And now, for your second idea:
[...]

Well, the height of the n-gon atop the 2n-p in that segmentochoron is h = sqrt[3 - 1/sin^2(pi/n)]/2. And the latteral displacement of the 2n-g clearly is 1/2 (half an edge length). This readily provides the tangens of half of the desired angle right from those 2 cathetes. You even could rewrite that as Sinus (provided that tangens), and then the cosinus of the doubled angle, i.e. of the desired one. This results in

cos(dichoral angle between the n-cupola) = (2 sin^2(pi/n) - 1)/(4 sin^2(pi/n) - 1)

esp.
n = 3 -> arccos(1/4) = 75.522488 degrees
n = 4 -> 90 degrees
n = 5 -> 144 degrees

That is, n = 4 provides the only convex solution, which then has a cell consist of 8 ops + 32 trips + 16+4 cubes, and 16 squippies. - Here the 8 ops form the one ring. The latteral squares of those form a 8x8 chessboard, whare alternate black fields get squippies, the remaining black ones get cubes. The white fields get digonal cupolae (trips) in the obvious orientations. The remainder here is a second orthogonal ring from 4 cubes. And indeed, the dihedral angle of a cube is 360 - 90 - 2*90 = 90 degrees.

But n = 5 would also provide a solution here, even so it isn't convex any more. There we use 10 dips + 50 trips + 25 cubes + 25 squippies + 5 stips. - Here the 10 dips form a first ring. The latteral squares thereof here form a larger 10x10 chessboard, again likewise been filled. The remainder here will be a ring of 5 stips (pentagrammal prisms). Indeed, the dihedral angle at their lacing edges is 360 - 144 - 2*90 = 36 degrees.

--- rk

And this n=5 case appears to be a Stott expansion of a previous star polychoron (I forgot what the name of it is -- basically, what you get if you glue 5 5-antiprism||5-gon's together like lunae -- was it called stawros, iirc?).
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### Re: Johnsonian Polytopes

Klitzing wrote:
[...]
Oh, hold on a sec. I think i've studied this case before... The conclusion I got was that the cubes and triangular prisms will be coplanar with the m-prisms, so they will merge into 2m-prisms, and therefore the result is nothing other than an n,12-duoprism. So this doesn't give us anything new.

Don't be afraid. It just has been a try.

And it is indeed amasing that it becomes Independent of n, but m has always to be 6.
And then that some cells become corealmic, is also not a usual behavior.
It is just that those resulting n,12-duoprisms are nothing new...

It's actually pretty normal behaviour, and it has also been studied before.

The reason this behaviour is not unusual, is because the dichoral angle between a duoprism's two rings are exactly 90°, and one of the pieces used in your augmentation is just a prism of the duoprism's cells, and prisms by definition have 90° dichoral angle between the lacing cells and the base cells, so after attaching the augments, the lacing cells must be corealmar with the cells in the orthogonal ring.

I studied this case some time ago when investigating the so-called bridged augmentations of duoprisms, where some given augment on an m,n-duoprism would make it non-convex, but by bridging it with suitable wedge-shaped segmentochora, it would become convex again. I found this case as the only bridged case that produces a CRF; I couldn't find any other combination of augments that would work. Sadly this didn't lead to new CRFs. (However, this is only the combinations of augments that occurred to me; there may be other combinations that work, I didn't find anything to exclude that possibility.)
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### Re: Johnsonian Polytopes

I think you got the right feeling, quickfur, both with the relation of the 2 specific finds (according to my second idea) to sidpith and to stawros. But both weren't exact.

First consider the case n=m=4:
we have the lace city of sidpith
Code: Select all
`    x4o   x4o                     x4o x4x   x4x x4o                                  x4o x4x   x4x x4o                     x4o   x4o    `

That of quawros (quadratic wedge rosette), i.e. the corresponding Stott contraction - in one of the 2 possible displays - is
Code: Select all
`    x4o               x4o x4x x4o               x4o    `

But my new figure would expand in contrast rather the inverse (contracted) case
Code: Select all
`    x4x               x4x x4o x4x               x4x    `
thus, in fact, it happens to be
Code: Select all
`    x4x   x4x                     x4x x4o   x4o x4x                                  x4x x4o   x4o x4x                     x4x   x4x    `
Note how thiis display shows up quite nicely the ring of 8 ops (running along the outline) and also the orthogonal ring of 4 cubes in the center (running orthogonal to the display plane).

Now consider the case n=5, m=5/2:
For stawros we had the lace city
Code: Select all
`         x5o                                                   x5o               x5o                              x5x                                                                            x5o       x5o    `
Note that the relavant single component here has been 5 times (in pentagrammal sequence)
Code: Select all
`x5o               x5o                              x5x         `

But here we likewise would use the the inverse usage for those components, i.e.
Code: Select all
`x5x               x5x                              x5o         `
i.e. in total (still for the contracted thing)
Code: Select all
`         x5x                                                   x5x               x5x                              x5o                                                                            x5x       x5x    `

Now consider first the Stott expansion of stawros (similar to going back from quawros to sidpith). It then would have the lace city
Code: Select all
`     x5o               x5o                                                  x5x                                                                                 x5x               x5x                                                                                                  x5o                         x5o                                        x5x       x5x                                                                                                                    x5o              `
But as you can see, the (pentagrammal) ring of 5 pips of stawros here would become a sequence of 10, outlining the outer x5o-pentagon twice. Thus that figure happens to become degenerate.

The same problem holds true in my recent n=5, m=5/2 figure, just that x5o and x5x are inversed here again
Code: Select all
`     x5x               x5x                                                  x5o                                                                                 x5o               x5o                                                                                                  x5x                         x5x                                        x5o       x5o                                                                                                                    x5x              `
i.e. the intended pentagonal ring of stips can be seen quite nicely in the center (orthogonal to the plane of display), but the intended ring of 10 dips would circle around at the outline twice!

But sure, there is a way out in both cases! We ought to blend out those pips (of the extended stawros), resp. the dips (in here), by identifying all completely coincident subelements, i.e. just cross-connecting the respective incident cells.

So, to conclude, what have we got in all the outlined stuff above?

Wrt. the 4-fold thingies:
Known components, being used
• there is 8-gon||cube, having for cell total: 1 cube, 2 squacues, 4 trips, and 4 tets
• there is op||4-gon, having for cell total: 1 op, 2 squacues, 4 trips, and 4 squippies
• there is op||cube, having for cell total: 1+4 cubes, 1 op, 2 squacues, and 4 trips
• there is sodip (= (4,8)-duoprism), having for cell total: 8 cubes and 4 ops
• there is tes (= (4,4)-duoprism), having for cell total: 4+4 cubes
Deduced stuff
• there is quawros (using 4 times that 8-gon||cube, connected at the squacues), having thus for cell total: 4 cubes, 16 trips, and 16 tets
• there is the Stott re-extended quawros (using in the first ring alternatingly 4 times that 8-gon||cube and 4 times op||cube, connected at the squacues, and in the second 1 sodip, interconnecting those rings at the ops), having thus for cell total: 4+4+16+4+4 cubes, 16+16 trips, and 16 tets (this in fact is nothing but sidpith, having for cell total: 32 cubes, 32 trips, and 16 tets)
• there is my new n=m=4 figure (using in the first ring alternatingly 4 times that op||4-gon and 4 times op||cube, connected at the squacues, and in the second 1 tes, interconnecting those rings at the cubes), having thus for cell total: 4+16 cubes, 4+4 ops, 16+16 trips, and 16 squippies

Wrt. the 5-fold thingies:
Known components, being used
• there is 10-gon||pip, having for cell total: 1 pip, 2 pecues, 5 trips, and 5 tets
• there is dip||5-gon, having for cell total: 1 dip, 2 pecues, 5 trips, and 5 squippies
• there is a reduced "10/2-p||5/2-gram" (blending out that 10/2-p by cross-connecting the incident cells), having for cell total: 2 5/2-cuploids, 5 trips, and 5 squippies
• there is dip||pip, having for cell total: 1 pip, 1 dip, 2 pecues, 5 trips, and 5 cubes
• there is a reduced "10/2-p||stip" (blending out that 10/2-p by cross-connecting the incident cells), having for cell total: 2 5/2-cuploids, 5 trips, and 5 cubes
• there is stardedip (= (5/2,10)-duoprism), having for cell total: 10 stips and 5 dips
• there is starpedip (=5/2,5)-duoprism), having for cell total: 5 stips and 5 pips
Deduced stuff
• there is stawros (using 5 times that 10-gon||pip, connecting at the pecues), having thus for cell total: 5 pips, 25 trips, and 25 tets
• there is the reduced Stott extended stawros (using in the first ring alternatingly 5 times that 10-gon||pip and 5 times dip||pip, connected at the pecues - but reducing the pairs of coincident pips, and in the second 1 stardedip, interconnecting those rings at dips), having thus for cell total: 10 stips, 25+25 trips, 25 tets, and 25 cubes
• there is the Stott contraction of my (un-reduced) n=5, m=5/2 figure (using 5 times that dip||5-gon, connected at the pecues), having thus for cell total: 5 dips, 25 trips, and 25 squippies
• there is the reduced version of my intended n=5, m=5/2 figure (using in the first ring alternatingly 5 times that dip||5-gon and 5 times that dip||pip, connected at the pecues - but reducing the pairs of coincident dips, and in the second 1 starpedip, interconnecting those rings at pips), having thus for cell total: 5 stips, 25+25 trips, 25 squippies, and 25 cubes
The important point here is for all that stuff to work: that the wedge angle of 2n-gon||n-p and that of n-gon||2n-p is the same!

--- rk
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### Re: Johnsonian Polytopes

Klitzing wrote:I think you got the right feeling, quickfur, both with the relation of the 2 specific finds (according to my second idea) to sidpith and to stawros. But both weren't exact.
[...]
The important point here is for all that stuff to work: that the wedge angle of 2n-gon||n-p and that of n-gon||2n-p is the same!
[...]

Interesting! So I got 2n-gon||n-prism mixed up with n-gon||2n-prism. It's interesting that they have exactly the same wedge angle. It makes sense in retrospect, but it's not something I would've expected just from being told their structure.
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### Re: Johnsonian Polytopes

More thoughts about CRFs with duoprismic symmetry: suppose you have a polychoron with a ring of n-prisms and an orthogonal ring of m-prisms, with some lacing edges between them. I'm thinking that if the polychoron is CRF, then it must correspond with some augmentation of some underlying m/d,n/e-duoprism, where d and e are resp. divisors of m and n. The reason is simple: suppose we cut the CRF around the ring of n-prisms, thus removing the ring of m-prisms and all the lacing cells to it. The result must be some duoprism containing n-prisms. Similarly, if we cut the CRF around the rings of m-prisms, removing all the cells from the orthogonal ring and the lacing edges, then it must be some duoprism containing m-prisms. So the CRF can be regarded as an augmentation of either one of these trimmed duoprisms.

The same analysis holds for the case where there's an intermediate layer of vertices between the two rings -- as long as this layer of vertices lie in the torus equidistant from the two rings of prisms. Then you just cut the polychoron along this torus, and you get two augmented duoprisms, which can each be trimmed again into plain j,k-duoprisms for some j and k.

But the thing is, we have already studied duoprism augmentations before, and my investigations into bridged augments has not revealed anything new -- one is x4o3o3x, and the other is the 6,n-duoprism -> 12,n-duoprism augmentation which has coplanar cells. The only other CRF augmentations we found were with n-prism pyramids, which have been completely enumerated, and with 2n-prism||n-gon wedges (n-gonal magnabicupolic rings), which, although I haven't fully enumerated them yet, are basically parallel to the n-prism pyramid augmentations, plus some additional cases due to the possibility of gyration and odd-spaced augments in the 2n,m-duoprism. Since none of these augmentations yield a CRF with orthogonal rings of prisms that are separated by one or more edges, it seems to indicate that either (1) we haven't thought of some novel augment shapes that would produce such a thing, or (2) there are no other CRFs of this kind than we've already found.

Conclusion (2) is tempered, of course, by the fact that it only applies if the intermediate layer of vertices lie on a torus equidistant from either ring. So this suggests two different directions for searching: (a) investigate unusual duoprism augment shapes, that is, not n-prism pyramids or 2n-prism||n-gon wedges -- this seems unlikely to yield a shape that would have a uniform orthogonal ring, though I can't 100% exclude that possibility yet; or (b) if a CRF with duoprismic symmetry exists, that has two rings of prisms separated by one or more edges, then it should have some intermediate vertices that are not equidistant from the two rings.

Other, more remote possibilities are rings of antiprisms, of which we know the grand antiprism is one example, but the other possibilities have not been fully explored yet -- though it appears to be harder than initially thought; or one ring of prisms with an orthogonal ring of antiprisms -- but this seems numerically unlikely as Wendy pointed out. Nevertheless, I feel hopeful that something must exist out there somewhere -- after all, spidrox sports rings of alternating prisms and antiprisms, and it is far more complicated than something with mere duoprismic symmetry, so maybe there are some CRFs to be found with similar heterogenous rings? (Haha, this is starting to sound like organic chemistry's hetero-rings ).
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