Johnsonian Polytopes

[Marek14]

So you're saying we already have the list of these verfs (which are polyhedra with Johnson solid verfs as faces)?

Well, no. We have to search for those. My idea was to check abstract polyhedra of up to 12 vertices, maybe 13 and pick the legal verfs from them. I don't know how many abstract polyhedra exist, but the number should be manageable.
The thing is that I don't think there are any nonrigid polyhedra the way there nonrigid polygons. That's because their faces are also rigid, so they are not allowed to be distorted (obviously there are nonrigid polyhedra if you only consider the edges).

[quickfur]

Well, then you've found a new direction to search for CRF polychora in.

Not quite - it would be polytopes in 5D and higher.

Ah, right.

Well, the 5D CRFs do get rather interesting in any case. There's the 600-cell 5-teddy, for example, with 600 4-teddies and 120 icosahedral pyramids. Wouldn't you just love to look at it from a 5Der's POV, eh? And the 24-cell 5-teddy, which may be the only CRF teddy in any dimension that contains non-simplex (n-1)-teddies for facets.

Hopf fibration

Right, thank you for the excellent explanation! After reading that I read through the Wikipedia page and was able to understand that too - I'd heard of the Hopf fibration before but never understood it until now.

The first time I read about the Hopf fibration, it was so abstract to me that I could barely understand what the article was saying, much less have any idea of how to visualize such a thing. The second time I read about Hopf fibration was after I had (re)discovered the duocylinder as the limiting shape of duoprisms, and had come to terms with the idea of orthogonal circles that are equidistant and non-intersecting. So I had this mental image of the fibres of the Hopf fibration being laid out in concentric tori that start from one circle and converge on the orthogonal circle. But it was only after I re-read Bowers' description of polytwisters and swirlprisms, that I finally realized that the fibres in those concentric tori aren't parallel to each other, but spiral around each other -- and they are all great circles of the 3-sphere. I think iit was while I was doing the structure renders for the BXD (bi-icositetradiminished 600-cell) that it all "clicked" for me, prompting me to re-read Bower's polytwisters (I didn't quite have the right understanding of them before that).

[quickfur]

So you're saying we already have the list of these verfs (which are polyhedra with Johnson solid verfs as faces)?

Well, no. We have to search for those. My idea was to check abstract polyhedra of up to 12 vertices, maybe 13 and pick the legal verfs from them. I don't know how many abstract polyhedra exist, but the number should be manageable.
The thing is that I don't think there are any nonrigid polyhedra the way there nonrigid polygons. That's because their faces are also rigid, so they are not allowed to be distorted (obviously there are nonrigid polyhedra if you only consider the edges).

Ah, but this is what I was getting at. The polyhedra themselves are rigid, but when you're trying to put the verfs together into these polyhedra, you don't know how many verfs to fit around a vertex (since the point is to find these polyhedra -- we wouldn't know how many of which verfs to put around the vertex and what the dihedral angles should be, until after we have already found said polyhedra). So the vertices are non-rigid: you can't build these polyhedra piecemeal without allowing for free deformation of the >3 degree vertices so that the dihedral angles can be resolved later once you have enough faces to fix the angles.

As for the number of abstract polyhedra... there are tons of them!! Most Johnson solids' verfs will be triangles, and there are a huge number of AP's that have triangular faces. Since you don't know in advance whether some AP structure will be convex or not, you'll have to generate huge amounts of candidate face lattices and then reject the non-convex ones one by one. And then among those remaining, you still have to find a deformation that will correctly fit in all the verfs (edge lengths, angles, etc) -- this is not much different to solving the non-rigid vertex problem.

So we've managed to shuffle the 4D non-rigid edges problem into the 3D non-rigid vertex problem, which is probably an improvement, I'll agree, but we still haven't solved the fundamental problem yet. How do we get the list of these polyhedra, and how do we know the list is complete? 'cos any missing polyhedra means potentially missing 4D CRFs when we get to the main part of the algorithm.

[Marek14]

So you're saying we already have the list of these verfs (which are polyhedra with Johnson solid verfs as faces)?

Well, no. We have to search for those. My idea was to check abstract polyhedra of up to 12 vertices, maybe 13 and pick the legal verfs from them. I don't know how many abstract polyhedra exist, but the number should be manageable.
The thing is that I don't think there are any nonrigid polyhedra the way there nonrigid polygons. That's because their faces are also rigid, so they are not allowed to be distorted (obviously there are nonrigid polyhedra if you only consider the edges).

Ah, but this is what I was getting at. The polyhedra themselves are rigid, but when you're trying to put the verfs together into these polyhedra, you don't know how many verfs to fit around a vertex (since the point is to find these polyhedra -- we wouldn't know how many of which verfs to put around the vertex and what the dihedral angles should be, until after we have already found said polyhedra). So the vertices are non-rigid: you can't build these polyhedra piecemeal without allowing for free deformation of the >3 degree vertices so that the dihedral angles can be resolved later once you have enough faces to fix the angles.

As for the number of abstract polyhedra... there are tons of them!! Most Johnson solids' verfs will be triangles, and there are a huge number of AP's that have triangular faces. Since you don't know in advance whether some AP structure will be convex or not, you'll have to generate huge amounts of candidate face lattices and then reject the non-convex ones one by one. And then among those remaining, you still have to find a deformation that will correctly fit in all the verfs (edge lengths, angles, etc) -- this is not much different to solving the non-rigid vertex problem.

So we've managed to shuffle the 4D non-rigid edges problem into the 3D non-rigid vertex problem, which is probably an improvement, I'll agree, but we still haven't solved the fundamental problem yet. How do we get the list of these polyhedra, and how do we know the list is complete? 'cos any missing polyhedra means potentially missing 4D CRFs when we get to the main part of the algorithm.

We do know how many verfs to fit around a vertex since the idea is to set an abstract polyhedron first and then try to concretize it. So when we try to fit the polygons together, we already know which structure precisely we're building.
Number of polyhedra - actually, I think that most Johnson solids' verfs are quadrangles, not triangles. Triangular verfs are surprisingly limited.
First of all, I'm not sure how many different abstract polyhedra there is for up to 12 vertices. I hope it's only millions or less -- that would be still well-accessible by computer. And it should be possible to compute 4D coordinates of the verf (has to be 4D because with Johnson polyhedra some of the polygons could be skew), either exactly or just numerically.
As for their convexity, that seems more complex than I thought -- what exactly does "convex" mean for a skew polyhedron that doesn't fit in a single hyperplane? Are skew polygons convex? But I guess there are two kinds of convexity that need to be checked: the local convexity of vertex which would be done while building the verfs, and the global convexity to prevent intersecting faces.

The main improvement from 4D edge to 3D vertices is that there is now only finite amount of possible distortions. Let's say that we start with an abstract polyhedron.

Step 1 -- lengths are assigned to the polyhedron's edges.
Step 2 -- check the faces to verify each matches a valid cell. If a single face does not (for example, if there's a triangle that would have a triangle, square and pentagon around), the length assignment is invalid.
Step 3 -- for each face with 4 or more edges check all possible skewings for that combination, in various orientation if necessary.
Step 4 -- now that each face is clearly defined, the computer has to actually build them into the structure of the shape. This will involve solving a series of equations since the relationships between the vertices are described in terms of distances and angles.
Step 5 -- once you have the polyhedron and the paint dries a bit, you try to find a point with distance 1 from each of its vertices. If this is not possible, the polyhedron cannot serve as a verf and is discarded.
Step 6 -- check the convexity. Here you can also check that you didn't end up with a degenerate 2D shape, like happened with that delirious tetrahedron of mine,
Step 7 -- the verf is ready to use.

Clearly, step 4 is hardest, but it should be in capabilities of computers.

I also suspect that the more outlandish and complicated structure of the abstract polyhedron, the less chance it will actually lead to a valid verf. Most of the possibilities will be thrown out in step 5 or 6. Number of abstract polyhedra will be further culled (though probably not that much) by the fact that anything that contains hexagon or larger polygon can be discarded outright.

[quickfur]

[...]
We do know how many verfs to fit around a vertex since the idea is to set an abstract polyhedron first and then try to concretize it. So when we try to fit the polygons together, we already know which structure precisely we're building.
Number of polyhedra - actually, I think that most Johnson solids' verfs are quadrangles, not triangles. Triangular verfs are surprisingly limited.

Interesting. So the dichoral angles based algo will not get very far then, if the 4D CRFs also mostly exhibit degree-4 edges.

First of all, I'm not sure how many different abstract polyhedra there is for up to 12 vertices.

Wait, why only 12? An icosahedral pyramid has 20 tetrahedra meeting at a vertex. Narrower shapes like triangular cupola may be able to fit even more around a vertex. (The 11 or 12 figure we got was for maximum number of cells around an edge, but we're talking about vertex figures here.)

[...] As for their convexity, that seems more complex than I thought -- what exactly does "convex" mean for a skew polyhedron that doesn't fit in a single hyperplane? Are skew polygons convex? But I guess there are two kinds of convexity that need to be checked: the local convexity of vertex which would be done while building the verfs, and the global convexity to prevent intersecting faces.

Maybe one way to check convexity would be to assemble the cells associated with the verf into an actual 4D figure (a partial polychoron - a bunch of cells around a single vertex) and check that the figure is convex. If all verfs give convex figures, then by extension the entire polychoron must be convex.

The main improvement from 4D edge to 3D vertices is that there is now only finite amount of possible distortions. Let's say that we start with an abstract polyhedron.

Step 1 -- lengths are assigned to the polyhedron's edges.
Step 2 -- check the faces to verify each matches a valid cell. If a single face does not (for example, if there's a triangle that would have a triangle, square and pentagon around), the length assignment is invalid.
Step 3 -- for each face with 4 or more edges check all possible skewings for that combination, in various orientation if necessary.

But still, you have the problem of which deformations to use to get a valid polyhedron. Just because the abstract polyhedron tells you there are 5 faces around the vertex, doesn't mean you know what dihedral angles to assign to them. So you still end up having to solve the non-rigid vertex problem. Although, I suppose you could assign verfs to faces first, so that gives you a fixed length for all the edges, then you can use that to compute the angles. Hmm. Maybe this will work after all. But how to handle the case when some verfs are non-planar?

Step 4 -- now that each face is clearly defined, the computer has to actually build them into the structure of the shape. This will involve solving a series of equations since the relationships between the vertices are described in terms of distances and angles.

Better make sure there are no systems of quadratics here!

Step 5 -- once you have the polyhedron and the paint dries a bit, you try to find a point with distance 1 from each of its vertices. If this is not possible, the polyhedron cannot serve as a verf and is discarded.

Maybe there's a way to make use of this fact to cull the number of abstract polyhedra that are generated or considered? Not sure how, though.

Step 6 -- check the convexity. Here you can also check that you didn't end up with a degenerate 2D shape, like happened with that delirious tetrahedron of mine,
Step 7 -- the verf is ready to use.

Clearly, step 4 is hardest, but it should be in capabilities of computers.

I also suspect that the more outlandish and complicated structure of the abstract polyhedron, the less chance it will actually lead to a valid verf. Most of the possibilities will be thrown out in step 5 or 6. Number of abstract polyhedra will be further culled (though probably not that much) by the fact that anything that contains hexagon or larger polygon can be discarded outright.

Hmm. I'm still not convinced that the abstract polyhedron approach will save us very much. But maybe we should put it in more concrete terms so that we can analyze it. How do we go about enumerating abstract polyhedra? You can start with n vertices, then generate all possible pairings (edges) between them, which is (n choose 2) EDIT: no that's wrong, it should be 2^(n^2) combinations. Then from those edges, generate all possible groupings (faces) between adjacent edges, which is (E choose k) EDIT: 2^(E^k) for E = number of edges, 3≤k≤5. That's a total of about (n choose 2) * (E choose k) 2^(n^2 + E^k) combinations to check (loose upper bound). And you have to loop from n=4 to at least 20, maybe more. I'm not sure how to count this correctly, but it does look like a very huge number of combinations to go through.

But still, once you assign lengths to edges, how to turn that into coordinates? If all you have is a graph with labelled edges indicating length, how do you turn that into an actual polyhedron? Edge lengths constrain the location of vertices to a fixed radius from adjacent vertices, so to solve for vertex coordinates, you could fix one vertex (say at (0,0,0)), and maybe the second vertex at (x,0,0) for some x, then try to solve for the rest. Since the constraints involve radius, this gives you a system of quadratic equations.

[quickfur]

[...] How do we go about enumerating abstract polyhedra? You can start with n vertices, then generate all possible pairings (edges) between them, which is (n choose 2) EDIT: no that's wrong, it should be 2^(n^2) combinations. Then from those edges, generate all possible groupings (faces) between adjacent edges, which is (E choose k) EDIT: 2^(E^k) for E = number of edges, 3≤k≤5. That's a total of about (n choose 2) * (E choose k) 2^(n^2 + E^k) combinations to check (loose upper bound). And you have to loop from n=4 to at least 20, maybe more. I'm not sure how to count this correctly, but it does look like a very huge number of combinations to go through.

We can perhaps somewhat constrain the enumeration by using additional requirements of abstract polyhedra: edges can only be shared between two faces, so this should eliminate a significant number of possibilities. Edges must cover all vertices, and faces must cover all edges, so any assignment that violates this can be rejected outright. What else can be done to reduce the number of combinations?

EDIT: the definition of abstract polytope also requires strong connectedness: not sure how to quickly check this, or how to generate only candidates that satisfy strong connectedness, but at least it can be used as a post-filter to discard invalid edge/face assignments.

[Marek14]

[...]
We do know how many verfs to fit around a vertex since the idea is to set an abstract polyhedron first and then try to concretize it. So when we try to fit the polygons together, we already know which structure precisely we're building.
Number of polyhedra - actually, I think that most Johnson solids' verfs are quadrangles, not triangles. Triangular verfs are surprisingly limited.

Interesting. So the dichoral angles based algo will not get very far then, if the 4D CRFs also mostly exhibit degree-4 edges.

First of all, I'm not sure how many different abstract polyhedra there is for up to 12 vertices.

Wait, why only 12? An icosahedral pyramid has 20 tetrahedra meeting at a vertex. Narrower shapes like triangular cupola may be able to fit even more around a vertex. (The 11 or 12 figure we got was for maximum number of cells around an edge, but we're talking about vertex figures here.)

[...] As for their convexity, that seems more complex than I thought -- what exactly does "convex" mean for a skew polyhedron that doesn't fit in a single hyperplane? Are skew polygons convex? But I guess there are two kinds of convexity that need to be checked: the local convexity of vertex which would be done while building the verfs, and the global convexity to prevent intersecting faces.

Maybe one way to check convexity would be to assemble the cells associated with the verf into an actual 4D figure (a partial polychoron - a bunch of cells around a single vertex) and check that the figure is convex. If all verfs give convex figures, then by extension the entire polychoron must be convex.

The main improvement from 4D edge to 3D vertices is that there is now only finite amount of possible distortions. Let's say that we start with an abstract polyhedron.

Step 1 -- lengths are assigned to the polyhedron's edges.
Step 2 -- check the faces to verify each matches a valid cell. If a single face does not (for example, if there's a triangle that would have a triangle, square and pentagon around), the length assignment is invalid.
Step 3 -- for each face with 4 or more edges check all possible skewings for that combination, in various orientation if necessary.

But still, you have the problem of which deformations to use to get a valid polyhedron. Just because the abstract polyhedron tells you there are 5 faces around the vertex, doesn't mean you know what dihedral angles to assign to them. So you still end up having to solve the non-rigid vertex problem. Although, I suppose you could assign verfs to faces first, so that gives you a fixed length for all the edges, then you can use that to compute the angles. Hmm. Maybe this will work after all. But how to handle the case when some verfs are non-planar?

Step 4 -- now that each face is clearly defined, the computer has to actually build them into the structure of the shape. This will involve solving a series of equations since the relationships between the vertices are described in terms of distances and angles.

Better make sure there are no systems of quadratics here!

Step 5 -- once you have the polyhedron and the paint dries a bit, you try to find a point with distance 1 from each of its vertices. If this is not possible, the polyhedron cannot serve as a verf and is discarded.

Maybe there's a way to make use of this fact to cull the number of abstract polyhedra that are generated or considered? Not sure how, though.

Step 6 -- check the convexity. Here you can also check that you didn't end up with a degenerate 2D shape, like happened with that delirious tetrahedron of mine,
Step 7 -- the verf is ready to use.

Clearly, step 4 is hardest, but it should be in capabilities of computers.

I also suspect that the more outlandish and complicated structure of the abstract polyhedron, the less chance it will actually lead to a valid verf. Most of the possibilities will be thrown out in step 5 or 6. Number of abstract polyhedra will be further culled (though probably not that much) by the fact that anything that contains hexagon or larger polygon can be discarded outright.

Hmm. I'm still not convinced that the abstract polyhedron approach will save us very much. But maybe we should put it in more concrete terms so that we can analyze it. How do we go about enumerating abstract polyhedra? You can start with n vertices, then generate all possible pairings (edges) between them, which is (n choose 2) EDIT: no that's wrong, it should be 2^(n^2) combinations. Then from those edges, generate all possible groupings (faces) between adjacent edges, which is (E choose k) EDIT: 2^(E^k) for E = number of edges, 3≤k≤5. That's a total of about (n choose 2) * (E choose k) 2^(n^2 + E^k) combinations to check (loose upper bound). And you have to loop from n=4 to at least 20, maybe more. I'm not sure how to count this correctly, but it does look like a very huge number of combinations to go through.

But still, once you assign lengths to edges, how to turn that into coordinates? If all you have is a graph with labelled edges indicating length, how do you turn that into an actual polyhedron? Edge lengths constrain the location of vertices to a fixed radius from adjacent vertices, so to solve for vertex coordinates, you could fix one vertex (say at (0,0,0)), and maybe the second vertex at (x,0,0) for some x, then try to solve for the rest. Since the constraints involve radius, this gives you a system of quadratic equations.

Icosahedral pyramid has 20 tetrahedra meeting at a vertex, true. So vertex figure of its apex is an icosahedron. How many vertices does it have? That's right, 12 I assume this is the maximum amount, but it might be a bit more.
Solving the problem of assembling faces into polyhedron: this would be done in step 4, not 3 -- step 3 is only for abstract enumeration. Step 4 tries to fit all the faces at once using a system of equations and whatnot. As it tries to fit whole polyhedron, and the polyhedron itself won't be nonrigid (because its faces are rigidly defined), the solution should be unique. Or... there won't be a solution, if the edge lengths or skew angles make it impossible. But there shouldn't be any nonrigid, continually variable solutions.

As far as sets of quadratic equations go... I'm afraid we won't be able to get around without them. All equations involved in the process seem to be quadratic -- either determining distance between two points or determining angle between two lines (which would be probably easiest to do through scalar product?). But if the exact solutions can't be worked out, we can just use numeric solving with sufficient preciseness.

Now, enumerating abstract polyhedra. I actually thought about this problem some years ago. I think that you can enumerate abstract polyhedra by using just these two steps:

Step 1: From a polyhedron with n vertices, get a polyhedron with n+1 vertices by putting a pyramid on one of its faces.
Step 2: From a polyhedron with n vertices and m edges, get a polyhedron with n vertices and m-1 edges by erasing an edge joining two vertices of degree 4 or higher and blending the two faces into one. Also, the resulting composite face can't have more than 1 vertex in common with any other face.

So, if you start with a tetrahedron, you use step 1 to get a triangular dipyramid, then step 2 to erase an equatorial edge and get a square pyramid. Since square pyramid has only 1 degree 4 node, you can't go any further, which makes 5 vertices complete.

For 6 vertices you start with the two polyhedra you just got and apply step 1: you get "tritetrahedron" or "augmented triangular dipyramid" (not a Johnson solid because it's nonconvex when all edge lengths are equal), a square pyramid augmented with a tetrahedron and an octahedron. Erasing edges then leads to more possibilities.

[Marek14]

I can actually prove that my algorithm leads to all abstract polyhedra (and streamline it a bit):

Any abstract polyhedron either contains a non-triangular face or not. If it doesn't, I can remove any one of its vertices and edges going from it and it's like removing a pyramid from a face.
If it has a non-triangular face, I can add an edge between two nonadjacent vertices of that face, splitting it in two. Eventually, this leads to a polyhedron with only triangular faces.

So, since these operations are reverse of my Step 1 and 2, and since every polyhedron can be eventually reduced to a polyhedron with lower number of vertices, it means that if we have complete set of polyhedra on n vertices, we can use Steps 1 and 2 to make a complete set of polyhedra on n+1 vertices.

In addition, we now see that we can decree that Step 1 shall be only used on polyhedra with no non-triangular faces OR on polyhedra with exactly one non-triangular face, but in that case the pyramid must be built on that face. This will result in better ordering of the polyhedra as we first generate only those n-vertex polyhedra which have maximum number of edges, and then successively lower that number.

Also, have a look here:
http://www.numericana.com/data/polyhedra.htm

I am not entirely sure if they count true abstract polyhedra or if they are restricted by geometry (not all abstract polyhedra can be realized geometrically with straight lines and planar faces). But if the numbers are close enough, we get less than 7 million abstract polyhedra on 12 or less vertices (with 13, it would jump to almost 100 million, though). 7 million should be definitely in computer's reach. I suspect the true number of verf-usable abstract polyhedra will be far less than that!

[Marek14]

And one more idea: Step 4 of search might be actually easier if combined with Step 5, i.e. if existence of center will be built in the polyhedron from start.

This means that instead of fitting the vertex figures in 4D space, we'll be fitting them on a 3D surface of hypersphere, eliminating one dimension. We can consider the vertices of the vertex figure as 4D vectors with specific angles prescribed for specific pairs of vectors (and if any two vectors are found to have angle less than 60 degrees, the verf can be thrown out because that will translate in two vertices with distance less than 1 -- and that cannot happen in convex case.

[Marek14]

And listen to this: If we have a polyhedron with a pyramidal vertex (vertex with only triangular faces around it that can be removed to leave a legal polyhedron), and there exists a valid arrangement for this polyhedron on the hypersphere, then you will still have a valid arrangement if you remove that pyramidal vertex (of course, the face resulting from this removal might not correspond to any polyhedron). But this also works in other direction: If a certain polyhedron with certain lengths cannot be put on the hypersphere, then adding a pyramidal vertex won't fix this -- and thus negative results can be used to eliminate certain polyhedra with more vertices.

In addition, working on hypersphere makes the skew polygons much easier. On sphere, they are no longer skew, merely distorted -- but they can be now described only in the sphere's two dimension. Similarly, when building a vertex figure on the hypersphere, skew polygons become merely distorted polygons, but they will still be lying in the hypersphere without protruding out.

[quickfur]

[...]
Icosahedral pyramid has 20 tetrahedra meeting at a vertex, true. So vertex figure of its apex is an icosahedron. How many vertices does it have? That's right, 12 I assume this is the maximum amount, but it might be a bit more.

Hehe you're right. I was thinking about 20 tetrahedra and forget the verf will only take vertices.

Solving the problem of assembling faces into polyhedron: this would be done in step 4, not 3 -- step 3 is only for abstract enumeration. Step 4 tries to fit all the faces at once using a system of equations and whatnot. As it tries to fit whole polyhedron, and the polyhedron itself won't be nonrigid (because its faces are rigidly defined), the solution should be unique. Or... there won't be a solution, if the edge lengths or skew angles make it impossible. But there shouldn't be any nonrigid, continually variable solutions.

As far as sets of quadratic equations go... I'm afraid we won't be able to get around without them. All equations involved in the process seem to be quadratic -- either determining distance between two points or determining angle between two lines (which would be probably easiest to do through scalar product?). But if the exact solutions can't be worked out, we can just use numeric solving with sufficient preciseness.

The problem is not with exact or inexact solutions, but with the lack of a general solution algorithm that will find all solutions of a quadratic system. But I just looked at System_of_polynomial_equations again, and it seems that there are some methods that can be used to attack the problem. Given that all our systems arise geometrically, we may not need to worry about pathological extremely-hard-to-solve systems. But there is still the possibility that a particular system might be very hard to solve, or require human intervention to solve, which is very bad if we're trying to get through millions of candidate verfs.

Now, enumerating abstract polyhedra. I actually thought about this problem some years ago. I think that you can enumerate abstract polyhedra by using just these two steps:

Step 1: From a polyhedron with n vertices, get a polyhedron with n+1 vertices by putting a pyramid on one of its faces.
Step 2: From a polyhedron with n vertices and m edges, get a polyhedron with n vertices and m-1 edges by erasing an edge joining two vertices of degree 4 or higher and blending the two faces into one. Also, the resulting composite face can't have more than 1 vertex in common with any other face.

So, if you start with a tetrahedron, you use step 1 to get a triangular dipyramid, then step 2 to erase an equatorial edge and get a square pyramid. Since square pyramid has only 1 degree 4 node, you can't go any further, which makes 5 vertices complete.

For 6 vertices you start with the two polyhedra you just got and apply step 1: you get "tritetrahedron" or "augmented triangular dipyramid" (not a Johnson solid because it's nonconvex when all edge lengths are equal), a square pyramid augmented with a tetrahedron and an octahedron. Erasing edges then leads to more possibilities.

Cool! This is certainly much better than enumerating all possible vertex pairs and edge subsets.

[quickfur]

And listen to this: If we have a polyhedron with a pyramidal vertex (vertex with only triangular faces around it that can be removed to leave a legal polyhedron), and there exists a valid arrangement for this polyhedron on the hypersphere, then you will still have a valid arrangement if you remove that pyramidal vertex (of course, the face resulting from this removal might not correspond to any polyhedron). But this also works in other direction: If a certain polyhedron with certain lengths cannot be put on the hypersphere, then adding a pyramidal vertex won't fix this -- and thus negative results can be used to eliminate certain polyhedra with more vertices. [...]

Hmm. I'm thinking, if we're going to be working on the hypersphere anyway, couldn't we just start constructing CRFs too?

Instead of computing all verfs beforehand, we can use a dynamic programming approach: start enumerating CRFs, and when the algorithm needs verfs, compute the possible verfs (based on the current cells being considered) on-the-fly. Cache the result of this computation somewhere, so that subsequent calls will just return the cached results.

[Marek14]

I don't know, I think that we'll need the verfs way too often. It would be better computing them beforehand.

Working on the hypersphere means that we don't need to consider the lengths of the chords at all -- we just need to consider the angles. For squares and larger, where's the possibility of skewing, we just need to know ALL the angles -- not just the ones related to sides, but also angles related to the diagonals. So in octahedron, both diagonal angles are 90, while for triangular dipyramid, they are 60 and 109,471 and for pentagonal dipyramid they are 63,4349 and 108. (I found the easy values, 60 and 108, by logic, but for the harder ones I had to measure them in Stella, and the only way I found to do that was to define a faceting containing face with angle of interest.)

This means that the total number of checks is 3 for triangular faces, 6 for quadragonal and 10 for pentagonal.

EDIT: This also means that we might actually not need to search for several skew verfs (like triangular dipyramid) directly -- instead they would be a special case whenever verf would have two tetrahedra and they would turn to have dihedral angle exactly 180 degrees. If we implement this, we'll only have to search for "primitive" verfs which can't be split, i.e. verfs where no diagonal is a valid chord, i.e. where the polyhedron can't be cut along any diagonal of that particular vertex. This means that the number of polygons we have to "try" will be reduced substantially and we can generate new verfs from "invalid" cases where some dihedral angles are exactly 180! So number of quadrangles to try is reduced from 45 to mere 14 and number of pentagons from 24 to 15. And majority of those is from the "weird" Johnson solids at the end of the line (most of the corona line's verfs are primitive except for two that occur in augmented sphenocorona).

The disadvantage is that some symmetrical verfs like octahedral one have more than one diagonal as valid chord, meaning that verfs containing them might be generated more than once. Also, some uniforms would have to be abandoned as primitive cells: octahedral verf would have to be built from two square pyramids, icosahedral from gyroelongated pentagonal pyramid and pentagonal pyramid, or even from metabidiminished icosahedron and two pentagonal pyramids, cuboctahedron from two square cupolas, rhombicuboctahedron from square cupola and octagonal prism, icosidodecahedron from two pentagonal rotundas and rhombicosidodecahedron from pentagonal cupola and diminished rhombicosidodecahedron.

[Keiji]

Hi quickfur,

After seeing the wiki edits you made referencing the magnabicupolic rings, I had a look through this topic, and as far as I can see you've not made any pretty renders of them!

Can we have some, please?

[quickfur]

Hi quickfur,

After seeing the wiki edits you made referencing the magnabicupolic rings, I had a look through this topic, and as far as I can see you've not made any pretty renders of them!

Can we have some, please?

Actually I did post some renders of them already, here.

[Keiji]

Aha, thank you I shall add a page for that polytope so I don't lose them again!

[quickfur]

Aha, thank you I shall add a page for that polytope so I don't lose them again!

Should we create separate pages for each CRF we discover? Or a page for each group of CRFs? That might be helpful, since this thread has grown unmanageably long, and it's kinda hard to find stuff in it. Sometimes when looking at old posts I don't even know if the results have subsequently been disproven, etc.. It would be nice if everyone can help keep the wiki updated on the latest known info, so that we don't waste time repeating what's already been done or basing something on a disproven result.

[Keiji]

The general rule is that anything with some interesting property should have a page.

So your 1,633 augmentations of duoprisms should not have pages, for example but things like the bicupolic rings, ursatopes and BXD get their own pages.

The even easier rule of thumb is, if you don't know whether it's worth adding, then just add it anyway.

Also feel free to dump algorithms or workings on wiki pages too, if it helps keep track of where things are. When projects are finished, those pages can always be tidied up to show the results and include the bits of workings that are still useful at the end.

[quickfur]

The general rule is that anything with some interesting property should have a page.

So your 1,633 augmentations of duoprisms should not have pages, for example but things like the bicupolic rings, ursatopes and BXD get their own pages.

And btw, there are at least another 1633 (probably more) duoprism augmentations: each n-gonal prism can be polygonally expanded into 2n-gonal prism || n-gon (i.e., n-gonal magnabicupolic ring), which means for every one of the 1633 pyramid augmentations, there's a corresponding augmentation for the m,2n-duoprism. And there are probably more than 1633, too, because of the orientation imposed by the bicupolic ring. The 8,8-duoprism, for example, can be augmented into the cantellated tesseract. And a whole bunch of other things.

The even easier rule of thumb is, if you don't know whether it's worth adding, then just add it anyway.

Also feel free to dump algorithms or workings on wiki pages too, if it helps keep track of where things are. When projects are finished, those pages can always be tidied up to show the results and include the bits of workings that are still useful at the end.

Well it would help if you & I aren't the only ones updating the wiki. (hint hint nudge nudge)

[quickfur]

Alright, I've found another series of CRFs: the hemidemi-600-cells.

The hemi-600-cell is made by bisecting the 600-cell, and discarding bisected edges (which would make the result non-CRF) and replacing them with pentagonal pyramids. Well, this shape has an icosidodecahedral cell, which is a hint that we can cut it up even more.

To get a handle on exactly which cuts should be made, I'll use vertices to denote cuts, because given a vertex on the 600-cell, you can uniquely identify a hyperplane that bisects the 600-cell; it makes a pentakis icosidodecahedron cross-section of the 600-cell. Deleting the bisect edges and replacing with pentagonal pyramids makes this cross-section an icosidodecahedron. So this isn't real bisection, but a modified bisection to make the result CRF.

Wedge #1

If we make two such bisections based on two adjacent vertices, we get a wedge-like shape with 2 pentagonal rotundae, 12 pentagonal pyramids, and 210 tetrahedra:



The rotundae are joined in ortho orientation (pentagon to pentagon) at a decagon; antipodal to this decagon is an edge surrounded by a cluster of 5 tetrahedra.

Wedge #2

If we make two bisections based on two vertices that lie on a great circle, separated by 2 edges, then we get a slightly narrower wedge with 2 pentagonal rotundae, 12 pentagonal pyramids, and 150 tetrahedra:



The rotundae are joined in gyro orientation (pentagon to triangle) at a decagon; antipodal to this decagon is a vertex surrounded by a cluster of 20 tetrahedra.

Wedge #3

If the two bisections are based on vertices separated by 3 edges on a great circle, then we can an even narrower wedge, having 2 pentagonal rotundae, 12 pentagonal pyramids, and 90 tetrahedra:



Notice that the rotundae are again joined in ortho orientation, and antipodal to the decagon is a cluster of 5 tetrahedra.

These are the wedges I've found so far; there may be more.

[quickfur]

I've found wedge #4, corresponding with bisecting the 600-cell twice orthogonal to two vertices separated by 4 edges on a great circle:



I've changed the 3D viewpoint here so that the structure of the thing is easier to see -- it's very flat in 4D so the side-view doesn't look very good. This wedge consists of two pentagonal rotundae, 12 pentagonal pyramids (10 of which form an alternating ring and 2 of which touch the antipodal vertex to the decagonal face), and 30 tetrahedra alternating in two concentric rings.

It's easy to see that if we delete the antipodal vertex (i.e., cut off a pentagonal antiprism pyramid) then we'll get a rotundic variant of Keiji's pentagonal gyrobicupolic ring: this one may be called the pentagonal gyrobirotundic ring. It's a truly beautiful shape!

The coordinates:

Code: Select all

<0, 0, 0, ±2>
<0, 0, 2, 0>
<0, 2, 0, 0>
<0, ±1/phi, 1, ±phi>
<0, 1/phi, -1, ±phi>
<0, ±1, phi, ±1/phi>
<0, 1, -phi, ±1/phi>
<0, phi, ±1/phi, ±1>
<1/phi, 0, phi, ±1>
<1/phi, 1, 0, ±phi>
<1/phi, phi, ±1, 0>
<1, 1/phi, phi, 0>
<1, phi, 0, ±1/phi>
<1, 1, 1, ±1>


where the golden ratio phi = (1+sqrt(5))/2.

P.S. Note that by construction, wedge #4 can be used to construct the other wedges, and, indeed, the entire 600-cell. Attaching two copies of wedge #4 by their pentagonal rotunda cells produces wedge #3 with 12 concave gaps, which may be filled in by 5 tetrahedra each. Attaching wedge #4 to wedge #3 produces wedge #2 with 12 gaps, again filled in by 5 tetrahedra each. Thus, at each stage, we add in 60 tetrahedra where the pentagonal pyramids meet at a concave angle. So 5 copies of wedge #4 (plus the requisite tetrahedra to close up the concave gaps) makes a hemi-600-cell; 10 copies of wedge #4 makes the entire 600-cell.

So one may say that a 4D watermelon in the shape of a 600-cell can be cut into 10 watermelon slices in the shape of wedge #4's. (The missing tetrahedra are just the bits that fall out when you cut up the melon, y'know.)

[Marek14]

Hmm, what's latin for "watermelon slice"?

[quickfur]

Hmm, what's latin for "watermelon slice"?

Googling for the Latin word for "watermelon" gave me incongruous answers, so I looked up the Greek instead: καρπούζι (karpouzi). Anglicised, it would be something like carpouzi or carpuzi. So maybe carpuzichoron for 600-cell wedges?

[Keiji]

It's easy to see that if we delete the antipodal vertex (i.e., cut off a pentagonal antiprism pyramid) then we'll get a rotundic variant of Keiji's pentagonal gyrobicupolic ring: this one may be called the pentagonal gyrobirotundic ring. It's a truly beautiful shape!

Mhm, I really like that Although I'd call it gyrobirotundular, because that formation sounds a little better.

Then "wedge #4" can be called the augmented pentagonal gyrobirotundular ring, unless there's some other way of augmenting a pentagonal gyrobirotundular ring that isn't done by putting a pentagonal antiprism pyramid on top?

[quickfur]

It's easy to see that if we delete the antipodal vertex (i.e., cut off a pentagonal antiprism pyramid) then we'll get a rotundic variant of Keiji's pentagonal gyrobicupolic ring: this one may be called the pentagonal gyrobirotundic ring. It's a truly beautiful shape!

Mhm, I really like that Although I'd call it gyrobirotundular, because that formation sounds a little better.

It sounds a bit more verbose, but since it appears to be one of its kind, I guess it's OK. When I first saw it, I was elated that perhaps this will lead to another infinite CRF family, but then I realized that that would require non-pentagonal CRF 3D rotunda, which we know aren't CRF. That structure of alternating pentagonal pyramids is certainly very compelling, though, as if it should work with other kinds of antiprisms as well. Now, if we could find CRF replacements for those pentagonal rotunda, perhaps by using CRF cell complexes instead of just single rotunda cells, we might be able to get something interesting with non-pentagonal birotundular rings.

Then "wedge #4" can be called the augmented pentagonal gyrobirotundular ring, unless there's some other way of augmenting a pentagonal gyrobirotundular ring that isn't done by putting a pentagonal antiprism pyramid on top?

You may be able to stick pentagonal pyramid pyramids on it, but I'm not certain the result will be CRF. I think likely not, but you can never be certain about these things until you try it.

[Keiji]

Klitzing's paper says 4.73, the square || square cupola, has cells: 4 tetrahedra + 2 square pyramids + 4 trigonal prisms + 1 cube.

Shouldn't that be 2 square cupolae, not square pyramids? 4.73 should be the square orthobicupolic ring.

[quickfur]

Klitzing's paper says 4.73, the square || square cupola, has cells: 4 tetrahedra + 2 square pyramids + 4 trigonal prisms + 1 cube.

Shouldn't that be 2 square cupolae, not square pyramids? 4.73 should be the square orthobicupolic ring.

Yeah I noticed that, but I forgot to ask him. I believe it's a typo -- it should be square cupola, not square pyramids, since otherwise it doesn't make sense that a square || square cupola segmentochoron has no square cupola cells!

Maybe he can confirm this himself, now that he's here on the forum.

[Klitzing]

Excellent!

Do you know how many CRF ursachora there are?

We know of 6 already (the tetrahedral, octahedral and icosahedral teddies and their expanded versions), but are there any other 3D bases that produce CRF ursachora?

In fact there are 8!
Consider xfo3ooxPzzz&#xt with z variously being o (unringed) or x (ringed), and P being 2, 3, 4, or 5.
Just that P=2 and z=o is somehow degenerate, a flat double cover of just 2 teddies.
The other one, you were missing, is an easy one too: the teddi prism.

You might consider the opposite extension as well, kind of oox3xfoP???&#xt. But unfortunately there aro no bistratic polyhedra with a base square, surrounded by 4 pentagons. Likewise for P=5: it would ask for a dodecahedron here (and even a higher-stratic extension of that partial complex runs into troubles). With resp. to P=3 we get nothing new, as xfo3oox4ooo&#xt = oox3xfo3oox&#xt. And for P=2 we obviously back to the former case (unconnected nodes can freely been put on either side).

--- rk

[Klitzing]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

Wrong Marek!
If we would like to re-prove the set of Blind polychora, then yes, by using my algorithm, the set of vers of allowed cells are: unit triangle, unit square, unit pentagon, sqrt2-triangle, and tau-triangle. Then we would have to assemble those polygons to solids (in order to get the verfs of the Blind polychora. Obviously, the sqrt2 edges only connect to each other, and the tau edges too. Therefore those would provide only deltahedra as potential verfs. For the unit edged ones OTOH we would get all those Johnson solids with triangles, squares, and/or pentagons. Thus, indeed, that task was done by Norman.

Buteven for the slightly larger set of CUCs to be enumerated, we not only have for faces of those to be assembled verfs some regular polygons: the archemedeans already provide semiregular polygons and other stuff, down to general triangles (of some specific edge lengths). Therefore even here this step 3 is an extension of the task Norman once completed: we are looking for kind a set similar to the Johnson solids, but with different edge lengths (the first secants, i.e. verfs, of different regular polygons differ in size!). - And finally, when including Johnson solids for building blocks in a CRF research, that class of allowed faces of those to be assembled CRF-verfs again gets larger.

Thus, in my eyes, we shouldnT do that effort twice, when running the CUCs research first and the CRF research afterwards. We should set up the set of all proposed CRF-verfs directly. And, for that CUCs-research we then would just have to use a subset therefrom.

--- rk

[Marek14]

[...]
How do I find all polyhedra that can be assembled from these polygons? Luckily, I don't have to! That was done almost 50 years ago, all we have to do now is take the list and find the verfs in there

Oh really? lol... and here I thought we had to start from scratch! My ignorance is showing.

That's what Johnson solids are... I even posted the list of their vertex figures few days ago. Lots of them are skew, though.

Wrong Marek!
If we would like to re-prove the set of Blind polychora, then yes, by using my algorithm, the set of vers of allowed cells are: unit triangle, unit square, unit pentagon, sqrt2-triangle, and tau-triangle. Then we would have to assemble those polygons to solids (in order to get the verfs of the Blind polychora. Obviously, the sqrt2 edges only connect to each other, and the tau edges too. Therefore those would provide only deltahedra as potential verfs. For the unit edged ones OTOH we would get all those Johnson solids with triangles, squares, and/or pentagons. Thus, indeed, that task was done by Norman.

Buteven for the slightly larger set of CUCs to be enumerated, we not only have for faces of those to be assembled verfs some regular polygons: the archemedeans already provide semiregular polygons and other stuff, down to general triangles (of some specific edge lengths). Therefore even here this step 3 is an extension of the task Norman once completed: we are looking for kind a set similar to the Johnson solids, but with different edge lengths (the first secants, i.e. verfs, of different regular polygons differ in size!). - And finally, when including Johnson solids for building blocks in a CRF research, that class of allowed faces of those to be assembled CRF-verfs again gets larger.

Thus, in my eyes, we shouldnT do that effort twice, when running the CUCs research first and the CRF research afterwards. We should set up the set of all proposed CRF-verfs directly. And, for that CUCs-research we then would just have to use a subset therefrom.

--- rk

Basically, that conversation was mostly a misunderstanding, because we were speaking about different things. In subsequent posts I outlined a general plan how to hunt for CRF vertex figures, whose main component is to not assemble the polyhedra in E3 but in S3 -- this will automatically take care of the condition that there is a vertex of polychoron that has unit distance to all of polyhedron's vertices, and it will also eliminate the problem nonplanar skew polygons -- they might be nonplanar, but they are still... what's the word? Cospherical? I also realized that the exact shape of quadrangle or pentagon can be controlled by lengths of diagonals and a way to reduce both scope of search and number of misses.