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This morning an idea occurred to me. Just as the icosahedron can be considered a "snub triangular antiprism", the snub disphenoid can be considered a "snub line antiprism". Now, the interesting thing is that the icosahedron can be constructed as a partitioning of an octahedron, or one may think of it as a distortion and truncation of an octahedron. Or, alternatively, an octahedron can be turned into an icosahedron by pulling apart some of its faces and inserting new pairs of triangles in the gaps. It appears that a snub disphenoid can also be constructed likewise from a tetrahedron.

A 24-cell has octahedral cells, and by replacing them with icosahedra, we may derive the snub 24-cell. What if we started with a 5-cell or 16-cell, and replace its cells with snub disphenoids? Will it be possible to close up the shape in a CRF way?

A 24-cell has octahedral cells, and by replacing them with icosahedra, we may derive the snub 24-cell. What if we started with a 5-cell or 16-cell, and replace its cells with snub disphenoids? Will it be possible to close up the shape in a CRF way?

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

There is a challenge for this generalization to work:

when you replace all octahedra of o4o3x3o with icosahedra to get o4s3s3s, you do not change the symmetry very much ([4,3]->[4,3][sup]+[\sup]+some extra symmetry for free).

Notice that here, [4,3] is the actual symmetry of the octahedra in the o4o3x3o, i.e. all symmetries that leave the o3x3o invariant, also leave the complete o4o3x3o invariant.

What this means is, when you change o3x3o into s3s3s, you don't need to care about how the ox3xo&#xt is oriented when you change it into a ofox3xofo&#xt, because the symmetry of ofox3xofo&#xt is almost the same as before.

However, when you change e.g. ox2xo&#xt into oBox2xoBo&#xt (for some weird number B), you do get a much smaller symmetry, and thus you do need to think about the orientation of the ox2xo&#xt's before expansion.

Of course, this is very well possible. All we need to do is think of an orientation for the tets in either x3o3o3o or o4o3o3x (I guess s4o3o3o will give such an orientation quite easilly, I will look into this)

when you replace all octahedra of o4o3x3o with icosahedra to get o4s3s3s, you do not change the symmetry very much ([4,3]->[4,3][sup]+[\sup]+some extra symmetry for free).

Notice that here, [4,3] is the actual symmetry of the octahedra in the o4o3x3o, i.e. all symmetries that leave the o3x3o invariant, also leave the complete o4o3x3o invariant.

What this means is, when you change o3x3o into s3s3s, you don't need to care about how the ox3xo&#xt is oriented when you change it into a ofox3xofo&#xt, because the symmetry of ofox3xofo&#xt is almost the same as before.

However, when you change e.g. ox2xo&#xt into oBox2xoBo&#xt (for some weird number B), you do get a much smaller symmetry, and thus you do need to think about the orientation of the ox2xo&#xt's before expansion.

Of course, this is very well possible. All we need to do is think of an orientation for the tets in either x3o3o3o or o4o3o3x (I guess s4o3o3o will give such an orientation quite easilly, I will look into this)

- student91
- Tetronian
**Posts:**317**Joined:**Tue Dec 10, 2013 3:41 pm

I realize that the "weirdness" of the snub disphenoid's coordinates as well as its low symmetry will pose a challenge to a straight generalization from the 24-cell -> snub 24-cell construction. But one way to think about this, is to take some regular polychoron with tetrahedral cells (5-cell or 16-cell, I'd say also 600-cell but then the result would be too complex to easily analyze), and break the symmetry in some way by selecting a particular axial orientation for the tetrahedra, or possibly a subset of the tetrahedra. Basically, we treat them as line antiprisms rather than tetrahedra with full symmetry. The snub disphenoid can then act as a kind of "elaborate line antiprism" in the hyperplanes of these chosen tetrahedra.

The tricky part, of course, is how to choose the orientation within the hyperplanes such that the remaining gaps can be closed up in a CRF way. I was hoping to find some kind of continuous deformation between the tetrahedron and the snub disphenoid, analogous to the partitioning of the octahedron that ranges from octahedron to icosahedron to (degenerate) cuboctahedron, that would result in tetrahedral (or other CRF) gaps that only change in orientation as you vary the partitioning ratio over its range.

The tricky part, of course, is how to choose the orientation within the hyperplanes such that the remaining gaps can be closed up in a CRF way. I was hoping to find some kind of continuous deformation between the tetrahedron and the snub disphenoid, analogous to the partitioning of the octahedron that ranges from octahedron to icosahedron to (degenerate) cuboctahedron, that would result in tetrahedral (or other CRF) gaps that only change in orientation as you vary the partitioning ratio over its range.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

student91 wrote:There is a challenge for this generalization to work:

when you replace all octahedra of o4o3x3o with icosahedra to get o4s3s3s, you do not change the symmetry very much ([4,3]->[4,3][sup]+[\sup]+some extra symmetry for free).

Notice that here, [4,3] is the actual symmetry of the octahedra in the o4o3x3o, i.e. all symmetries that leave the o3x3o invariant, also leave the complete o4o3x3o invariant.

What this means is, when you change o3x3o into s3s3s, you don't need to care about how the ox3xo&#xt is oriented when you change it into a ofox3xofo&#xt, because the symmetry of ofox3xofo&#xt is almost the same as before.

However, when you change e.g. ox2xo&#xt into oBox2xoBo&#xt (for some weird number B), you do get a much smaller symmetry, and thus you do need to think about the orientation of the ox2xo&#xt's before expansion.

Of course, this is very well possible. All we need to do is think of an orientation for the tets in either x3o3o3o or o4o3o3x (I guess s4o3o3o will give such an orientation quite easilly, I will look into this)

Hmm, x3o3o3o (hex) equals to xo3oo3ox&#x and the digonal antiprismatic lacing tets here are xo .. ox&#x. Thus you might be interested to look into something of the sort of xoBo3oAAo3oBox&#xt, where that B is the one needed in J84 (snub disphenoid - "snadow") and A as is required in here to close as CRF - if possible...

--- rk

- Klitzing
- Pentonian
**Posts:**1377**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Just calculated the heights and lengths for the Ansatz xoBo3oAAo3oBox&#xt:

Provided unit edge sizes snadow itself (xoBo .... oBox&#xt) already determines the segment heights as well as B. That outer segment heights (between vertex layers) are 0.578369 and the inner segment height is 0.411123 (both for snadow itself and for that 4D candidate). Further B=1.289169.

But then it would follow (for unit lacings under that o3o3o tower symmetry) that A=-0,14460<0 and thus this Ansatz will contradict to CRF.

--- rk

Provided unit edge sizes snadow itself (xoBo .... oBox&#xt) already determines the segment heights as well as B. That outer segment heights (between vertex layers) are 0.578369 and the inner segment height is 0.411123 (both for snadow itself and for that 4D candidate). Further B=1.289169.

But then it would follow (for unit lacings under that o3o3o tower symmetry) that A=-0,14460<0 and thus this Ansatz will contradict to CRF.

--- rk

Last edited by Klitzing on Fri May 06, 2016 8:29 am, edited 1 time in total.

- Klitzing
- Pentonian
**Posts:**1377**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Hmm. If we can find a lace tower containing snadow that has integer (possibly negative) edges, we could use (partial) Stott expansion to make it CRF.

I just realized a fatal oversight in my original guess, though. In the 24-cell -> snub 24-cell construction, the edges of the octahedron are partitioned into 1 vertex each, so that 12 edges -> 12 vertices of the icosahedron. However, there is no analogous construction from the tetrahedron to snadow, because the tetrahedron has only 6 edges, yet snadow has 8 vertices. So there is no actual analogy here; the snub line antiprism construction is a different analysis of the octahedron -> icosahedron transformation, and doesn't apply directly to the snub 24-cell construction. So it's unlikely that a parallel construction with snadow will yield CRF holes in the result.

Still, the snub (X) antiprism construction seems very compelling, and I can't help wondering if there is some construction in which such an analogy would produce a CRF. If we can find a way to make it work, it could give us a handle on building a CRF from the snub square antiprism too. Thinking along these lines, apart from the snub 24-cell construction which now seems unlikely to be fruitful, I'm wondering about existing CRFs in which the octahedron functions as a triangular antiprism lacing cell in some kind of sandwich-like construction. The bisected 24-cell, for example, can be thought of as o4o3x || o4x3o with some lacing cells being octahedra serving as triangular antiprisms (o3x || x3o). Is there an analogous CRF in which there are icosahedral cells which function as snub triangular antiprisms in the same positions? I.e., replace the o3x || x3o with icosahedra, possibly modifying the top/bottom cells to remain CRF? Such a CRF would have 8 icosahedra as lacing cells. If it exists, we may be able to extract some kind of analogy or hint as to what kind of modifications are necessary in transforming o4o3x || o4x3o to this CRF, and perhaps it could give us a pathway to a similar construction involving tetrahedra -> snadow? Or, indeed, square antiprisms -> snub square antiprisms (e.g., in x4o3o || o4x3o).

I just realized a fatal oversight in my original guess, though. In the 24-cell -> snub 24-cell construction, the edges of the octahedron are partitioned into 1 vertex each, so that 12 edges -> 12 vertices of the icosahedron. However, there is no analogous construction from the tetrahedron to snadow, because the tetrahedron has only 6 edges, yet snadow has 8 vertices. So there is no actual analogy here; the snub line antiprism construction is a different analysis of the octahedron -> icosahedron transformation, and doesn't apply directly to the snub 24-cell construction. So it's unlikely that a parallel construction with snadow will yield CRF holes in the result.

Still, the snub (X) antiprism construction seems very compelling, and I can't help wondering if there is some construction in which such an analogy would produce a CRF. If we can find a way to make it work, it could give us a handle on building a CRF from the snub square antiprism too. Thinking along these lines, apart from the snub 24-cell construction which now seems unlikely to be fruitful, I'm wondering about existing CRFs in which the octahedron functions as a triangular antiprism lacing cell in some kind of sandwich-like construction. The bisected 24-cell, for example, can be thought of as o4o3x || o4x3o with some lacing cells being octahedra serving as triangular antiprisms (o3x || x3o). Is there an analogous CRF in which there are icosahedral cells which function as snub triangular antiprisms in the same positions? I.e., replace the o3x || x3o with icosahedra, possibly modifying the top/bottom cells to remain CRF? Such a CRF would have 8 icosahedra as lacing cells. If it exists, we may be able to extract some kind of analogy or hint as to what kind of modifications are necessary in transforming o4o3x || o4x3o to this CRF, and perhaps it could give us a pathway to a similar construction involving tetrahedra -> snadow? Or, indeed, square antiprisms -> snub square antiprisms (e.g., in x4o3o || o4x3o).

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Sure, you are right, we could Stott expand the previous Ansatz, i.e. consider xoBo-3-xaax-3-oBox-&#xt instead. Then we'd get a=x+A=1-0.14460=0.85540, which is positive, but most probably not much more useful when considering appropriate fillings.

We further could consider be means of that Ansatz any xoBo-P-oAAo-P-oBox-&#xt (always using upright snadows) or its Stott expansion. But then P=4 would already mean we are no longer within polychora, because the top cell already became a flat tiling (square tiling). So the only other possibility here is P=2. And that one clearly is solvable, providing A=o, so that we could use the Stott expansion as the non-degenerate case. But that then is nothing but the mere snadow-prism. Nothing very exciting.

--- rk

We further could consider be means of that Ansatz any xoBo-P-oAAo-P-oBox-&#xt (always using upright snadows) or its Stott expansion. But then P=4 would already mean we are no longer within polychora, because the top cell already became a flat tiling (square tiling). So the only other possibility here is P=2. And that one clearly is solvable, providing A=o, so that we could use the Stott expansion as the non-degenerate case. But that then is nothing but the mere snadow-prism. Nothing very exciting.

--- rk

- Klitzing
- Pentonian
**Posts:**1377**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

quickfur wrote:[...]I just realized a fatal oversight in my original guess, though. In the 24-cell -> snub 24-cell construction, the edges of the octahedron are partitioned into 1 vertex each, so that 12 edges -> 12 vertices of the icosahedron. However, there is no analogous construction from the tetrahedron to snadow, because the tetrahedron has only 6 edges, yet snadow has 8 vertices. So there is no actual analogy here; the snub line antiprism construction is a different analysis of the octahedron -> icosahedron transformation, and doesn't apply directly to the snub 24-cell construction. So it's unlikely that a parallel construction with snadow will yield CRF holes in the result.

Still, the snub (X) antiprism construction seems very compelling, and I can't help wondering if there is some construction in which such an analogy would produce a CRF. If we can find a way to make it work, it could give us a handle on building a CRF from the snub square antiprism too. Thinking along these lines, apart from the snub 24-cell construction which now seems unlikely to be fruitful, I'm wondering about existing CRFs in which the octahedron functions as a triangular antiprism lacing cell in some kind of sandwich-like construction. The bisected 24-cell, for example, can be thought of as o4o3x || o4x3o with some lacing cells being octahedra serving as triangular antiprisms (o3x || x3o). Is there an analogous CRF in which there are icosahedral cells which function as snub triangular antiprisms in the same positions? I.e., replace the o3x || x3o with icosahedra, possibly modifying the top/bottom cells to remain CRF? Such a CRF would have 8 icosahedra as lacing cells. If it exists, we may be able to extract some kind of analogy or hint as to what kind of modifications are necessary in transforming o4o3x || o4x3o to this CRF, and perhaps it could give us a pathway to a similar construction involving tetrahedra -> snadow? Or, indeed, square antiprisms -> snub square antiprisms (e.g., in x4o3o || o4x3o).

I think I might have found a way to see 24-cell->snub 24-cell as a deformation of xo3o3x&#xt to xofo3ofox&#xt.

When you read the wikipedia page about the snub 24-cell, there is a part in which they describe the 24-cell as consisting of four interlocking rings of six ocahedra each.

This means that if you would have a description of ico according to this kind of swirly-whirly symmetry, you would have something that would look like f(xo3o3x&#xt), where f is the function that copies the described octahedron to every place dictated by the swirly-whirly symmetry group. A deformation of the form xo3ox&#xt->xofo3ofox&#xt then gives f(xofo3ofox&#xt), which does seem to exactly coincide with the snub 24-cell. (Notice that the f-function is horribly ill-defined, but just think of it as a set of rotations of 4-space that place the oct at the desired places.)

Now how is this useful?

Well, the 16-cell allows for a similar decomposition into four rings of four tets in the orientation xo2ox&#xt.

(You can see this by viewing hex as the snub hypercube in [4,2,4]-symmetry. The hypercube decomposes into two orthogonal rings of cubes. After the alternation, these two rings of cubes morph into rings of tets. The remaining 8 tets can easily be put into two other rings.)

Now take the according swirly-whirly-function g. This function then takes whatever is put into it, and places it into four rings of four.

This means that g(xo2ox&#xt) would give the 16-cell. Now think of what g(xoBo2oBox&#xt) would give.

This construction would give you four rings of four snadows, in direct analogy with the construction of sadi from ico using the {3}-ap => snub {3}-ap morphism.

So it is possible to see ico->sadi as resulting from {3}-ap => snub {3}-ap, and then construct something similar for similar polytopes. However, I really doubt this construction to give a CRF, and I am not sure yet how to use these swirly-whirly-funtions. I will look into it a bit more.

(I thought there were some swirlprisms that might be able to use these functions as well... what is a swirlprism actually?)

- student91
- Tetronian
**Posts:**317**Joined:**Tue Dec 10, 2013 3:41 pm

Swirlprisms once were found by "Dinogeorge" George Olshevsky. He and "Hedrondude" Jonathan Bowers in the sequel found several more uniform ones. E.g. look into his page at misc (somewhere down). Essentially those are uniform polychora with swirl-symmetry, which have prismatic cells only (here subsuming antiprismatic ones).

As you can see from the there shown vertex figure polyhedra, none of those polychora is convex (as a polychoral convexity would imply the polyhedral convexity of its vertex figure as well).

--- rk

As you can see from the there shown vertex figure polyhedra, none of those polychora is convex (as a polychoral convexity would imply the polyhedral convexity of its vertex figure as well).

--- rk

- Klitzing
- Pentonian
**Posts:**1377**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Your "swirly-whirly" function is just the symmetry subgroup of the Hopf fibration of the 3-sphere. It features prominently in 4D geometry, especially in the uniform polytopes, because they can be thought of as discrete approximations of the surface of the 3-sphere, and their high level of symmetry makes them isomorphic to discrete subgroups of the Hopf fibration.

I think your idea is worth investigating, though I'm a bit doubtful about whether it will produce CRFs containing snadow, because the high level of symmetry generated by these subgroups impose a lot of restrictions on what dichoral angles are possible on the surface elements. So unless there is some hidden connection I'm not aware of between snadow and the Hopf fibration, I'd be quite surprised if such a construction could close up in a CRF way. My idea about using modified bistratic CRFs to find snadow-containing CRFs was based on the idea that a lower-symmetry construction would (hopefully!) grant more leeway in what cell shapes can close up in a CRF way.

But then again, you never know... before CRFebruary we all thought that J91 and J92 were too strange to close up in a CRF way, yet look at how many CRFs we have found since then that contain them! While I don't expect a similar large number of snadow-based CRFs, I'm hopeful that there should be at least a small number other than the snadow-prism. Among the "crown jewel" Johnson solids, snadow is one of the more symmetrical ones, so it would seem to suggest that it's more likely to have some way to close up in a CRF way than stranger cells like sphenocorona or sphenomegacorona.

I think your idea is worth investigating, though I'm a bit doubtful about whether it will produce CRFs containing snadow, because the high level of symmetry generated by these subgroups impose a lot of restrictions on what dichoral angles are possible on the surface elements. So unless there is some hidden connection I'm not aware of between snadow and the Hopf fibration, I'd be quite surprised if such a construction could close up in a CRF way. My idea about using modified bistratic CRFs to find snadow-containing CRFs was based on the idea that a lower-symmetry construction would (hopefully!) grant more leeway in what cell shapes can close up in a CRF way.

But then again, you never know... before CRFebruary we all thought that J91 and J92 were too strange to close up in a CRF way, yet look at how many CRFs we have found since then that contain them! While I don't expect a similar large number of snadow-based CRFs, I'm hopeful that there should be at least a small number other than the snadow-prism. Among the "crown jewel" Johnson solids, snadow is one of the more symmetrical ones, so it would seem to suggest that it's more likely to have some way to close up in a CRF way than stranger cells like sphenocorona or sphenomegacorona.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:Your "swirly-whirly" function is just the symmetry subgroup of the Hopf fibration of the 3-sphere. It features prominently in 4D geometry, especially in the uniform polytopes, because they can be thought of as discrete approximations of the surface of the 3-sphere, and their high level of symmetry makes them isomorphic to discrete subgroups of the Hopf fibration.

I think your idea is worth investigating, though I'm a bit doubtful about whether it will produce CRFs containing snadow, because the high level of symmetry generated by these subgroups impose a lot of restrictions on what dichoral angles are possible on the surface elements. So unless there is some hidden connection I'm not aware of between snadow and the Hopf fibration, I'd be quite surprised if such a construction could close up in a CRF way. My idea about using modified bistratic CRFs to find snadow-containing CRFs was based on the idea that a lower-symmetry construction would (hopefully!) grant more leeway in what cell shapes can close up in a CRF way.

But then again, you never know... before CRFebruary we all thought that J91 and J92 were too strange to close up in a CRF way, yet look at how many CRFs we have found since then that contain them! While I don't expect a similar large number of snadow-based CRFs, I'm hopeful that there should be at least a small number other than the snadow-prism. Among the "crown jewel" Johnson solids, snadow is one of the more symmetrical ones, so it would seem to suggest that it's more likely to have some way to close up in a CRF way than stranger cells like sphenocorona or sphenomegacorona.

Well, J91 and J92 can be still described without using cubic equation, right? Are there any other crown jewels that can...?

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

I see that my old idea of polynomial degree as a measure of polytope complexity still survives. I'm no longer so sure of its utility, though. The first problem is that just because we don't know of any lower-degree description of some given polytope P, doesn't mean that no such description exists. For example, if we were to construct an icosahedron from first principles without prior knowledge of its connection with the golden ratio, we might be faced with a degree 5 or higher system of equations to solve (it would be unlikely for us, without prior knowledge, to end up with just the right orientation with golden ratio coordinates). Absence of proof isn't proof of absence.

Second, I'm not sure if polynomial degree alone fully captures the idea behind complexity measure. There is some correlation, but perhaps fixating on polynomial degree in itself is somewhat missing the point. The point behind the construction of, say, a snub disphenoid, is that it requires a certain chain of dependencies between surface elements in order to close up. Take the two triangles sharing the top (or bottom) edge, for example. By themselves they are free to vary from 0 to 360 deg in dihedral angle. But the distance between their tips, let's call it A, is a parameter that the next section of the polyhedron depends on: the configuration of faces underneath that edge have certain degrees of freedom, one of which must match A in order for the polyhedron to close up. The degrees of freedom of this configuration in turn depends on other sub-sections of the surface, which by themselves can freely vary but are restricted by other elements they are connected to. So there is a dependency structure between these degrees of freedom. For some polyhedra (and polytopes), many of these dependencies can be resolved locally. However, for things like snub disphenoid, one or two degrees of freedom cannot be resolved locally, and must be solved globally as a single system.

In some as-yet ill-defined way, polytopes whose dependency chains can be resolved locally are "simpler" than polytopes whose dependency chains must be solved holistically. And between subsections of surface whose dependency chains can be solved locally, there's also the question of how small is the "radius of dependency", that is, how much of the total surface must be taken into consideration when resolving a certain parameter X. In 3D, when you have 3 faces around a vertex, the radius of dependency is just that vertex and the surrounding faces, since that's all that's necessary to fix all dihedral angles at the vertex. When there are 4 or more faces around the vertex, you may have to look at a larger section of the surface in order to resolve all parameters. For things like snub disphenoid, almost everything has to be considered simultaneously before solution is possible. So the snub disphenoid is somehow, intuitively, "more complex" than something like, say, a dodecahedron where all vertices are order 3 and all angles can be resolved locally. Sphenocorona, from my present failed attempts to solve for its coordinates, seems to be even harder in some sense, because there are more parameters to be resolved, yet none of them can be fixed without considering basically the entire polyhedron simultaneously.

Coming back to whether something can be closed up in a CRF way in 4D, it seems that closing up a polychoron in 4D itself already imposes quite a lot of restrictions, because the extra lateral space available means that more CRF cells must appear in said extra lateral space, so it doesn't have the freedom to vary as it would have in 3D. I'm not sure 100% how this connects to polyhedron complexity, but I suppose in some crude, ill-defined way you could say that polyhedra that require global resolution of parameters require enough freedom in certain regions in order to form that shape in the first place, but when we're trying to close up a shape in 4D, we already need to "use up" some degrees of freedom, so there may not be enough leeway to also fit in one of the "more complex" shapes like sphenocorona or snub disphenoid.

The idea did occur to me, though, that rather than use the crown jewel Johnsons as cells, perhaps a more likely place to find 4D crown jewels would be to look for analogous constructions using simpler 4D cells. That's why yesterday I was musing about whether it's possible to insert a triangular prism into an icosahedral dipyramid such that the resulting deformation will close up in a CRF way. In some abstract, ill-defined sense, one might think of it as using the 4D geometry itself to produce a crown jewel shape, rather than start with a 3D crown jewel already "using up" some degrees of freedom, then try to close it up in 4D which requires even more degrees of freedom which there may not be "enough" of.

Nevertheless, I still can't help wondering if all of this is just a figment of our imagination (or lack thereof!), and that perhaps there is a way to close up a CRF containing snub disphenoid cells, just that we haven't thought of it yet.

Second, I'm not sure if polynomial degree alone fully captures the idea behind complexity measure. There is some correlation, but perhaps fixating on polynomial degree in itself is somewhat missing the point. The point behind the construction of, say, a snub disphenoid, is that it requires a certain chain of dependencies between surface elements in order to close up. Take the two triangles sharing the top (or bottom) edge, for example. By themselves they are free to vary from 0 to 360 deg in dihedral angle. But the distance between their tips, let's call it A, is a parameter that the next section of the polyhedron depends on: the configuration of faces underneath that edge have certain degrees of freedom, one of which must match A in order for the polyhedron to close up. The degrees of freedom of this configuration in turn depends on other sub-sections of the surface, which by themselves can freely vary but are restricted by other elements they are connected to. So there is a dependency structure between these degrees of freedom. For some polyhedra (and polytopes), many of these dependencies can be resolved locally. However, for things like snub disphenoid, one or two degrees of freedom cannot be resolved locally, and must be solved globally as a single system.

In some as-yet ill-defined way, polytopes whose dependency chains can be resolved locally are "simpler" than polytopes whose dependency chains must be solved holistically. And between subsections of surface whose dependency chains can be solved locally, there's also the question of how small is the "radius of dependency", that is, how much of the total surface must be taken into consideration when resolving a certain parameter X. In 3D, when you have 3 faces around a vertex, the radius of dependency is just that vertex and the surrounding faces, since that's all that's necessary to fix all dihedral angles at the vertex. When there are 4 or more faces around the vertex, you may have to look at a larger section of the surface in order to resolve all parameters. For things like snub disphenoid, almost everything has to be considered simultaneously before solution is possible. So the snub disphenoid is somehow, intuitively, "more complex" than something like, say, a dodecahedron where all vertices are order 3 and all angles can be resolved locally. Sphenocorona, from my present failed attempts to solve for its coordinates, seems to be even harder in some sense, because there are more parameters to be resolved, yet none of them can be fixed without considering basically the entire polyhedron simultaneously.

Coming back to whether something can be closed up in a CRF way in 4D, it seems that closing up a polychoron in 4D itself already imposes quite a lot of restrictions, because the extra lateral space available means that more CRF cells must appear in said extra lateral space, so it doesn't have the freedom to vary as it would have in 3D. I'm not sure 100% how this connects to polyhedron complexity, but I suppose in some crude, ill-defined way you could say that polyhedra that require global resolution of parameters require enough freedom in certain regions in order to form that shape in the first place, but when we're trying to close up a shape in 4D, we already need to "use up" some degrees of freedom, so there may not be enough leeway to also fit in one of the "more complex" shapes like sphenocorona or snub disphenoid.

The idea did occur to me, though, that rather than use the crown jewel Johnsons as cells, perhaps a more likely place to find 4D crown jewels would be to look for analogous constructions using simpler 4D cells. That's why yesterday I was musing about whether it's possible to insert a triangular prism into an icosahedral dipyramid such that the resulting deformation will close up in a CRF way. In some abstract, ill-defined sense, one might think of it as using the 4D geometry itself to produce a crown jewel shape, rather than start with a 3D crown jewel already "using up" some degrees of freedom, then try to close it up in 4D which requires even more degrees of freedom which there may not be "enough" of.

Nevertheless, I still can't help wondering if all of this is just a figment of our imagination (or lack thereof!), and that perhaps there is a way to close up a CRF containing snub disphenoid cells, just that we haven't thought of it yet.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

"The first problem is that just because we don't know of any lower-degree description of some given polytope P, doesn't mean that no such description exists."

No, that is not really the case. I'm trying to distinguish "constructable" and "nonconstructable" polytopes, where "constructable" in Euclidean sense means that all distances are numbers that require nothing worse than taking square roots. And that's an objective property, it's not dependent on the specific description. Snub disphenoid contains nonconstructible numbers (like its various heights), while J91/J92 do not.

Why is that important? Because all regular polytopes in 3+ dimensions are constructible. And so must be uniform polytopes of their families, parts of those polytopes, etc. Snub disphenoid isn't there, and that's why it doesn't work too well.

No, that is not really the case. I'm trying to distinguish "constructable" and "nonconstructable" polytopes, where "constructable" in Euclidean sense means that all distances are numbers that require nothing worse than taking square roots. And that's an objective property, it's not dependent on the specific description. Snub disphenoid contains nonconstructible numbers (like its various heights), while J91/J92 do not.

Why is that important? Because all regular polytopes in 3+ dimensions are constructible. And so must be uniform polytopes of their families, parts of those polytopes, etc. Snub disphenoid isn't there, and that's why it doesn't work too well.

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

An order 4 polynomial can also have roots that require "no worse than square roots", albeit nested ones at that. But we can't reject nested square roots, because that would exclude pentagons in 2D, for example. But if we allow nested square roots, then we'd have to allow sphenocorona, because it involves degree 4 polynomials (or was it degree 8? I don't remember now -- but it's an even number). But I'm highly doubtful that sphenocorona is "constructible" in the sense you're describing.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

But speaking of constructible numbers... it has always intrigued me that the golden ratio can be considered the Fibonacci constant (solution to x^2 - x - 1 = 0), and it is so prevalent in the pentagonal polytopes and associated with a large number of CRFs that we have found. The snub cube is intimately related to the so-called Tribonacci constant (solution to x^3 - x^2 - x - 1 = 0), but where are the other polytopes that are intimately connected with this constant? The only other one I can think of is the snub dodecahedron, which sorta mixes up the golden ratio with the Tribonacci constant (well, more precisely, involves the golden ratio in a cubic polynomial). But where are the tribonacci-based polytopes besides these two? Are there any in 4D worth speaking of, CRF or otherwise?

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Actually, we have been discussing this before. This then led to the conclusion that one needs to consider the biggest prime number that any exponent is divisible by.quickfur wrote:[…]But if we allow nested square roots, then we'd have to allow sphenocorona, because it involves degree 4 polynomials (or was it degree 8? I don't remember now -- but it's an even number). But I'm highly doubtful that sphenocorona is "constructible" in the sense you're describing.

However, putting x^4+ax^2+bx+c=0 in wolfram does give a solution for x involving cube roots, so this definition is still not what we want. I think Mareks "constructible" notion is what we do need to formalize.

Furthermore, I remember Marek has once calculated that the number of restrictions a vertex of a 4-polytope has is such that all vertices must be locally-solvable. If we are now able to somehow proof that 1 CVP3-vertex implies that all vertices are CVP3, we just have to proof that non-(duo)prismatic vertices containing CVP3 shortchords do not exist. The algebra we need for this is quite complicated. Maybe I am able to look into it, but I am not sure.

- student91
- Tetronian
**Posts:**317**Joined:**Tue Dec 10, 2013 3:41 pm

Well, the general quartic does require solving cube roots, that's for sure. But specific instances may not always require them. Just like how the general quintic has no solution in radicals, but specific instances still may.

But on second thoughts, Marek's constructibility notion makes sense. If we can somehow link that to the solvability of a 4D vertex, then it may give us a way to prove/disprove the existence of crown jewels outside of the EKF's. But somehow I think the issue may be more complicated than we suspect. The Johnson solid prisms always exist, because in some ill-defined intuitive sense, the degree-(>3) vertices all lie within the same hyperplane, while the prism product lies in the orthogonal space. Does that mean that CVP-3 vertices can only happen within a hyperplane? Does it mean that the CVP-(>3)-ness of a vertex can be "made to work" by restricting connections to the orthogonal space? But what about crown jewels with "simple" (CVP-2 or lower) cells, but whose coordinates may require a "non-constructible" solution, such as the lace city I posted in the crown jewels thread? Is there a way to prove that that lace city cannot be CRF? Or is that a different case than a CRF that contains cells that require CVP-(>3) vertices?

But on second thoughts, Marek's constructibility notion makes sense. If we can somehow link that to the solvability of a 4D vertex, then it may give us a way to prove/disprove the existence of crown jewels outside of the EKF's. But somehow I think the issue may be more complicated than we suspect. The Johnson solid prisms always exist, because in some ill-defined intuitive sense, the degree-(>3) vertices all lie within the same hyperplane, while the prism product lies in the orthogonal space. Does that mean that CVP-3 vertices can only happen within a hyperplane? Does it mean that the CVP-(>3)-ness of a vertex can be "made to work" by restricting connections to the orthogonal space? But what about crown jewels with "simple" (CVP-2 or lower) cells, but whose coordinates may require a "non-constructible" solution, such as the lace city I posted in the crown jewels thread? Is there a way to prove that that lace city cannot be CRF? Or is that a different case than a CRF that contains cells that require CVP-(>3) vertices?

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Yes, if I remember it right, I found that a vertex of polychoron with rigid faces (which is required for a CRF polychoron) must be rigid as well. There are enough constraints.

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

@Marek: Link to relevant discussion?

I discovered yesterday that my attempt at constructing a 4D crown jewel doesn't work, because having edge length constraints outside of a 3D hyperplane essentially fixes everything to be rigid. If we can prove that such constraints apply in general, we can eliminate an entire class of polytopes from our search for CRFs, which would be helpful in finding the real crown jewels.

I discovered yesterday that my attempt at constructing a 4D crown jewel doesn't work, because having edge length constraints outside of a 3D hyperplane essentially fixes everything to be rigid. If we can prove that such constraints apply in general, we can eliminate an entire class of polytopes from our search for CRFs, which would be helpful in finding the real crown jewels.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Well, looking through here (viewtopic.php?f=32&t=1927&p=20187), it looks like snub disphenoid (and sphenocorona) are actually constructible. Though they yield weird numbers.

The rigidity discussion seems to be here:

viewtopic.php?f=32&t=1978&p=22215

The rigidity discussion seems to be here:

viewtopic.php?f=32&t=1978&p=22215

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

Thanks!! Haha, I actually participated in that discussion, but completely forgot about it later! Well, so much for trying to find CVP3 polychora without using CVP3 cells, then! But still, that doesn't fully close the case about CRFs that may contain CVP3 cells. And there's also the exceptional case of some of the equations constraining a vertex being dependent. Under what conditions would they be dependent? Perhaps that might lead us in the direction of a CVP3 4D crown jewel (if one exists!).

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Weird, the Wikipedia-page on the snub disphenoid has a similar construction, and that one does include cube roots. This means something has to be wrong somewhere. Best way to check this I think is by looking if all distances work out to be 1, when calculated the normal way.Marek14 wrote:Well, looking through here (viewtopic.php?f=32&t=1927&p=20187), it looks like snub disphenoid (and sphenocorona) are actually constructible. Though they yield weird numbers.

About the general theory: I think it couldn't be too difficult to get some general result about non-constructible shortchords around vertices in R

Now what I thus would really like, is to be able to represent a polytope without relying on some coordinate representation. I was thinking that the distance between all vertices would be sufficient.

However, I am as of yet unable to really do calculations, as in this approach you lose the dimensional information, which then has to be artificially enforced upon it again.

So basically, how can I have a general notion of dimensionality, combined with an abstraction that does give coordinate-invariance?

I will look further into it, but any help is appreciated.

- student91
- Tetronian
**Posts:**317**Joined:**Tue Dec 10, 2013 3:41 pm

It's perfectly possible to have a coordinate-agnostic approach of dealing with polytopes, if you treat its combinatoric structure abstractly, as an abstract polytope with a containment hierarchy (a poset) among its i-dimensional elements, for 0 <= i <= n. To endow this abstract structure with geometric properties (since abstract polytopes include all kinds of stuff that are not necessarily faithfully representable in any kind of space), you can then impose local constraints, such as positive angle defect around vertices / edges, etc., shortchord lengths, dihedral, dichoral angles, and also global constraints like finite closure (rather than extending indefinitely or self-intersecting without ever returning to the starting point). You can also work with vectors between vertices abstractly, by inferring properties from things like angle defect, edge lengths, polygon shapes, etc.. In fact, if you look at a book like Coxeter's classic Regular Polytopes, he only uses coordinates occasionally, but most of the time derives results from geometric reasoning directly working with polytope elements.

P.S. Yeah, I think something is wrong with snub disphenoid coordinates that doesn't involve cube roots. I'm pretty sure it's a CVP 3 polyhedron. We should check whether those coordinates with only square roots actually yield a real snub disphenoid and not something else. (OTOH, the equation I did get for snub disphenoid coordinates are technically not degree 3, but degree 6. I'm kinda doubtful that you can solve a degree 6 polynomial with only square roots, but it seems remotely plausible because it's an even degree. The cube roots come from a u=x^2 substitution to make the order 6 polynomial possible to solve via cubic polynomial formulas. This is one reason that I feel a bit shaky about using polynomial degree to measure complexity -- nested square roots in theory can solve very high degree polynomials (though obviously not all), but does that really mean the polynomial is "constructible" in some sense, or is it just masking the fact that it's really cubic/quartic/etc. in essence? If a substitution intended to simplify calculations turn a supposed CVP 2 polyhedron into a CVP 3 one, then something somewhere is fundamentally flawed with our definition of CVP.)

P.S. Yeah, I think something is wrong with snub disphenoid coordinates that doesn't involve cube roots. I'm pretty sure it's a CVP 3 polyhedron. We should check whether those coordinates with only square roots actually yield a real snub disphenoid and not something else. (OTOH, the equation I did get for snub disphenoid coordinates are technically not degree 3, but degree 6. I'm kinda doubtful that you can solve a degree 6 polynomial with only square roots, but it seems remotely plausible because it's an even degree. The cube roots come from a u=x^2 substitution to make the order 6 polynomial possible to solve via cubic polynomial formulas. This is one reason that I feel a bit shaky about using polynomial degree to measure complexity -- nested square roots in theory can solve very high degree polynomials (though obviously not all), but does that really mean the polynomial is "constructible" in some sense, or is it just masking the fact that it's really cubic/quartic/etc. in essence? If a substitution intended to simplify calculations turn a supposed CVP 2 polyhedron into a CVP 3 one, then something somewhere is fundamentally flawed with our definition of CVP.)

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Well, x^6 - 1 = 0 is a trivial example of a sixth-degree equation solvable with square roots. Still, I might have made some mistake in my original calculation...

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

Marek14 wrote:Well, x^6 - 1 = 0 is a trivial example of a sixth-degree equation solvable with square roots. Still, I might have made some mistake in my original calculation...

Yes, but this is a reducible equation. It is equivalent to (x-1)(x+1)(x^2-x+1)(x^2-x-1)=0. So, every factor can be solved separately and with basic operations and square roots.

To test this, I tried to calculate the snub disphenoid coordinates by vectors.

There are eight 3d vertices, so 24 unknowns.

With symmetry, this can be reduced to four groups of points, with three unknowns.

1st layer: an edge with the most distant points to the center(0,±1/2, a+b)

2nd layer: two opposite points nearer to the center and orthogonal to the first (±r, 0, b)

3rd layer: two points orthogonal to the second and at similar position (0, ±r, -b)

4th layer, the last two points which are also joined in an edge (±1/2, 0, -a-b)

With the distance between points, the unknowns can be solved by a system:

1-2: |(0±r, ±1/2-0, a+b-b)^T|=1 ====> r²+1/4+a²=1² (for every ± combination)

1-3: |(0,r-1/2, a+b-(-b)^T|=1 =====> (r-1/2)²+(a+2b)^2(for equal signs combination)

2-3: |(r,r,b-(-b)^T|=1=====>2r²+(2b)²=1===>r²+2b²=1/2

I don't know how to solve this system, so I put on Wolfram:

Simplifying the answer, each variable is a function with square roots of the root of this equation:

x³+24x²-99x-54=0 near 4,01

More exactly:

-8+97/k+k, where k= (24sqrt(237)i-881)^1/3

And the functions are:

a=1/2 sqrt(x/3)

b=7/8sqrt(x/3)-7sqrt(3)/144 sqrt(x³)-sqrt(3)/432 (m^5)

r=-3/4+7m/24+m²/72

P.S. I really don't know where do these numbers come from, but surely they work. I've plotted the points into a program and the edge length was marked as 0,999999999999999. If the calculations are right, trying to eliminate them will change the 1/2 into a function of x, so I think the minimum polynomial is 3rd degree. And sorry by the plain text. I couldn't insert equations.

- JMBR
- Mononian
**Posts:**6**Joined:**Wed Mar 30, 2016 12:58 pm

Marek14 wrote:Well, x^6 - 1 = 0 is a trivial example of a sixth-degree equation solvable with square roots. Still, I might have made some mistake in my original calculation...

However, your result concerning the sphenocorona surely is correct! I asked Wolfram for coordinates of the sphenocorona, and he gave me exact, constructible coordinates .

So according to wolfram, the sphenocorona can be given as oaox2xxbo&#xt, with a=(6+sqrt(6)+y)/15≈1.70545388569283371, b=(9-sqrt(6)+y)/15≈1.57885525332174329772195, and the newly introduced y=2sqrt(213-57sqrt(6))≈17.1323185426093275640265628819748750.

This means that it is a little bit more likely that a polytope with sphenocoronae exists than a polytope with snub disphenoids.

Still working on a coordinate-agnostic approach though. I am experiencing problems exactly at the point where I try to impose local constraints. If you e.g. want to impose local angle defects, you have to find a way to calculate the ditopal angles of the surtopes, solely from the representation you have. Therefore, in such an approach you do need to include enough beforehand to derive different sets of polytopes (non-abstract, constructible, ones with a defineable dimension, etc.).

So actually, I want a generalization to be used to prove things, rather than a representation that does not encompass all possible things that I think should be called "polytopes".

I was thinking of defining a polytope as a set of vertices S, together with a distance-function f:S×S -> R, that gives the distance of two points s

However, this does not yet seem to be enough to define a polytope. I think when we take (S,f,r), where S and f are as before, and r is some as-of-yet-ill-defined notion of curvature (with r=0 for Euclidean space), any polytope given as such is unique up to isomorphism. (where an isomorphism is a function g:S->S with f(s

- student91
- Tetronian
**Posts:**317**Joined:**Tue Dec 10, 2013 3:41 pm

student91 wrote:Marek14 wrote:Well, x^6 - 1 = 0 is a trivial example of a sixth-degree equation solvable with square roots. Still, I might have made some mistake in my original calculation...

However, your result concerning the sphenocorona surely is correct! I asked Wolfram for coordinates of the sphenocorona, and he gave me exact, constructible coordinates .

So according to wolfram, the sphenocorona can be given as oaox2xxbo&#xt, with a=(6+sqrt(6)+y)/15≈1.70545388569283371, b=(9-sqrt(6)+y)/15≈1.57885525332174329772195, and the newly introduced y=2sqrt(213-57sqrt(6))≈17.1323185426093275640265628819748750.

This means that it is a little bit more likely that a polytope with sphenocoronae exists than a polytope with snub disphenoids.

Still working on a coordinate-agnostic approach though. I am experiencing problems exactly at the point where I try to impose local constraints. If you e.g. want to impose local angle defects, you have to find a way to calculate the ditopal angles of the surtopes, solely from the representation you have. Therefore, in such an approach you do need to include enough beforehand to derive different sets of polytopes (non-abstract, constructible, ones with a defineable dimension, etc.).

So actually, I want a generalization to be used to prove things, rather than a representation that does not encompass all possible things that I think should be called "polytopes".

I was thinking of defining a polytope as a set of vertices S, together with a distance-function f:S×S -> R, that gives the distance of two points s_{1}and s_{2}.

However, this does not yet seem to be enough to define a polytope. I think when we take (S,f,r), where S and f are as before, and r is some as-of-yet-ill-defined notion of curvature (with r=0 for Euclidean space), any polytope given as such is unique up to isomorphism. (where an isomorphism is a function g:S->S with f(s_{1},s_{2})=f(g(s_{1}),g(s_{2})))

Well, we want to limit ourselves to convex polytopes, which means that edges, faces and cells are direct consequence of the vertices. So the basic constraints for the distance function are:

f(a,a) = 0

f(a,b) >=1 for a!=b

triangular inequality

and, for polychora, each a must have f(a,x) = 1 for at least 4 different vertices x.

So, the first pass would generate the edges. But generating faces seems to be a bit complicated. A face search starts with three vertices a,b,c so f(a,b) = f(b,c) = 1. If f(a,c) = 1, you have a triangle. If f(a,c) is a shortchord of another polygon, you check whether there are other vertices that complete that polygon (if f(a,c) is sqrt(2), you're looking for a vertex d with f(a,d) = f(c,d) = 1, f(b,d) = sqrt(2)).

But this approach isn't perfect because a polychoron can contain a polygon without it being a face. 16-cell contains vertices that form a square, but it doesn't actually contain square faces.

I think you will need some sort of coordinates, unfortunately. With coordinates, you can create cells by taking possible quartets of vertices and drawing a hyperplane through them. If the hyperplane separates the remaining vertices, it's not valid, otherwise you can take all vertices lying on that hyperplane and analyze whether they form a CRF polyhedron...

Though the distance matrix could be easily transformed into coordinates.

- Marek14
- Pentonian
**Posts:**1102**Joined:**Sat Jul 16, 2005 6:40 pm

A polytope, in its most abstract, modern sense, is just a lattice (i.e., a kind of partially-ordered set of j-dimensional faces) that satisfies certain constraints. This definition is very all-encompassing, and includes things that maybe some of us wouldn't regard as a polytope. But if we add further constraints, such as convexity and/or faithful-representability in an n-dimensional space of curvature c, we would have a pretty good definition for our purposes.

Now, using coordinates is the easiest way to deriving properties like measuring ditopal angles, edge lengths, etc., but strictly speaking this is not necessary, because the face lattice fully defines the polytope up to isomorphism. For CRF polytopes, this then boils down to checking whether a given face lattice can be realized in a CRF way or not. Using explicit coordinates is one way to do such a check, but there are other ways. For example, CRF requires all 2-faces to be regular polygons, so given a face lattice, you can assume that all faces are regular polygons, then compute the angle defect around a given j-face. This, along with some trigonometry, would tell you the angles/lengths of adjacent j-faces, and you can check whether these angles are permissible (regular polygon shortchords, etc.). You wouldn't want to do this merely by vertices or edges, because as Marek said, some CRFs can contain regular polygons or parts of regular polygons without having them as actual faces. For example, if you diminished 1 vertex of an icosahedron, the result is still CRF, but its edges will outline many truncated pentagons (i.e., pentagons with the vertex that got deleted, so now they are 1:phi trapezoids). If you have the face lattice, however, this wouldn't be a problem, because these partial pentagons aren't actual faces of the diminished icosahedron; they are merely internal cross-sections.

As for how to obtain a face lattice from a given set of vertices: that's the well-known convex hull problem, for which there are a number of well-known algorithms. For our purposes, I'd say the best one is the so-called "double description" method, for which you can obtain implementations in C from Komei Fukuda, or C++ from a Russian guy in a project called "Skeleton". This is because some of the other methods don't work very well with polytopes that contain non-simplex facets. Algorithms like the one used in qhull will automatically split "degenerate" facets into simplices, sometimes by perturbing the vertices, because otherwise the algorithm wouldn't work properly. But this is very bad for us, because we want non-simplex facets in our CRFs! From my experience, the double-description algorithm is the best one that can handle the kind of polytopes we're interested in, even though from a performance POV it may not be the fastest or most accurate.

Generally, I highly recommend working with the full face lattice when programming polytope algorithms, because working with just vertices and edges often isn't enough, or requires a lot of tedious patchwork to make up for the lack of direct face information. Convex hull algorithms do require explicit coordinates, but you could just generate the face lattice directly without any reference to coordinates, and work from there instead.

Now, using coordinates is the easiest way to deriving properties like measuring ditopal angles, edge lengths, etc., but strictly speaking this is not necessary, because the face lattice fully defines the polytope up to isomorphism. For CRF polytopes, this then boils down to checking whether a given face lattice can be realized in a CRF way or not. Using explicit coordinates is one way to do such a check, but there are other ways. For example, CRF requires all 2-faces to be regular polygons, so given a face lattice, you can assume that all faces are regular polygons, then compute the angle defect around a given j-face. This, along with some trigonometry, would tell you the angles/lengths of adjacent j-faces, and you can check whether these angles are permissible (regular polygon shortchords, etc.). You wouldn't want to do this merely by vertices or edges, because as Marek said, some CRFs can contain regular polygons or parts of regular polygons without having them as actual faces. For example, if you diminished 1 vertex of an icosahedron, the result is still CRF, but its edges will outline many truncated pentagons (i.e., pentagons with the vertex that got deleted, so now they are 1:phi trapezoids). If you have the face lattice, however, this wouldn't be a problem, because these partial pentagons aren't actual faces of the diminished icosahedron; they are merely internal cross-sections.

As for how to obtain a face lattice from a given set of vertices: that's the well-known convex hull problem, for which there are a number of well-known algorithms. For our purposes, I'd say the best one is the so-called "double description" method, for which you can obtain implementations in C from Komei Fukuda, or C++ from a Russian guy in a project called "Skeleton". This is because some of the other methods don't work very well with polytopes that contain non-simplex facets. Algorithms like the one used in qhull will automatically split "degenerate" facets into simplices, sometimes by perturbing the vertices, because otherwise the algorithm wouldn't work properly. But this is very bad for us, because we want non-simplex facets in our CRFs! From my experience, the double-description algorithm is the best one that can handle the kind of polytopes we're interested in, even though from a performance POV it may not be the fastest or most accurate.

Generally, I highly recommend working with the full face lattice when programming polytope algorithms, because working with just vertices and edges often isn't enough, or requires a lot of tedious patchwork to make up for the lack of direct face information. Convex hull algorithms do require explicit coordinates, but you could just generate the face lattice directly without any reference to coordinates, and work from there instead.

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:The idea did occur to me, though, that rather than use the crown jewel Johnsons as cells, perhaps a more likely place to find 4D crown jewels would be to look for analogous constructions using simpler 4D cells. That's why yesterday I was musing about whether it's possible to insert a triangular prism into an icosahedral dipyramid such that the resulting deformation will close up in a CRF way. In some abstract, ill-defined sense, one might think of it as using the 4D geometry itself to produce a crown jewel shape, rather than start with a 3D crown jewel already "using up" some degrees of freedom, then try to close it up in 4D which requires even more degrees of freedom which there may not be "enough" of.

Nevertheless, I still can't help wondering if all of this is just a figment of our imagination (or lack thereof!), and that perhaps there is a way to close up a CRF containing snub disphenoid cells, just that we haven't thought of it yet.

I sort of see what you're getting at here, it's like how the snub disphenoid can be derived from the pentagonal bipyramid, or the sphenocorona or hebesphenomegacorona from the icosahedron; from the shape-shiftyness of higher-order apices (or pyramids) . this might seem obvious, but take the pentagonal pyramid for example:

if you then change the value for f, the o moves up or down accordingly, and as far as I know, the x-(f)-o angle is then solely dependent on the length of (f) (for which I'll have to figure out a nice expression )

- Code: Select all
`o o | o (o) | o`

x A | x x | x o

where A>f>B

o

x f o

o

x B

o

now you can create the snub disphenoid by splitting the pentagonal bipyramid at the rightmost o and decreasing the f value, until the lengt between the two o's is 1:

or create the sphenocorona, by taking the lace city of ike, deleting one x from the lace city and increasing the values of f until the length between two opposite o's has become 1 (from the top view, this'd look like crimping an f-edge until it is 1:

- Code: Select all
`o o o`

x f o x A

o o o

o | x

o C o | o D o

o o | o o

o o | o

f | B o

x (x)| x

f | B o

o o | o

(o) (o) |

f | x

x x | x x

f | D

o o | o o

or, what is even cooler, is that you take two pentagonal pyramids, as in ike, and decrease the value of f:

- Code: Select all
`o o`

f

x x

becomes:

o o

C x

x

becomes:

o o

x x x which is coplanar

this way, you could create the hebesphenomegacorona from ike, by taking the lace city, splitting the rightmost x, and decreasing the values of f, until the distance between the two x's has become 1:

- Code: Select all
`o o | o o | x x`

f | C x | B

x x | x | x x

f | C x | A

o o | o o | o o

there are two cases in which squippy actually does something:

- Code: Select all
`o | o o | o //squippy -> triangular bipyramid`

o u o | o x | o D

o

o | o o | o oct -> pentagonal bipyramid

o u o | o C | o f x

o | o o | o

if we now want to use an analogous method to create 4D shapes, we'd first have to consider the behaviour of 4D pyramids, of which the tetrahedal pyramid is rigid, but the octahedral, cubic, and icosahedral pyramids allow for freedom!

take for example the octahedral pyramid:

- Code: Select all
`o4o`

o4o x4o o4o

changing the x4o node, will break the octahedron, and result in a shallower, or higher bipyramid, as a result of which the dichoral angle changes:

- Code: Select all
`o4o o4o`

o4o A4o as A approaches o, this becomes a square :S

o4o

o4o B4o

o4o as B approaches some value, the distance between the leftmost and rightmost o4o might become one, which may result in something cool?

if we now take the 16-cell, split the rightmost o4o, and start expanding, what might happen?

- Code: Select all
`o4o | o4o o4o`

o4o x4o o4o | o4o A4o

o4o | o4o o4o

this might actually result in a sort-of-cool CRF for some value of A, analogous to how oct can transform into the pentagonal bipyramid

I'd of course have to calculate the right value for A, but I really have to get to my homework now

- student5
- Dionian
**Posts:**37**Joined:**Tue Feb 18, 2014 2:48 pm

Yes, that is exactly what I had in mind, but I didn't know how to represent it precisely. So thanks for having actual, precise notation for that.

Though as for whether that last lace city you posted will be CRF, seems to be much more constrained, because in 4D many more things have to line up with each other than in 3D. So I'm not sure if there are enough degrees of freedom left to make all edges unit length. It would be very interesting to find out which of these constructions have unit-edged (i.e. CRF) solutions, though!

Though as for whether that last lace city you posted will be CRF, seems to be much more constrained, because in 4D many more things have to line up with each other than in 3D. So I'm not sure if there are enough degrees of freedom left to make all edges unit length. It would be very interesting to find out which of these constructions have unit-edged (i.e. CRF) solutions, though!

- quickfur
- Pentonian
**Posts:**2484**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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