Marek14 wrote:I don't feel too well toght and part is a CRF idea that won't let me sleep, so I share it in a misguided attempt to fall asleep
I was thinking about various tetrahedra as vertex figures and one of them was this: a tetrahedron with two opposite edges of length sqrt(2) and other four 1.
This is a shape where four square pyramids fit to one vertex. I can imagine two more points of this figure (opposite points of the square faces), but I can't go further while trying to sleep. Is there a way to add more edges and end up with a CRF figure?
quickfur wrote:Klitzing wrote:I rather would try to lift the prove of completeness of Normans original research directly, by means of a brute force computer research sticking allowed polyhedra face to face, with lots of freedoms of deformation until those complexes would close finally. - Just could run into problems with the sheer number of to be found polychora, so that such a program wont end within a human lifetime...
I've been thinking about that, but the major difficulty is the sheer number of 3D CRFs, which means a giant combinatorial explosion, coupled with the complexity of representing partial CRF complexes in an efficient way that would allow algorithmic detection of when something can be closed up to make a CRF polychoron.
Marek14 wrote:I posted list of 3D verfs yesterday, so now next step should be building 4D verfs, right?
Klitzing wrote:Marek14 wrote:I posted list of 3D verfs yesterday, so now next step should be building 4D verfs, right?
Dont forget list2, the pairings, including all possible orientations:
- the [3,4,5,4] of srid is clear, no orientation to be encountered.
- but with respect to the johnsonian diminished srid versions or its gyrated versions, the question arises, which [3,4,5,4]-vertex was taken, that is, how the johnson solid has to be oriented, when applying that specific verf.
This is why not only verfs as such are relevant, but those pairings.
--- rk
Marek14 wrote:Actually, thinking about it now, we don't actually NEED the list of various ways a polyhedron can fit a verf. We only need those verfs themselves.
Why? Well, verf can be imagined in various ways, but let's go with the most "solid" one -- let's interpret the verf as a chunk cut off of the polychoron, with full edges and parts of faces (full faces for triangles).
Fitting two verfs next to each other means you overlap an edge and blend the faces around it. It's like interlocking of lego blocks. And as for cells, every verf around a cell will contribute something to its outline and since all faces are regular and co-hyperplanarity is easy to check, we'll automatically get CRF cells as a result; we don't have to enter them. So instead of trying all polyhedra in all possible orientations, we might just add other verfs around and some polyhedron or other will appear.
As an aside, I wonder how versatile the various gyrate/diminished rhombicosidodecahedra are. They should figure in a family of srahi (small rhombated 120-cell) derivatives since you can cut a 5g||10p cupola and either leave it or glue it back gyrated. In addition to more traditional diminishings by removing rhombicosidodecahedron||truncated dodecahedron caps.
Klitzing wrote:[...]
Here is a way how to set up that brute force algorithm:
1. enlist all to be used 3d solids: Archimedeans, Prismatics, Jonsonians.
2. enlist all vertex figures of those as pairs: verf + solid it comes from (possibly even counting its orientation!) Note, all those verfs are rigid!
3. try to build up all solids out of those verfs (and multiply those by all those pairings), resulting in a list of all possible vertex surrounding complexes. Note, again all those constructed potential verfs of 4d solids, as being built from rigid faces, and additionally being closed, will be rigid as well!
4. now gradually build them up: take any edge emanating from an already completely surrounded vertex to an uncompletely surrouded vertex. Try to find all counterparts out of the list of step 4, which would complete that vertex. Thus the complexes iteratively would grow. You might run into a blind end, but else you should get it finally closed: thus you would have constructed a biesty you were looking for.
EDIT1: Suppose a good try for this algorithmus wood be to calculate all 4d CUCs. Here list1 wood be smaller already. But the main benefit wood be at list2: no different orientations have to be taken in account! So this procedure wood be a good test for the program, while its application to search for the 4d CRFs = 4d CUHs then would just become a question of memory space and calculation time.
--- rk
Marek14 wrote:We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.
quickfur wrote:Marek14 wrote:We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.
But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.
Marek14 wrote:quickfur wrote:Marek14 wrote:We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.
But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.
You mean past 3 polygons per edge -- there must be always at least 4 per vertex.
But no, that won't make the vertex become nonrigid. You're concentrating on that one edge and forgetting that the edge can't exist alone -- you also need the rest of the vertex figure and that will put additional constraints. In case of 10-polygon edge, there will have to be at least 5-6 other polygons at the same vertex...
quickfur wrote:Klitzing wrote:[...]
Looks crazy to me if xfo3ooxPooo&#x would have true teddis for P=3, 5; but not for P=4.
You're right, I made another mistake. The icosidodecahedron is o3x5o, not x3o5x.
Unless I made a mistake in my calculations for the octahedron/cuboctahedron case, it's pretty crazy.
# x4o3x
< 1, 1, (1+sqrt(2)), sqrt(2*phi)>
< 1, 1, -(1+sqrt(2)), sqrt(2*phi)>
< 1, -1, (1+sqrt(2)), sqrt(2*phi)>
< 1, -1, -(1+sqrt(2)), sqrt(2*phi)>
<-1, 1, (1+sqrt(2)), sqrt(2*phi)>
<-1, 1, -(1+sqrt(2)), sqrt(2*phi)>
<-1, -1, (1+sqrt(2)), sqrt(2*phi)>
<-1, -1, -(1+sqrt(2)), sqrt(2*phi)>
< 1, (1+sqrt(2)), 1, sqrt(2*phi)>
< 1, (1+sqrt(2)), -1, sqrt(2*phi)>
< 1, -(1+sqrt(2)), 1, sqrt(2*phi)>
< 1, -(1+sqrt(2)), -1, sqrt(2*phi)>
<-1, (1+sqrt(2)), 1, sqrt(2*phi)>
<-1, (1+sqrt(2)), -1, sqrt(2*phi)>
<-1, -(1+sqrt(2)), 1, sqrt(2*phi)>
<-1, -(1+sqrt(2)), -1, sqrt(2*phi)>
< (1+sqrt(2)), 1, 1, sqrt(2*phi)>
< (1+sqrt(2)), 1, -1, sqrt(2*phi)>
< (1+sqrt(2)), -1, 1, sqrt(2*phi)>
< (1+sqrt(2)), -1, -1, sqrt(2*phi)>
<-(1+sqrt(2)), 1, 1, sqrt(2*phi)>
<-(1+sqrt(2)), 1, -1, sqrt(2*phi)>
<-(1+sqrt(2)), -1, 1, sqrt(2*phi)>
<-(1+sqrt(2)), -1, -1, sqrt(2*phi)>
# x4o3f
< 1, 1, (1+phi*sqrt(2)), 0>
< 1, 1, -(1+phi*sqrt(2)), 0>
< 1, -1, (1+phi*sqrt(2)), 0>
< 1, -1, -(1+phi*sqrt(2)), 0>
<-1, 1, (1+phi*sqrt(2)), 0>
<-1, 1, -(1+phi*sqrt(2)), 0>
<-1, -1, (1+phi*sqrt(2)), 0>
<-1, -1, -(1+phi*sqrt(2)), 0>
< 1, (1+phi*sqrt(2)), 1, 0>
< 1, (1+phi*sqrt(2)), -1, 0>
< 1, -(1+phi*sqrt(2)), 1, 0>
< 1, -(1+phi*sqrt(2)), -1, 0>
<-1, (1+phi*sqrt(2)), 1, 0>
<-1, (1+phi*sqrt(2)), -1, 0>
<-1, -(1+phi*sqrt(2)), 1, 0>
<-1, -(1+phi*sqrt(2)), -1, 0>
< (1+phi*sqrt(2)), 1, 1, 0>
< (1+phi*sqrt(2)), 1, -1, 0>
< (1+phi*sqrt(2)), -1, 1, 0>
< (1+phi*sqrt(2)), -1, -1, 0>
<-(1+phi*sqrt(2)), 1, 1, 0>
<-(1+phi*sqrt(2)), 1, -1, 0>
<-(1+phi*sqrt(2)), -1, 1, 0>
<-(1+phi*sqrt(2)), -1, -1, 0>
# x4x3o
< 1, (1+sqrt(2)), (1+sqrt(2)), -sqrt(2/phi)>
< 1, (1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
< 1, -(1+sqrt(2)), (1+sqrt(2)), -sqrt(2/phi)>
< 1, -(1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
<-1, (1+sqrt(2)), (1+sqrt(2)), -sqrt(2/phi)>
<-1, (1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
<-1, -(1+sqrt(2)), (1+sqrt(2)), -sqrt(2/phi)>
<-1, -(1+sqrt(2)), -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), 1, (1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), 1, -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), -1, (1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), -1, -(1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), 1, (1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), 1, -(1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), -1, (1+sqrt(2)), -sqrt(2/phi)>
<-(1+sqrt(2)), -1, -(1+sqrt(2)), -sqrt(2/phi)>
< (1+sqrt(2)), (1+sqrt(2)), 1, -sqrt(2/phi)>
< (1+sqrt(2)), (1+sqrt(2)), -1, -sqrt(2/phi)>
< (1+sqrt(2)), -(1+sqrt(2)), 1, -sqrt(2/phi)>
< (1+sqrt(2)), -(1+sqrt(2)), -1, -sqrt(2/phi)>
<-(1+sqrt(2)), (1+sqrt(2)), 1, -sqrt(2/phi)>
<-(1+sqrt(2)), (1+sqrt(2)), -1, -sqrt(2/phi)>
<-(1+sqrt(2)), -(1+sqrt(2)), 1, -sqrt(2/phi)>
<-(1+sqrt(2)), -(1+sqrt(2)), -1, -sqrt(2/phi)>
quickfur wrote:Marek14 wrote:quickfur wrote:Marek14 wrote:We shouldn't have a problem with nonrigid vertices because the vertex figures are built from a finite, discontinuous set of polygons (including skew polygons). That should mean they can't vary that much.
But what if one of the polygons is, say, an isosceles triangle with a very narrow base? Then we can fit many copies of it around a vertex (e.g. to make a shape like a 10-gonal pyramid), and once you get past 3 polygons per vertex, it becomes non-rigid.
You mean past 3 polygons per edge -- there must be always at least 4 per vertex.
But no, that won't make the vertex become nonrigid. You're concentrating on that one edge and forgetting that the edge can't exist alone -- you also need the rest of the vertex figure and that will put additional constraints. In case of 10-polygon edge, there will have to be at least 5-6 other polygons at the same vertex...
I think you misread me. I wasn't talking about the 4D case; I was talking about the preliminary step of generating the 3D verfs. So you have a bunch of polygons (which are verfs of CRF polyhedra) and you want to assemble them together to form verfs of potential 4D CRFs. How do you find all such verfs? You have to assemble these polygons into polyhedra, of course. But how do you find all polyhedra that can be made from these polygons? One way is the reuse the algorithm to brute force search all such polyhedra -- except you're running it for the 3D case now. So you make verfs of the polygons (verfs of the verfs), i.e., take the chords of the polygons, and then try to assemble possible verfs from those line segments. The radius of the polygon is restricted, of course, since it must be smaller than the length of edges emanating from the vertex corresponding with the verf.
But this is still not enough to fully define the polygon. A short chord (say from a narrow isosceles triangle) allows you to make a big polygon as verf. So that's equivalent to making a large n-gonal pyramid from the isosceles triangles. When n>3, the apex of the pyramid is non-rigid; you can vary the dihedral angles of the isosceles triangles. So how do we know which angles will lead to a closed 4D verf? We have to solve the non-rigid peak (vertex) problem for the 3D case.
Keiji wrote: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?
Marek14 wrote:[...]
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
Keiji wrote:Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?
quickfur wrote:Marek14 wrote:[...]
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.
Marek14 wrote:quickfur wrote:Marek14 wrote:[...]
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.
quickfur wrote:Keiji wrote:Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?
I believe you need even dimensions to get the higher-dimensional equivalent of the Hopf fibration (which is what the BXD is based on). But this is just a wild guess, so I might be completely wrong.
And teddies aren't required for these things; all the regular polychora (except perhaps the 5-cell) exhibit this spiralling structure. The bitruncated 24-cell does, too (and probably the bitruncated 5-cell as well). The tesseractic family mainly exhibit the 2-ring structure of the duoprisms (which is a subset of this spiralling symmetry -- think the first ring and the equatorial ring). The 24-cell family exhibits the 8-ring structure you see in the BXD. The 120-cell family exhibits more spiralling rings -- 12 of them, IIRC. But we just don't notice it because there are too many other interesting symmetries that obscure it.
quickfur wrote:Marek14 wrote:quickfur wrote:Marek14 wrote:[...]
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.
Wait, what? I thought we were talking about taking Johnson solids' vertex figures and putting them (the verfs) together to make verfs for potential CRFs. Or did you mean dual Johnson solids? I'm confused.
Keiji wrote:quickfur wrote:Keiji wrote:Another point to ponder: knowing the existence of the BXD (bixylodiminished hydrochoron) as well as the higher-dimensional ursatopes, does this mean there are equivalents to the BXD in 5D and up, one for each ursatope, which are also chiral, CRF, vertex- and cell-transitive? Or perhaps they would only exist in even dimensions 6D, 8D etc, due to the spiralling structure required?
I believe you need even dimensions to get the higher-dimensional equivalent of the Hopf fibration (which is what the BXD is based on). But this is just a wild guess, so I might be completely wrong.
And teddies aren't required for these things; all the regular polychora (except perhaps the 5-cell) exhibit this spiralling structure. The bitruncated 24-cell does, too (and probably the bitruncated 5-cell as well). The tesseractic family mainly exhibit the 2-ring structure of the duoprisms (which is a subset of this spiralling symmetry -- think the first ring and the equatorial ring). The 24-cell family exhibits the 8-ring structure you see in the BXD. The 120-cell family exhibits more spiralling rings -- 12 of them, IIRC. But we just don't notice it because there are too many other interesting symmetries that obscure it.
That all makes sense, but see my bolded conditions above - the regular polytopes aren't chiral.
Also, it's interesting that the duoprisms are a special case with 2 rings - what are the possibilities with respect to number of rings? Does it have to be even, is there a maximum, are there polychora with 6 or 10 rings?
Marek14 wrote:quickfur wrote:Marek14 wrote:quickfur wrote:Marek14 wrote:[...]
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.
Wait, what? I thought we were talking about taking Johnson solids' vertex figures and putting them (the verfs) together to make verfs for potential CRFs. Or did you mean dual Johnson solids? I'm confused.
Yes, taking vertex figures of Johnson solids and building CRF verfs out of them.
quickfur wrote:Well, then you've found a new direction to search for CRF polychora in.
Hopf fibration
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