General Approach--can 3D methods be generalized?

[Marek14]

And final base configuration, (3,5,3,5) corresponding to icosidodecahedron and related polyhedra. Same as in (3,4,5,4), both diagonal squares are (5+Sqrt[5])/2.

Solutions are:

3535-3333: hyperbolic solution.

3535-3344: Blend of 334-045 acrochoron and 334-530 acrochoron joined in pentagonal cupola.

3535-3355: Blend of 335-055 acrochoron and 335-530 acrochoron joined in pentagonal rotunda.

3535-3553: Blend of 355-530 acrochoron and 333-550 acrochoron joined in pentagonal rotunda.

3535-3504: Blend of 350-045 acrochoron and 334-550 acrochoron joined in diminished rhombicosidodecahedron, or blend of 355-030 acrochoron and 334-050 acrochoron joined in truncated dodecahedron.

3535-3003: Blend of 350-030 acrochoron and rhodomesohedral rotunda joined in truncated dodecahedron.

3535-4444: standard prismatic solution.

3535-4466: Blend of 346-065 acrochoron and expanded/truncated rhodomesohedral rotunda joined in truncated icosidodecahedron.

3535-4004: hyperbolic solution.

3535-5555: flat solution. Surrounding of icosidodecahedron with dodecahedra and tridiminished icosahedra.

3535-6666: hyperbolic solution.

[Marek14]

OK, it took a while, but this should be all verfs in shape of quadragonal pyramid.

[johannes@itap.physik.uni-stuttgart.de]

Recently, Marek has determined all the vertex figures CRF polychora based on
tetraehdra (type abc-def) and those based on quadragonal pyramids (44 cases,
type 3bcd-efgh). I have completed the study for the 24 cases based on
pentagonal pyramids (type abcde-fghij).

Here is the table of the possible bases:

3 3 3 3 3 0.17323532263 117.356 143.738 143.738 117.356 161.4829 2.18927 2.70949 2.70949 2.18927 2.92234 J88
3 3 3 3 3 0.18411909463 96.1983 166.441 121.743 121.743 166.441 1.66196 2.95819 2.28917 2.28917 2.95819 J84
3 3 3 3 3 0.19150678129 109.4712 144.7356 144.7356 109.4712 169.4712 2.00000 2.72474 2.72474 2.00000 2.97474 J50
3 3 3 3 3 0.19556348419 109.4712 164.2596 118.892 135.992 152.1911 2.00000 2.94375 2.22474 2.57886 2.82676 J87
3 3 3 3 3 0.19683051116 109.4712 158.572 127.552 127.552 158.572 2.00000 2.89631 2.41421 2.41421 2.89631 J10
3 3 3 3 3 0.19699313021 114.645 144.144 144.144 114.645 164.2578 2.12550 2.71573 2.71573 2.12550 2.94374 J85
3 3 3 3 3 0.19864310695 128.496 157.148 111.735 157.148 128.496 2.43369 2.88227 2.05547 2.88227 2.43369 J89
3 3 3 3 3 0.19954254423 171.755 86.7268 171.646 117.356 117.356 2.98408 1.41436 2.98408 2.18927 2.18927 J88
3 3 3 3 3 0.20088090366 118.892 159.892 118.892 143.479 143.479 2.22474 2.90857 2.22474 2.70545 2.70545 J86
3 3 3 3 3 0.20660038687 148.434 133.591 133.591 148.434 124.702 2.77806 2.53426 2.53426 2.77806 2.35396 J90
3 3 3 3 3 0.20727639431 128.496 141.31 141.31 128.496 149.565 2.43369 2.67132 2.67132 2.43369 2.79330 J89
3 3 3 3 3 0.20886393771 143.479 135.992 135.992 143.479 131.442 2.70545 2.57886 2.57886 2.70545 2.49278 J86
3 3 3 3 3 0.20965059100 138.1898 138.1898 138.1898 138.1898 138.1898 2.61803 2.61803 2.61803 2.61803 2.61803 ico
3 3 3 3 4 0.24291936021 164.25964 109.47122 171.75464 109.52403 159.89240 2.94375 2.00000 3.71415 2.57886 2.90857 J87
3 3 3 3 4 0.27332082997 171.64574 129.445 154.722 137.24 143.738 2.98408 2.45300 3.56620 3.27168 2.70949 J88
3 3 3 3 4 0.27368887139 159.187 159.187 132.624 159.095 126.964 2.90211 2.90211 3.04151 3.61803 2.51578 J24
3 3 3 3 4 0.27607141638 145.222 145.222 169.428 125.264 153.635 2.73205 2.73205 3.55189 3.00000 2.97454 J22
3 3 3 3 4 0.27721017186 166.81137 133.59119 154.41883 136.33594 148.43399 2.96044 2.53426 3.56227 3.25297 2.77806 J90
3 3 3 3 4 0.28136914499 157.14815 157.14815 141.34110 152.97558 133.97281 2.88227 2.88227 3.20259 3.54293 2.67132 J89
3 3 3 3 4 0.28253596482 164.25739 144.14362 145.44063 145.44063 144.14362 2.94374 2.71573 3.42641 3.42641 2.71573 J85
3 3 3 3 4 0.28553475395 153.962 153.962 151.33 144.736 141.595 2.84776 2.84776 3.35729 3.41421 2.81610 J23
3 3 3 3 4 0.28565364345 153.235 153.235 153.235 142.983 142.983 2.83929 2.83929 3.38298 3.38298 2.83929 snucub
3 3 3 3 5 0.35293273058 159.187 159.187 174.434 142.6227 158.682 2.90211 2.90211 3.84358 3.61803 2.99293 J25
3 3 3 3 5 0.35886935933 164.172 164.172 164.172 152.93 152.93 2.94315 2.94315 3.77584 3.77584 2.94315 Snudod

1-5. column: edges
6. column: solid angle
7-11. column: dihedral angle between nth and n+1sr face
12-16. column: square of length of diagonal between nth and n+2nd vertex (2: square, 2.618: pentagon, 3: hexagon, 3.414: octagon, 3.618: decagon.

Trivially, the prismatic solution 3333a-44444 exists for all cases.

The other solutions are:

33333-(10)5335 for the 10.row (J90), 12. row (J86) and the 13.row (icosahedron).

33333-34(10)(10)4 exist for the icosahedron only.

Furthermore

33333-33333 (icosahedral pyramid)
33333-44444 (icosahedral prism)
33333-55555 (tridiminished icosahedra on icosahedron faces)
33333-66666 (vertex figure of truncated hexacosichoron)

[Marek14]

Very interesting! I see you've managed to acquire better data for the "crown jewels" than me -- could you please look at the quadrangle pyramids with "irregularly" skewed bases to see if there are any I've overlooked?
This includes:
(3,3,3,3) vertex in J84
(3,3,3,3) vertex in J88
(3,3,3,4) vertex in J86/J87
(3,3,4,4) vertex in J86
(3,3,4,4) vertex in J88
(3,3,4,4) vertex in J89
(3,3,4,4) vertex in J90

As for your solutions, they look very interesting. 33333-05335 (two pentagonal rotundas, two pentagonal pyramids and one tetrahedron around an edge) fitting to J86 or J90 is unexpected. I presume only one specific arrangement of these cells work?

So, what next? Triangular bipyramids have the problem that there additional convexity constraints that must be checked. How about triangular prisms as verfs? Those have two triangles and three quadrangles and it might be interesting how the quadrangles fit together.

[johannes@itap.physik.uni-stuttgart.de]

I recently determined some limits on the number of edges and faces of the CRF
polychora vertex figures.

There is a program named plantri (http://cs.anu.edu.au/~bdm/plantri/) with
which you can enumerate all polyhedra in three dimensions. With the qualifiers
of plantri you can limit the number of edges of the faces and the number of
edges of the polyhedra, the degree of the vertices and so on.

If you compute all the solid angles of the Johnson polyhedra you find that the
maximal number of faces is 22 for the regular tetrahedron. However, we know
that no more than 20 tetrahedra fit together in 4d to form an icosahedron. The
maximal number of vertices is much more difficult to determine. It could in
principle go up to 30=(12-2)*3, if all vertices are simple. However, in the
end I obtained 13, and my believe is that it is 12 namely the icosahedron.

What other limits do we have? The dihedral angles of the Johnson polyhedra
determine the maximal degree of the vertices, i.e. the number of faces that
can meet at a vertex. The minimal diehdral angle is 31.7175 from the
pentagonal cupola which would lead to a degree of 11. But since the vertex is
asymmetric, the maximal number of tiles has to be even, namely 10. But we find
that there is no polyhedron which fits into the gap between two pentagonal
cupolas. Thus this vertex does not exist. I have enumerated all possible edge
configurations and found further 10-degree vertices, but it was possible to
show that none of them exists. In the end the maximal degree was found to be 9
for 4 configurations which I could not eliminate.

How else could one reduce the possible vertex figures? Well by computing the
total solid angle: The smallest triangle has solid angle 00.4387, the
quadrangle 0.08774, the pentagon 0.173235 in units of the surface of a
sphere. Thus #3*0.04387+#4*0.08774+#5*0.173235 < 1.0.

So the number of possible vertex figures of a certain type is lower than given
in the following table:

#vert 4 5 6 7 8 9 10 11 12 13
1 2 7 33 249 2473 29846 394498 5528006 79919023 <- sum
#face
4 1 1
5 2 1 1
6 7 1 2 2 2
7 25 2 7 9 5 2
8 149 2 11 39 55 35 7
9 944 8 71 248 379 235 3
10 ^ 5 76 590 1976 2930 389
11 | 38 748 5290 16401 11684 3
12 sum 14 558 8309 50226 109398 409
13 219 7776 91966 449409 15892
14 50 4442 106558 1008926 213923
15 1404 78684 1400693 1296312
16 233 36528 1282828 4157926
17 9714 780953 7698651
18 1249 306470 8609942
19 70454 5875223
20 7595 2384890

Obviously, this estimate is completely useless. But it shows that it is
impossible by starting from the combinatorical vertex figure to determine the
allowed vertex figures in 4d. Because each number means only a type! There are
more than 100 vertex figures of type (4,4), 44 of type (5,5), 32 of tpye (6,6)
and maybe several hundred of the biypyramidral type (5,6) with
(#vert,#faces). I have determined the latter but not yet fully analysed them.
Although I could show that there are no infinite series for the bipyramidal
case (an the triangular pyramidal case (6,5), there seem to be vertex figures
with polygons with up to 169 edges!

To my opinion, it would be interesting to list the combinatorial content of all vertex figures of all known CRF polyhedral. I started such work, but I have not yet finished the segmentochora. Compared to the above list one finds that there is a lot of space at the top. The maxmium vertex figure is still the icosaedron with 12 vertices and 20 faces.

[Marek14]

Very interesting. I think that the basic algorithm for CRF enumeration is not that hard, just requires a lot of space:

1. Start with a polyhedron set in 4D space and mark all its faces as "unpaired". These polyhedra will be Lv. 1 open-CRF (oCRF) polychora.
2. To make a Lv. n+1 oCRF from Lv. n, add a polyhedron to an unpaired face and mark that face as "paired".
3. Check gaps between faces. If a gap has two identical polygons, attempt to pull them together (bending cells into 4D). If successful (and convex), mark that face as paired and this becomes a new Lv. n oCRF that will be checked from step 3 (this one) onward.
4. For every gap, check whether a new polyhedron can be inserted into it. If at least one gap has nothing to fit into, the oCRF is invalid and eliminated from further considerations. If all gaps can be fitted with something, it's retained (the fittings will be added later). If there are no gaps (i.e. all faces are paired), the oCRF is closed and a CRF.
5. If you have a valid oCRF or CRF at this stage, perform combinatoric comparison with others Lv. n oCRFs found so far to eliminate duplicates.

This would have to be expanded a lot (especially step 3), but theoretically, it could allow for complete enumeration. Note that this needs 4D to work: it won't enumerate Johnson solids in 3D because the vertices in 3D are not fully determined by the polygons around them; in 4D, the cells around a vertex DO determine that vertex completely since there can be no continuous changes of angles.

[johannes@itap.physik.uni-stuttgart.de]

Sorry, I just see that my matrices of the combinatorial vertex figures got mixed up completely. I will post them again when I have time.

My crown jewel data are from antiprism, which again are from the NIST data base, also used in Mathematica. The data of triangular vertices, however, are computed directly. There are errors in antiprism and NIST which I corrected. Especially J65, J66, J68, J88 and J87 are wrong.

My results were derived numerically, and a comparison to Marek's results show that very high precision does not seem to be required.

Johannes

[Marek14]

Looking at the possible 6-vertex verfs, there seems to be these seven:

12 edges/8 faces: Tri-tetrahedron (a tetrahedron with pyramid built on two of its faces). 8 triangles. Vertex degrees: 3,3,4,4,5,5. Could be split (in two ways) into triangular bipyramid and tetrahedron.
12 edges/8 faces: Octahedron. 8 triangles. Vertex degrees: 4,4,4,4,4,4. Could be split (in three ways) into two square pyramids.

11 edges/7 faces: Augmented square pyramid (a square pyramid with tetrahedron on one of its faces). 6 triangles + 1 square. Vertex degrees: 3,3,3,4,4,5. Could be split in square pyramid + tetrahedron, or formed in two ways as a blend of triangular bipyramid + tetrahedron.
11 edges/7 faces: Diagonal square wedge (diagonally oriented line connected to base square). 6 triangles + 1 square. Vertex degrees: 3,3,4,4,4,4. Could be formed as a blend of two square pyramids.

10 edges/6 faces: Pentagonal pyramid. 5 triangles + 1 pentagon. Vertex degrees: 3,3,3,3,3,5. Could be formed as a blend of square pyramid and tetrahedron in five ways.
10 edges/6 faces: Broken triangular prism (triangular prism with one square face broken in two triangles). 4 triangles + 2 squares. Vertex degrees: 3,3,3,3,4,4.

9 edges/5 faces: Triangular prism. 2 triangles + 3 squares. Vertex degrees: 3,3,3,3,3,3.

[johannes@itap.physik.uni-stuttgart.de]

I want to draw your attention also to this page:

https://www.uwgb.edu/dutchs/symmetry/polynum0.htm

where there are plots of the combinatorial polyhedra.

Johannes

[johannes@itap.physik.uni-stuttgart.de]

And here are the combinatorical vertex figures again.

Johannes

[Marek14]

Hm, last update from 1999. I wonder if that profesor is still alive...

[JMBR]

Hello again. If I can remember it well, one of the steps of enumerating johnson solids involved the vertex angle defect.
In 2 dimensions, the exterior angles of every non-self-intersecting polygon sum up to 2π because it seems like they can be deformed into a circle. The case in 3 dimensions sums up to 4π, equivalent to a sphere. My question is: is there an equivalent of that in 4 dimensions? The complete theorem is the Gauss-Bonnet theorem, but when I tried to find such relation, I drowned in heavy calculus. So I was curious to know if someone found this result.

[Marek14]

Hello again. If I can remember it well, one of the steps of enumerating johnson solids involved the vertex angle defect.
In 2 dimensions, the exterior angles of every non-self-intersecting polygon sum up to 2π because it seems like they can be deformed into a circle. The case in 3 dimensions sums up to 4π, equivalent to a sphere. My question is: is there an equivalent of that in 4 dimensions? The complete theorem is the Gauss-Bonnet theorem, but when I tried to find such relation, I drowned in heavy calculus. So I was curious to know if someone found this result.

If it's equivalent to hypersphere, then it would be 2π^2.

[JMBR]

But if that's the case, there's some kind of angle defect which has this sum. What would it be? Dihedral angles and vertex solid angles are not comensurable with π, except for the cube (which can tile the space alone) and, maybe, truncated octahedron. https://en.wikipedia.org/wiki/Platonic_solid#Geometric_properties
That led me to question if this invariant even exists, which it does because of that theorem.

[mr_e_man]

This has been at the back of my mind for months, as I'm trying to reduce the infinite list of polyhedra to a finite list:

The last three are generally valid and they all correspond to n-gonal antiprism joined to n-gonal prism with cupolas and truncated polyhedra filling the gaps.
333n-6644: n-gonal antiprism, two triangular cupolas, truncated tetrahedron, n-gonal prism.
333n-8844: n-gonal antiprism, two square cupolas, truncated cube, n-gonal prism.
333n-0044: n-gonal antiprism, two pentagonal cupolas, truncated dodecahedron, n-gonal prism.

I finally got around to proving that, for n≥7, these vertex types can't be used in CRF polychora. The vertex thus described gets a sharp part of a cupola, so an adjacent vertex gets a blunt part of a cupola, which takes up a lot of space; there's no way to complete that vertex.

I could give more details, if anyone wants.

[mr_e_man]

If there's a trivalent edge with an n-gon prism and antiprism (n≥7) and a triangular prism (or J7 or J26 or J49 at a 90° edge), then both vertices must be what Marek denotes 333n-4444. It follows that the n-gon must be surrounded by a ring of (possibly augmented) triangular prisms. Furthermore (with possible exceptions when n=8 or 10), the n-gon prism must have n-gon antiprisms on both sides, and an n-gon antiprism must have n-gon prisms on both sides; the result is a known uniform polychoron, an antiprismatic prism. So nothing new can be made from such an edge configuration.

Similarly, if there's a trivalent edge with an n-gon prism and antiprism and a square pyramid, then both vertices must be what Marek denotes 333n-3344. It follows that the n-gon must be surrounded by a ring of square pyramids and (possibly augmenting or elongated etc.) tetrahedra. Furthermore (again with possible exceptions when n=8 or 10), the n-gon antiprism must have an n-gon antiprism on its other side, attached to the squippy and tet triangles. This already closes up everything; the result is a known segmentochoron, n-gon prism || dual n-gon. Nothing new here either.

[mr_e_man]

viewtopic.php?p=22993#p22993

Unfortunately, now I can't find any solutions for these despite the fact that I know that 3333-4444 must always have a solution (it's a prism) and 3333-3464 should have as well. Until someone helps me here, I'll assume that the non-quadratic polyhedra only have solutions that all others have...

So, what do we get? Well, pentagonal dipyramid (and presumably snub disphenoid and sphenomegacorona) have the same three solutions as triangular dipyramid: A solution for 3333-4444 (prism) and two solutions for 3333-3464 (surrounding of degree-4 vertex with two square pyramids and two triangular cupolas with two variants based on which pair of edges has squares built on them and which has triangle and hexagon).

How did you conclude that 3464 works in general? It's not obvious to me.

The other solutions are:

33333-(10)5335 for the 10.row (J90), 12. row (J86) and the 13.row (icosahedron).

33333-34(10)(10)4 exist for the icosahedron only.

As for your solutions, they look very interesting. 33333-05335 (two pentagonal rotundas, two pentagonal pyramids and one tetrahedron around an edge) fitting to J86 or J90 is unexpected. I presume only one specific arrangement of these cells work?

The base has a specific line of symmetry. The rotundas and pyramids can be arranged symmetrically in only one way. But it's possible that the complete configuration has less symmetry than the base.

[Marek14]

http://hi.gher.space/forum/viewtopic.php?p=22993#p22993

Unfortunately, now I can't find any solutions for these despite the fact that I know that 3333-4444 must always have a solution (it's a prism) and 3333-3464 should have as well. Until someone helps me here, I'll assume that the non-quadratic polyhedra only have solutions that all others have...

So, what do we get? Well, pentagonal dipyramid (and presumably snub disphenoid and sphenomegacorona) have the same three solutions as triangular dipyramid: A solution for 3333-4444 (prism) and two solutions for 3333-3464 (surrounding of degree-4 vertex with two square pyramids and two triangular cupolas with two variants based on which pair of edges has squares built on them and which has triangle and hexagon).

How did you conclude that 3464 works in general? It's not obvious to me.

I think it was in the message before that:
"So, a working hypothesis is that when lateral edges have 3464 or 4444 configuration, the quadrangle can be arbitrarily skewed and the shape will still fit. The fact that sums of angles of both pairs of opposite faces is 180 degrees is probably related."
But note that this was seven years ago, and I haven't looked at this problem for quite some time

[mr_e_man]

Success!

I think I've proven that there are no CRF polychora involving n-gons, for n=7,9,11,12,..., other than the known ones: antiprismatic prisms, biantiprismatic rings (or wedges), and duoprisms (possibly augmented with square or pentagonal prism pyramids or magnabicupolic rings). Any new polychora must have only 3,4,5,6,8,10-gons. It follows also that there are only finitely many CRF polychora outside of these three known infinite families.

But I might have made a mistake somewhere. I sketched hundreds of spheric multihedra, and did some angle calculations for each one, as I tried to build up the verf in all possible ways (starting from n.3.3.3 or n.4.4). Pictures can be misleading; for example, they might suggest that two (incomplete) vertices are distinct, that two polygons don't touch each other, when in fact they may share a vertex. Numerical errors are another problem, but I was careful to use many digits, and to investigate any numbers that got too close to each other.

This really should be automated. We need a program that finds all convex polyhedra with faces from a given set of polygons, perhaps with an input multihedron (incomplete polyhedron) that any output polyhedron should contain as a subset. The first version of the program doesn't need to prove that certain polyhedra definitely do exist; it only needs to prove that certain types of polyhedron definitely don't exist, to narrow down the search space. It should work in Euclidean space, to find the Johnson solids, as well as in spheric space, to find potential polychoron verfs.

This would have to be expanded a lot (especially step 3), but theoretically, it could allow for complete enumeration. Note that this needs 4D to work: it won't enumerate Johnson solids in 3D because the vertices in 3D are not fully determined by the polygons around them; in 4D, the cells around a vertex DO determine that vertex completely since there can be no continuous changes of angles.

I don't see the difference. A vertex in 3D is not fully determined by the polygons around it; and an edge in 4D is not fully determined by the polyhedra around it. You can build out from a vertex in 3D to complete the polyhedron and make the vertex rigid; and you can surround a vertex in 4D with polyhedra to make the edges rigid.

The infinite number of polygon types is not a problem. We can use a generic n.3.3.3 tetragon with angles 90°, 150.2223°, and a generic n.4.4 trigon with angles 90°, 128.5714°, as long as we're only adding up angles in faces around a vertex and comparing to 360°, and not calculating dihedral angles. The angles in an actual specific n.3.3.3 or n.4.4 (with n≥7) are at least as large as those given; if a sum involving the latter exceeds 360°, then a sum involving the former also exceeds 360°, which rules out infinitely many cases at once. The program can output a complete polyhedron (just the basic connective structure, not coordinates or metrical values) where all vertices have angle sums less than 360°; I think this happens rarely enough that we can further analyze the results by hand.

The program should also check the multihedron's topology: It should be orientable, any complete vertex should have at least 3 faces around it, any two faces should share no more than 2 vertices (and in case of 2 exactly, they should also share an edge), etc.

A later version of the program should work with dihedral angles to check convexity. Here we'd need a finite set of polygons, because a polyhedron involving n.3.3.3 or n.4.4 may be concave (or non-existent in the relevant 3D space) for n=7 but convex for some larger n.

In my manual work, I didn't always complete the polyhedron before moving through 4D to an adjacent vertex. This allowed me to quickly eliminate some cases that would have taken longer if I stayed in 3D.

I don't want to post my whole proof here, but I might post parts of it, to give you the general idea of the process.

[quickfur]

We need a program that finds all convex polyhedra with faces from a given set of polygons, perhaps with an input multihedron (incomplete polyhedron) that any output polyhedron should contain as a subset.

I have built a database of all 3D CRF polyhedra (excluding members of known infinite families, from which only a select number of the most common members are included). It contains the full face lattice of every polyhedron, and distinguishes between orientations of the chiral ones. As such, it's suitable for querying facts about 3D CRFs via SQL, and can probably serve as the base upon which you could write such a program.

If you're interested I can post the file somewhere, either in the native SQLite3 format or in a text-based export dump that can be imported into any other RDBMS that supports SQL.

[mr_e_man]

I imagine the program, for enumerating CRF polychora, being split into three parts. The first two parts are strictly 3D, enumerating polyhedra in spheric space. The third part is 4D, and uses those polyhedra as verfs. The full CRF polyhedron data would be needed only in the third part.

I described the first part in the previous post. It's mostly combinatorial, fitting polygons together in all possible ways. The only numerical calculations involved are additions (and comparisons).

The second part takes the results from the first part, and checks whether they actually exist as convex polyhedra. This is much more involved numerically. It would use trigonometric functions (or at least square roots) and multiplication, and it would have to search the "space" of possible dihedral angles to find whether a solution exists. I think we should use some form of interval arithmetic, so that all rounding errors are completely accounted for, and so that the "space" can be searched with a finite number of steps. (This could prove that a solution definitely doesn't exist, or that a solution probably does exist. I'm not sure it can prove that a solution definitely does exist; though, the Intermediate Value Theorem comes to mind, if the "space" can be reduced to 1 dimension, which would happen if all of the polyhedron's vertices are tetravalent.)

The third part is purely combinatorial. It takes the results from the first two parts, and matches the spheric polygons to the CRF polyhedra they were taken from, thus producing 4D vertex configurations. Then it tries to fit these together to make polychora. (This fitting would require that the dichoral angles, given to a face by the surrounding vertices, be all equal. They correspond to dihedral angles in the second part.)

Again, in my manual work, I didn't always keep these three parts separate.

[quickfur]

A practical consideration to keep in mind for any brute-force program that tries to enumerate all CRFs is that there needs to be a reliable way to discard CRFs that belong to known large families. E.g., the 600-cell has 314,248,344 non-adjacent diminishings, probably a lot more if we consider some adjacent diminishings that lead to CRFs (plus any corresponding diminishings of the 600-cell family uniform polytopes that share the same underlying structure). The duoprisms have (at least) 11,956,959 augmentations, of which 11,921,273 come from the augmentations of the 10,10-duoprism alone. It would be wise to avoid having the program "stuck" for a very long time enumerating members of these families before it gets to the more interesting stuff (the unique stuff, the crown jewels that we all seek). It would also be wise to prune these cases so that the interesting stuff don't get buried in the "noise" of 600-cell diminishings.

Another consideration is some way of pruning branches of the combinatorial tree that are known to not contain any interesting CRFs. Combinatorial explosion is a thing, and if left untamed, could lead to the program running for several lifetimes before it yields any interesting results. Ideally, we'd like to see at least some results within our lifetimes.

[mr_e_man]

That is easily solved by providing an input which is "interesting".

For example, you could input the sphenocorona's verf 4.3.3.3 C to the first two parts of the program. The output should include the sphenocorona-prism verf (which is a pyramid of 4.3.3.3 C), and an augmentation of that, but also some other "interesting" things. The latter, when input to the third part, would produce an output of either some new crown jewel polychora involving the sphenocorona, or nothing (thus proving that they don't exist).

Or you could take some non-constructible polyhedron found in the first two parts, such as those shown here (with, say, n=5), and input that to the third part, to find some non-constructible polychoron. (I haven't actually proven that they're non-constructible (by square roots etc.); that would require some algebra.)

If the first two parts of the program have been completed, in the sense that all potential polychoron verfs have been found, then you could search those results for something involving 4.3.3.3 C, instead of running the program again with that specific input.

If the first two parts have not been completed, then the third part would need to call them as it builds the polychoron, instead of merely using a fixed list of their output.

So, I think we should first look at the "interesting" CRF polyhedra (snic, snid, J84-90, in my opinion); the program would give us all crown jewels containing these, if they exist. Then with those out of the way, the first two parts could be completed more quickly (having fewer polygons available), and we could search those results for anything non-constructible (or otherwise "interesting"), and input that to the third part.

The numerous 600-cell diminishings are results of the third part; they're complete polychora. I expect there would be far fewer results from the first two parts, maybe thousands instead of billions.

[johannes@itap.physik.uni-stuttgart.de]

We need a program that finds all convex polyhedra with faces from a given set of polygons, perhaps with an input multihedron (incomplete polyhedron) that any output polyhedron should contain as a subset.

I have built a database of all 3D CRF polyhedra (excluding members of known infinite families, from which only a select number of the most common members are included). It contains the full face lattice of every polyhedron, and distinguishes between orientations of the chiral ones. As such, it's suitable for querying facts about 3D CRFs via SQL, and can probably serve as the base upon which you could write such a program.

If you're interested I can post the file somewhere, either in the native SQLite3 format or in a text-based export dump that can be imported into any other RDBMS that supports SQL.

I would certainly be interested. In both formats. How can I get that database?

[quickfur]

We need a program that finds all convex polyhedra with faces from a given set of polygons, perhaps with an input multihedron (incomplete polyhedron) that any output polyhedron should contain as a subset.

I have built a database of all 3D CRF polyhedra (excluding members of known infinite families, from which only a select number of the most common members are included). It contains the full face lattice of every polyhedron, and distinguishes between orientations of the chiral ones. As such, it's suitable for querying facts about 3D CRFs via SQL, and can probably serve as the base upon which you could write such a program.

If you're interested I can post the file somewhere, either in the native SQLite3 format or in a text-based export dump that can be imported into any other RDBMS that supports SQL.

I would certainly be interested. In both formats. How can I get that database?

Here, I made a zip file that contains the SQLite3 database and a text file of the database dump. The schema is hopefully self-evident; if anything is unclear, please ask.

[mr_e_man]

The program can output a complete polyhedron (just the basic connective structure, not coordinates or metrical values) where all vertices have angle sums less than 360°; I think this happens rarely enough that we can further analyze the results by hand.

I was wrong about that.

I basically finished the first part of my program, last year, in September or October.
The problem is that, when applied to a pair of n.3.3.3 tetragons, it gives more than a hundred results.

Some of them have angle sums being exactly 360°. It would be difficult to exclude these. (It could be done by hard-coding known combinations of angles, such as 70.5288° + 54.7356° + 54.7356° = 180°, and 79.1877° + 37.3774° + 31.7175° + 31.7175° = 180°, and 116.5651° = 37.3774° + 79.1877°, etc.) Anyway, I'm not sure I want to exclude them, because these decompositions produce such "flat" vertices. If 360° sums are excluded then we can't use the shorter list of polygons.

Some of them have all angle sums strictly less than 360°, but cannot exist for other reasons. These would be excluded by the second part of the program (calculating dihedral angles), which I haven't worked on yet.
Previously I said it would need a finite set of polygons. But interval arithmetic can handle, for example, a generic n.3.3.3 tetragon with n≥24, which would have one angle in the interval [90°, 92.1687°], and the other angle in the interval [171.3376°, 180°].

[student5]

Sooooo..... It's been a looong time!

However, very interesting things have been happening, esp. mr_e_man's program seems to be promising.

As said, there is some limiting that needs to be done, so that it doesn't get lost in diminishings of the 600-cell family. So far a lot of polychora discovered in 2014 turned out to be EKFs:
Based on the discovery index

EKF:
D4.10/D4.11 family: EKF of ex and diminishings
D4.8.x family: Two EKF's of ex joined together and diminishings
D4.7: Can be augmented undocumented but probably some EKF? diminishing of some sorts?

Bilbiro'ing: (Taking two diametrically opposing pieces of a polychoron, chopping off the middle vertices, sliding them together, then filling in the non-CRF parts) <- this may produce bilbro's, but is more generic
D4.4.1 is like a trigonally-expanded ex, such as the J92 hexachoron: Direct EKF diminishing, but then, some weird bilbiro-like cut-and-paste operation happens. Note that this is different from this shape on Klitzing's site
D4.5.4 is a bilbiro'd version of D4.5.3

Thawro'ing: (Chopping off a full (n-1)-sphere of points and filling in with a single face/polyhedron) <- so far always producing thawro's
D4.5.1 here

Code: Select all

o3x5x (-x)3x5x    x3o5x
x3f5o    o3f5o    o3f5o
F3o5x <- f3o5x <- f3o5x
x3x5x    o3x5x    o3x5x

o3x5o
o3o5f
o3x5x     o3x5x
x3f5o     x3f5o
F3o5x     F3o5x    /- D4.5.1
f3f5o <-> x3x5x --<     this gets replaced in the thawroing process
o3F5o     o3F5o    \- D4.5.3+
f3x5x     f3x5x
x3o5f     x3o5f
F3o5f     F3o5f --\
V3x5o     V3x5o    |-- remove for D4.5.4 (bilbiro-ing)
o3o5V     o3o5V    |
F3o5f     F3o5f --/
x3o5f     x3o5f
f3x5o     f3x5o
o3F5o     o3F5o
f3f5o     x3x5x --<   this gets replaced in the thawroing process

x3x5x
o3F5o
f3x5x
x3o5F


Note that the un-expanded lace tower is a diminishing of rox:

Code: Select all

x3o5o
o3x5o
x3o5x <-
F3o5o
o3f5o <-
f3o5x <-
o3x5x <-
...


which makes me wonder if

Code: Select all

xofo3ofox5xoxx&#xt

is CRF? There was also talk of possibly finding a wedge shape like this that has thawros.

D4.3.1 and D4.3.2 are really cool: I couldn't find anything EKF-like from them. Very far shot, but could be seen as part of rox: ..xF..(o)..Fx..3..o..(o)..o..5..xo..(f)..ox..&#xt But that'd be easier found using a brute force algorithm I think.
Lace tower for reference:

Code: Select all

x5o3x
o5o3F
f5o3o
o5o3F
x5o3x


Anyways, these crown jewels are thus far always found using a subsymmetry, I think a program should also do that, but I think that's obvious

[quickfur]

Sooooo..... It's been a looong time!

Indeed, it has been a long time! Been many years since I last seriously dabbled in discovering new CRFs...

[...] So far a lot of polychora discovered in 2014 turned out to be EKFs:

I'd say that practically all of the discoveries in 2014 are EKFs or some modification/combination of them.

[...]
which makes me wonder if

Code: Select all

xofo3ofox5xoxx&#xt

is CRF? There was also talk of possibly finding a wedge shape like this that has thawros.

Don't know, could be worth taking a look. Not this week, though, I'm quite busy. But maybe sometime after that.

D4.3.1 and D4.3.2 are really cool: I couldn't find anything EKF-like from them.

I see it as an EKF (i.e., modified Stott expansion) of some faceted prism of an icosahedral family polyhedron.

[...]
Anyways, these crown jewels are thus far always found using a subsymmetry, I think a program should also do that, but I think that's obvious

Hmm. I wonder if it might be more profitable to look for crown jewels that are not derived from a subsymmetry -- if any such are found, they'd be crown jewels of a new kind that we've not yet seen so far! Or, if none can be found, we can eliminate that possibility from the equation, perhaps find some proof of non-existence.

[quickfur]

[...]
I'd say that practically all of the discoveries in 2014 are EKFs or some modification/combination of them.

Hmm, I take that back. There are a few that aren't related to EKFs: icosahedron || cube is one of them (though it was discovered by Klitzing before 2014).

There's also D4.16, which is related to one of the smaller wedges that contain icosahedral/diminished icosahedral cells. And maybe a few others, like the ursachora. But the vast majority of discoveries from 2014 are EKFs.

[student5]

D4.3.1 and D4.3.2 are really cool: I couldn't find anything EKF-like from them.

I see it as an EKF (i.e., modified Stott expansion) of some faceted prism of an icosahedral family polyhedron.

In this lace tower configuration here, I found it difficult to "see" the bilibiro's, but they are on the sides:

Code: Select all

x2x    x2o       x2o
o2F    o2f       o2f
f2o <- f2(-x) <- f2o
o2F    o2f       o2f
x2x    x2o       x2o


Ah yes, now I see it: The bilbs are expanded along the square, so the original shape is a dodecahedral-prism kind of thing:

Code: Select all

x5o3x    x5o3o                   x5o3o
o5o3F    o5o3f                   o5o3f
f5o3o <- f5o3(-x) <- f5(-x)3x <- o5x3o
o5o3F    o5o3f                   o5o3f
x5o3x    x5o3o                   x5o3o


Notice that we have to swap twice to get back to a non-faceted something. However, this is not anything as far as I can see... The ikes don't show up, because the faceting doesn't work out... I can't find it as an ekf now...