Johnsonian Polytopes

[Marek14]

[...] No, these were closer to 2^100 but rectified 120-cell (using only one type of diminishing) beats them - there are proven 2^156 different CRFS with estimations up to 2^500

These crazy huge numbers make me think that there must be a lot of "unique" CRFs in 4D... unique as in, not obviously derived from uniform polytopes and not in the set of segmentotopes. There's got to be a large number of these (though probably not as large as the uniform diminishings). We should've found some of them by now!

The problem is: how to search for them? Making a program to play with a 4D analogue of Geomag?

[Mrrl]

The problem is: how to search for them? Making a program to play with a 4D analogue of Geomag?

Good news here: every vertex of convex polychoron is rigid. That is if yot know all cells that meet in this vertex then you can compute all bichoral angles

[quickfur]

The problem is: how to search for them? Making a program to play with a 4D analogue of Geomag?

Good news here: every vertex of convex polychoron is rigid. That is if yot know all cells that meet in this vertex then you can compute all bichoral angles

And the maximum number of CRF polyhedra meeting at a vertex is 20 (tetrahedron has smallest dihedral angle, and 600-cell has the most tetrahedra meeting per vertex).

Actually, I wonder if it's easier to use edges to search for CRFs. The maximum number of CRF polyhedra per edge is 5, I believe. I think you should be able to calculate dichoral angle if you know all the polyhedra sharing an edge.

[Marek14]

My guess is that both are needed. I have posted vertex data for polyhedra (and part of edge data) before. The vertex data would have to be updated a bit, since 3D vertices are not necessarily rigid (compare elongated triangular, square and pentagonal pyramids). From these, it's theoretically possible to put together all possible vertices of CRF. Of course, there's a lot of them... If I take a 600-cell vertex, I could replace any tetrahedra with triangular dipyramids or elongated triangular pyramids or dipyramids, and these combinations would probably all have to be tried by hand...

[Mrrl]

Yes, my idea is to enumerate all possible vertices generated by CRF cells (with standard limitation: only 3,4,5,6,8,10-gons as ridges), then find compatible edges and build a graph of connections... don't know what will be next.

[Marek14]

Yes, my idea is to enumerate all possible vertices generated by CRF cells (with standard limitation: only 3,4,5,6,8,10-gons as ridges), then find compatible edges and build a graph of connections... don't know what will be next.

I think we need a more general concept than just polychora. We need a system that will accept partial polychora with "active zones" - unpaired faces. Two partial polychora can then be joined via their active zones.

The list of partial polychora will be big, but it can be generated in parts, starting from 1 vertex upwards, and a computer program never has to understand more than 1 stratum (partial polychora with the same amount of full vertices).

In the next part, for each partial polychoron in the set, each of its partial vertices will be completed in all possible ways. Newly-created partial polychora will be checked for duplicates. If a full polychoron results, it will be sent to results list, if completing one vertex also completes other vertices, but not the whole polychoron, the partial polychoron will be sent to the correct strata (after checking for duplicates).

This way, the full list would be eventually built.

First step is to catalogue all possible vertex figures...

[quickfur]

[...] I think we need a more general concept than just polychora. We need a system that will accept partial polychora with "active zones" - unpaired faces. Two partial polychora can then be joined via their active zones. [...]

To simplify things, this "active zone" can just be a generalized polyhedron (i.e., a connected set of 2-faces in 4D, not necessarily lying flat on a hyperplane). The completed part of the polychoron will be kept separately.

If we can somehow canonicalize these "boundary polyhedra", we may be able to do "macroscopic" cut-n-paste operations, e.g., compute some incomplete configurations of polyhedra, and then if two incomplete configurations have the same boundary polyhedron, they can be glued together to form a CRF. Or better yet, if two boundary polyhedra share a common subset, they can be glued together to partially close up the polychoron (and during the gluing operation we can reject combinations that will self-intersect or produce impossible vertex figures, etc.).

But I don't know if there's a feasible way to compute equality or subset equality of boundary polyhedra... maybe some kind of sorting algorithm, but you may have to spend a lot of time transforming the thing into canonical orientation before they can be correctly compared.

[Marek14]

I guess adding polyhedra one at a time would be better for exhaustive computation. So far we have some infinite families and some "finite, but uncomfortably large" families. The question si what the number of "families" themselves is (using that word in an imprecise sense).

[quickfur]

If we're going for exhaustive computation, we probably want to make the computation incremental, so that we don't wait for 2 days running the computation only to discover there was a bug that produces wrong results.

Due to the uncomfortably large sizes of some of the families (e.g. augmented duoprisms and 600-cell family), we probably want to do a breadth-first search, so that the computation produces smaller polychora first before getting to the larger ones.

One way to do it is to use a size limitation: say the first run we set the limit to 10, so any polychora that are not closed by the time 10 cells are added will be saved into a partial results file, to be continued by the next iteration. Next time we set the limit to 20 and load the partial results file as our starting point, so we don't repeat the previous computation unnecessarily. This way, we can slowly work our way up the list to larger polychora, and also gives a chance to verify the correctness of the results by hand.

[Marek14]

Exactly. And if you add OFF to the list of outputs, I will be glad

[wendy]

The o3x3o5o should be a profitable source to look at. The second section, icosahedron-first, is x3o5x (or rID).

This means that any diminished x3o3o5o, gives rise to a matching rectified-diminished figure, which has a combination of faces, including square-pyramids, x3o5x, and the usual faces of octahedra and icosahedra.

The truncate x3x3o5o, for removal of the icosahedron, gives h3o5x also, which has sides of sqrt(3).

With o3o3x5o, you can also get the identical set also, but is also possible to mount various cupola on the o3x5o rID on adjacent dodecahedra too.

However, the x3o3o5o supports a kind of diminishing that leads to a pentagonal antiprism. This leads to a ring of 10 of these, the GAP is an example of one having two such rings. There are three other examples. Except for the three-ring example, there are instances that support both kinds of diminished examples.

[Marek14]

The o3x3o5o should be a profitable source to look at. The second section, icosahedron-first, is x3o5x (or rID).

This means that any diminished x3o3o5o, gives rise to a matching rectified-diminished figure, which has a combination of faces, including square-pyramids, x3o5x, and the usual faces of octahedra and icosahedra.

The truncate x3x3o5o, for removal of the icosahedron, gives h3o5x also, which has sides of sqrt(3).

With o3o3x5o, you can also get the identical set also, but is also possible to mount various cupola on the o3x5o rID on adjacent dodecahedra too.

However, the x3o3o5o supports a kind of diminishing that leads to a pentagonal antiprism. This leads to a ring of 10 of these, the GAP is an example of one having two such rings. There are three other examples. Except for the three-ring example, there are instances that support both kinds of diminished examples.

As I posted before, the rectified-diminished figures also support removing icosahedron with one more layer of vertices, and removind pentagonal prismatic pyramids from the vertices.

[Mrrl]

I've started the search with CRFs that have only trianlular ridges. There are 7 possible cells for them, they have 7 types of vertex and 9 different values dihedral angles. Now I make a list of possible edfe configuration (that has 3,4 or 5 ridges). I afraid that it will be very long list, but most of configurations give no possible extensions to the vertex. Largest family of trianglar CRFs that I know is a set of augmented snub 24-cells, and it is not very large (about 2^14 different bodies). Problem is that it's difficult to filter out congruent CRFs before end of their construction.
As for "boundary polyhedra", it's a good idea (kind of dynamic programming technique), but how are you going to continue unfinished polychora so that there will be good chances of repeating of them? It may be that most boundaries will be unique...

Update: number of edge configurations for triangular case is about 1000 (between 750 and 1500 - it's difficult to check all symmetries manually).

[quickfur]

Has anyone started on designing an exhaustive search algorithm for finding CRF's? Seems awfully silent here suddenly.

One potential issue is how to prevent potentially unmanageably long running times if the algorithm starts hitting, say, the 600-cell diminishings. I suppose we could just ignore polychora with certain "blacklisted" vertex figures (derived from the 600-cell), but we might miss a few CRFs that aren't 600-cell diminishings this way (they could just happen to look like a 600-cell diminishing locally but have a different structure globally).

We will also need a way of uniquely identifying CRFs, so that we can use a database to keep track of what has been discovered and what hasn't. This may need to be a database of searched vertex figure combinations (or some such) so that the program knows what has been tried and what hasn't, and can continue from wherever it left off. This can also be the bases of a SETI-like distributed project for finding all these CRFs.

[Deedlit]

Do the "diminishings" of the 600-cell include the "special cuts" of the 600-cell? (i.e. take a subset of non-adjacent vertices of the 600-cell, and slice off each vertex). These have been enumerated, see for example http://www.liga.ens.fr/~dutour/SpecialCuts/index.html, and each instance is an isomorphism class of Johnsonian polychora.

[Mrrl]

Do the "diminishings" of the 600-cell include the "special cuts" of the 600-cell? (i.e. take a subset of non-adjacent vertices of the 600-cell, and slice off each vertex). These have been enumerated, see for example http://www.liga.ens.fr/~dutour/SpecialCuts/index.html, and each instance is an isomorphism class of Johnsonian polychora.

Not exactly. You can cut off adjacent vertices (and get diminished icosahedra as faces). But you can't cut off three vertices of the same ridge.

[quickfur]

Do the "diminishings" of the 600-cell include the "special cuts" of the 600-cell? (i.e. take a subset of non-adjacent vertices of the 600-cell, and slice off each vertex). These have been enumerated, see for example http://www.liga.ens.fr/~dutour/SpecialCuts/index.html, and each instance is an isomorphism class of Johnsonian polychora.

Well that is helpful. So that takes care of the simple case where the cuts produce only icosahedral cells.

However, as mrrl mentioned, there are also special cases where two adjacent vertices can be cut, resulting in cells with diminished icosahedral faces. So the number of possibilities is actually quite great. However, we currently haven't fully enumerated these cases yet. There are at least two kinds of cases possible: one is when a pair of vertices is removed, resulting in two gyroelongated pentagonal prism (aka diminished icosahedron) cells, and another when three or more vertices along a great circle are removed (ala grand antiprism), resulting in two gyroelongated pentagonal prism cells plus one or more pentagonal antiprisms.

There may be other cases as well.

All of these possibilities compound with each other as long as the resulting cells are disjoint, so there are a very great number of possibilities here.

[wintersolstice]

In case it hasn't been posted, the aumentations of the tesseract(which can also be made by diminishing a 24-cell) are

Augmented tesseract (aka septadiminished 24-cell, elongated cube pyramid)
Parabiaugmented tesseract (aka parasexadimished 24-cell, elongated bicube pyramid)
Metabiaugmented tesseract (aka Metasexadimished 24-cell)
Paratriaugmented tesseract (aka Paraquintidimished 24-cell)
Metatriaugmented tesseract (aka Metaquintidimished 24-cell)
Paraquadriaugmented tesseract (aka Paraquadridimished 24-cell)
Metaquadriaugmented tesseract (aka Metaquadridimished 24-cell)
Verquadriaugmented tesseract (aka Verquadridimished 24-cell)
Paraquintiaugemented tesseract (aka Paratridimished 24-cell)
Metaquintiaugemented tesseract (aka Metatridimished 24-cell)
Parasexaaugemented tesseract (aka Parabidimished 24-cell)
Metasexaaugemented tesseract (aka Metabidimished 24-cell)
Septaaugemented tesseract (aka dimished 24-cell)

a guide to what they mean:

for

2

para - two on opposite sides
meta - two ajacent

3

para - two (of the three) opposite
meta - none of them opposite

4

para - two opposite pairs
meta - one opposite pairs (the other pair not opposite)
ver - none of them opposite

if you were to augment 5 of them there's 3 left, so base on those 3 left for
same for 6 and 7

I used the Latin prefixes

EDIT:

when two cube pyramids are on ajacent cells of the tesseract the 4-pry at the face (where those two cells meet) lie in a plane, but they can be blended together to make an octahedron. the tesseract has 24 faces so completly augment it and it has 24 octahedra: 24-cell!!

[SharkRetriver]

Does augment mean to add on a pyramid to a cell or face?
Also, is there augmented 5-cell?

[quickfur]

Does augment mean to add on a pyramid to a cell or face?

Yes.

Also, is there augmented 5-cell?

There is one, which is obtained by glueing two 5-cells together. I don't think there is any other convex augmentation, because the dihedral angle of the 5-cell implies that augmenting two adjacent cells will produce a non-convex polytope, and all cells on the 5-cell are adjacent.

[wintersolstice]

It is possible to remove 48 vertices from the 600-cell (two 24-cells) the result is vertex transitive, regular faced, but not uniform (it's scaliform)

the vertex figure on a grand antiprism means you can't remove a vertex with losing it's regular face, I don't know about multiple vertices though.

btw I think I've worked something out

The grand antiprism can be made by alternating the decagonal octagol triate! (I think )

[quickfur]

It is possible to remove 48 vertices from the 600-cell (two 24-cells) the result is vertex transitive, regular faced, but not uniform (it's scaliform)

Oh? Which two 24-cells, though? How is the second 24-cell oriented relative to the first? I assume it's not the dual 24-cell 'cos I don't think a 600-cell contains dual 24-cells.

[...]btw I think I've worked something out

The grand antiprism can be made by alternating the decagonal octagol triate! (I think )

I still have trouble understanding just what a brick product does. If you have a precise definition, I'd be very happy if you could share it.

[quickfur]

[...]Also, is there augmented 5-cell?

There is one, which is obtained by glueing two 5-cells together. I don't think there is any other convex augmentation, because the dihedral angle of the 5-cell implies that augmenting two adjacent cells will produce a non-convex polytope, and all cells on the 5-cell are adjacent.

And in fact, now that I think of it, it also has 8 regular cells just like the tesseract (8 tetrahedra), although it is not a regular figure in itself because it's not vertex-transitive.

[wintersolstice]

I'm not really sure how brick product works but I nearly understand powertopes (not sure how brick product relates to it.

With the 600-cell, it is the convex hull of 5 24-cells compounded together. remove the vertices of one and you get sadi (snub dis 24-cell) remove another (I don't think it matter which one ) you get Bidex (bi icosatetra dimished 600-cell) which has no tetrahedra but has 48 tridiminished icosahedra. (24 are from underneath the vertices that are removed from sadi, the others are diminishes if the 24 icosihedra in sadi. 3 5.g pyramid removed from each as well as removing the last 120 tetrahedra)

if it helps to study crown-jewels here is a description of the constructions for each of the "crown-jewels" (elementary Johnson solids)

Snub disphenoid

this is simply made by replacing the square faces of a square antiprism with a pair of triangles, the rest of the shape is stretched and squashed accordingly (to make it regular faced)

Snub square antiprism

truncate a square anti prism then alternate it or take a square and put a trangle at each edge than put two triangles between each "edge traingle (on the vertices) and take two of these structures and interlock them.

Sphenocorona

take an icosahedron and position it sao that so that it is standing on an edge and has a parallel edge on the the top, there are two opposite edges which are horizontal and in the middle of the shape and two which are vertical remove the pair of triangle at the vertical set and then cut what's left in half (you now have eight triangles (the result is open ended it is not closed up) this is the "corona"
put two squares that meet on the horizontal edge pair an then fill the empty space with 4 triangles (what you will have joined to "corona" is a "wedge-like structure made from two lunes [where a lune is two triangles either side of a square] this structure is call "spheno")

Augmented sphenocorona

this is just simply a sphenocorona with a square pyramid joined to one of the squares

Sphenomegacorona

this is "spheno" joined to "megacorona" the latter is the remainder of the icosahedron when "corona" is reemoved (or by putting those four triangles removed back on the "corona")

Hebesphenomegacorona

above I described "megacorona", "hebespheno" is three lunes joined together (one is sandwiched between the other two) if you joined "spheno" and "hebespheno" you get an elongated 5-gon bipyramid

Disphenocingulum

"cingulem" is a belt of 12 triangles similar to the 12 traingles in a 6-gon antiprism
join a "spheno" to either end so it's like a gryoelongation of two "spheno" stuctures

btw the structures above may need to be stretched and squashed before being joined together

Bilunabirotunda

take a square and join a triangle to every edge of it then fold an opposite pair of traingles to nearly 90 degrees with the square. now take two such structures amd join the points of the two triange pairs, put a pentagon on either side of each pair of joined points (so there are two vertices with 3,5,3,5 configuration)
fold the pentagons (there are two above an two below assuming that the triangles joined are point are the horizontal and the squares are vertical) now fold the other triangles( the vertical set) so that they meet the penatgo pairs on th etop and bottom (so there are 4 vertices with a 3,5,5 configuration) doing this corectly all 4 vertices of both squares will be part of a 3,4,3,5 vertex configuration)

Triangular hebesphenorotunda

start with a hexagon base and put a triangle on every other edge and put a "lune" (see above) placed on it's side on the other edges so that it alternates between lune and triangle, for eacj triangle (joined to the hexagon) fold it (and the lunes) so that each triangle and the two triangles from either side (for the lunes) go round the same vertex. put a pentagon at these intersection (of the three triangles round a vertex), put a traingle on the top of each square (part of the lunes) and fold them up, put a triangle on the top, the pentagons 5 edges will be joined to, the to triangles on two lunes (the bottom two) the two triangles placed on top of each lunen(the next two) and finally to the triangle at the top (the top edge) the point of the triangle place on top of the lunes will join to the vertices of the top triangle

I'm using these constructions to see if I can found analogies for 4D cases (now I'm sharing my secret) though I haven't found any "crown-jewels" yet

[wintersolstice]

these might be useful, these 4 polychora are CRF but are vertex-transitive but not uniform (they're scaliform) they could be included in "cut'n'paste" manipulations, but since these are scaliform they don't count as Johnson polychora

Truncated tetrahedral cupoliprism

this is one of the segmentochora:
Truncated tetrahedron || Inv Truncated tetrahedron

Bi-icosatetra diminished hexacosichoron

(I described this in an earlier post)

Swirlprismatodimished rectified hexacosichoron

removing 120 vertics from rectified 600-cell

Prismatorhombatosnub icosatetrachoron

I'm not sure about this one????

but I can give it's cell count:

24 Icosahedra

96 triangular prisms

96 triangle cuplola

24 Truncated tetrahedra

[quickfur]

[...]Swirlprismatodimished rectified hexacosichoron

removing 120 vertics from rectified 600-cell
[...]

Just out of curiosity, which 120 vertices should be removed? Every vertex is shared by 2 icosahedra, and removing each produces a pentagonal prism and two gyroelongated pentagonal pyramids (diminished icosahedra), but I can't think of any way of removing 120 vertices and still have the result be vertex-transitive.

[wintersolstice]

[...]Swirlprismatodimished rectified hexacosichoron

removing 120 vertics from rectified 600-cell
[...]

Just out of curiosity, which 120 vertices should be removed? Every vertex is shared by 2 icosahedra, and removing each produces a pentagonal prism and two gyroelongated pentagonal pyramids (diminished icosahedra), but I can't think of any way of removing 120 vertices and still have the result be vertex-transitive.

I think it safe to say that with 6 times as many vertices as the 600-cell itself (720 compared with 120) you remove an entire 600-cell worth of vertices leaving only 5 of them.

You can't remove two ajacent vertices without losing the regular faces.

The 10^30+ dimishes can be thought of as augmenting this shape

another example: if you diminish the 24-cell 8 times (no two vertices removed are ajacent) you get the tesseract which vertex-transitive

and that's the rectified 16-cell!

"ajacent vertices" btw are vertices joined by an edge (that's what I call them!)

does that help?

While I'm talking about dimished rectates

there is the dimished and bidiminished 5-cell when I can I'll get the element counts up

If you take any other rectate, I think it's safe to say that the vertex figure is not regular faced so they can't be dimished (at least not with pyramids anyway)

[Marek14]

Well, first of all, I believe the word you meant to use is "adjacent".

Second of all, not all of the diminishings can be considered augmentations of this shape. Think of bidiminishings: when you diminish one vertex from a certain 600-cell, the second vertex can be from another one. Of course, if you will continue, you will encounter the final (undiminishable) result sooner than after 120 diminishings.

Funny question: how many such "maximal diminishings" exist?

[wintersolstice]

Second of all, not all of the diminishings can be considered augmentations of this shape. Think of bidiminishings: when you diminish one vertex from a certain 600-cell, the second vertex can be from another one. Of course, if you will continue, you will encounter the final (undiminishable) result sooner than after 120 diminishings.

But that's only if the 720 vertices are grouped into "fixed sets" of 120 vertices so that a vertex is only part of one 600-cell but surely you can regroup them so that the second vertex can be taken from the same 600-cell rather than a different set?

If it were possible to choose less than or equal than 120 vertices that can't be contained inside a single 600-cell (given that no two are adjacent)regardless of how the vertices are grouped, then it would be true. I've no idea?

btw I didn't discover spidrox

[wintersolstice]

I noticed it said in the wiki than a 5,n-duoprism can be augented with pentagonal prism pyramid for any number up to 38.
I don't think it's right.

I did a calculation for find the angle between the pentagonal pryamid cell and the pentagonal prism base and the angle is 18 degrees. now if you allow aumentations to be adjacent the angles will add as follows 180(n-2) (the angle between the 5-prisms) + 18 +18 which for n=11+ is greater than 180 degrees which means it's concave, with n=10 it equals 180 degrees but the 5-pyramid can be blended with the 5-pyramis of the adjacent pyramid to make a pentagonal bipyramid so it can still be valid.

if you want to go further by disallowing adjacent augmentations then you only do 180(n-2) + 18 which for 21+ is more than 180 degrees (180 exactly for n=20)

Now, that's just my calcualtion but I did it twice and can't find an error

Edit

sorry you need to divide 180(n-2) by n