List of uniform honeycombs

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Re: List of uniform honeycombs

Postby polychoronlover » Thu Jun 01, 2017 6:04 am

wendy wrote:A number of stary hyperbolic tilings of finite density are known. But there are not a lot of star-subgroups up there, and we mostly rely on Coxeter's theorm that starry symmetries are nade from regular ones.


There's a theorem that symmetries of nonconvex Wythoffian figures can always be described by Dynkin diagrams for convex figures? I'm not surprised; this seems true in practice but it's nice that there's a proof.

I wonder how well this extends to compound polytopes. It's probably still true if the compound is formed by replacing each instance of an element or element-figure (whose symmetry is given by the Dynkin diagram with one node removed) and replacing it by a compound that preserves the symmetry of each component. Such compounds are the only ones I consider to be "true extensions" of Wythoffian polytopes. For example, the rhombihedron (compound of 5 cubes) doesn't count because the triangular symmetry around each vertex of a cube degrades into chiral triangular symmetry in the compound.

There is a similar subclass of uniform polytopes that I call "locally uniform", where each element is not only uniform but also has the property that symmetries of the original shape can map a vertex from one element onto every other vertex of the same element, while still mapping the element onto itself. This filters out almost all polytopes whose vertex configurations can't be made by kaleidoscopical construction; it excludes the antiprisms and snubs in 3 dimensions and almost everything from Bowers' "miscellaneous" categories in 4 and 5 dimensions.
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Re: List of uniform honeycombs

Postby wendy » Thu Jun 01, 2017 8:27 am

Wythoffian simply means that it has a construct from a Dynkin symbol. Wythoff discovered the mirror-edge reflections that the marked nodes of the Dynkin symbol means.

There are a number of non-wythoffian hyperbolic and euclidean tilings, and a number of star-tilings as well, although the star-tilings are all wythoffian.

For example, there is a tiling of truncated cubes, 16 at a corner. It derives from o8o3x4x, but the cell o8o3x is perfectly flat (ie concentric with the tiling), and can be used as a mirror. This leads to space filled with truncated cubes o3x4x, the vertex figure is a octagonal tegum oxo8ooo&#tq. There is a tiling of rhombo-cuboctahedra x4o3x, and triangle prisms x2x3o, sixteen of each at a vertex: the resulting tiling has a vertex figure of oxqxo8ooooo&#tq (the #tq means a tower with lacing of sqrt(2), rather than 1).

The usual form for non-wythoff star-tilings in the euclidean plane is the pleat-fold. You take some sort of tiling like LPA2, which is an alternating layers of the oct-tet tiling and a triangular prism. But instead of making the oct-tet on top of the triangle prisms, like ---o----, you make the common plane into a pleat, like o====. The oct-tet tiling then faces into the paper, while the prisms face outwards.

There's an ASN x4x4x, whose vertex figure is a pentagram (5/2), with an array of 4,3,4,3,3.

The group o5/2o5o5/2o3*a (d4) and o5/2o5o5/2o3o3*a (d8), both produce stars, as well as the regular x5/2o5o3o3o, and x5o5/2o5o3o. I have not really looked into o4o3o3o3o3o3*a to see if it makes stars yet, but the edges on it are quite long.
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Re: List of uniform honeycombs

Postby wendy » Thu Jun 01, 2017 8:51 am

The proof of Coxeter's theorm, is that when you divide a symmetry group into two parts with a mirror, these are also symmetry groups, since the mirrors reflect what's on one side. The final destination here is usually a simplex-group that has its angles as fractions of a half-circle 'pi/n'.

The 'Wythoffian' group uses simplex mirror groups, like triangles, tetrahedra, etc. You can show that a vertex is equidistant from each of two walls, by bisecting the angle between the walls. Likewise, the intersection of two bisectors will give a set of points where A=B, C=D, or A=B=C. And so forth. The Mirror-edge construction relies on the transport of a vertex over a mirror, eg v----|----i v = vertex, | is mirror, i is image of v, -------- is the line.

While you can construct mirror-edge polytopes in nearly any symmetry, you can only guarrantee that the thing is uniform in the simplex groups. Moreover, the various matricies I wrote for Richard rely on the simplex groups. These are 'wythoff symmetries', and we usually use a 'dynkin' graph for it. A 1:q rectangle supports mirror edge figures, but they are not uniform.

WME are 'wythoff mirror-edge figures'. These consist of all of the available dottings of a Dynkin symbol.

WMM are 'wythoff mirror-margins', that is, figures whose faces (N-1) reflect into each other through the margins (N-2). The margin angles and cell inspheres are all identical.

WSN are the 'wythoff snubs'. All of these can be derived from deriving the omnitruncate of some WME. This group includes the antiprisms, the snubs in 3 and 4d, and snub tilings. Unlike the WME and WMM, these are generally irregular, because you are trying to position N vertices of a simplex so that the N(N-1)/2 edges are all equal. When N-1=2, this is always possible, but in 4D, you have N=4, and edges = 6.

Then you have the laminate tilings. These are made by taking uniform layers, like a band of triangles or a band of squares, and alternating them. The non-wythoff LPC1 is a tiling of strips of squares (P = prismatic layers), and a 1-dimensional coordinate swap C1. (this gives the triangles).

Many, but not all, of the hyperbolic tilings are of these kinds. There's an infinite class of tiling in H3 not listed here, the borrochoral group.

Then you list
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Re: List of uniform honeycombs

Postby Klitzing » Fri Jun 02, 2017 2:12 pm

wendy wrote:For example, there is a tiling of truncated cubes, 16 at a corner. It derives from o8o3x4x, but the cell o8o3x is perfectly flat (ie concentric with the tiling), and can be used as a mirror. This leads to space filled with truncated cubes o3x4x, the vertex figure is a octagonal tegum oxo8ooo&#tq. There is a tiling of rhombo-cuboctahedra x4o3x, and triangle prisms x2x3o, sixteen of each at a vertex: the resulting tiling has a vertex figure of oxqxo8ooooo&#tq (the #tq means a tower with lacing of sqrt(2), rather than 1).

The first verf there is wrong. It rather is oxo8ooo&#kt. Here k=x(8,2) is the chord of an octagon.

The second verf, oxqxo8ooooo&#qt, was known to me as polyhedron (octagonal ball). But never thought about it as being used as a verf. Thought about it. Found x4o3x *b8o. That then describes half of the rhombicuboctahedra at least. The corresponding square tiling of order 8 should work in a similar way as a mirror. Thus obtaining the other rhombicuboctahedra too. But when thinking about the triangular tilings of order 8, then those have a different curvature. So you could cap these, say by a full triangular prisms layer - as your verf would imply - but thereafter? I would not assume, that this so far constructed hyperbolic honeycomb could be made uniform, i.e. having a single vertex type only.

Or am I missing something?

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Re: List of uniform honeycombs

Postby wendy » Sat Jun 03, 2017 8:12 am

The first style is in the original lacing form, where a lacing-edge of q is a vertex-figure of x4x. But we use real lengths now, so it should be a(8) here. So you are right.

The second vertex-figure (octagon-ball = oxqxo8ooooo&#tq), is indeed correct. The polar caps are a prismatic layer of x3o8o, of which you see only the eight prisms at the end of a prism. The middle two layers, can be variously read as x4o8o *b3x, where the equatorial plane is both a tiling of x4o8o, and a mirror.

But the octagon ball consists of two cuboctahedra, rotated by 45 degrees around an axis through the square-centres. So there are two intersecting o4o8x *b3x, whose octagons appear as the octagonal layers of the rCO x4o3x, and there is a common dividing symmetry that turns one of the four-fold axies into an eight-fold one.

The lines of longitude are also x4o8o, these are variously the square faces of x4o3x (particularly the set of 12, the set of 8 are the equatorial planes), and the square faces of the triangular prisms.

The first one involves a lot of similar crossing symmetries of x8o3o4o. The vertex-figure is comprised of a pair of octahedra, being the vertex figure of x8o3o4o, some octagons appear in full as the faces of tC, the others are eight-triangles of x3o8o. You can feed a second o8o3o4x whose vertices are the centres of cells. This one never crosses the layer of triangles, and is rather like the horizontal planes in x4o3o4o, reflected in the vertical.
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Re: List of uniform honeycombs

Postby Klitzing » Sat Jun 03, 2017 11:56 am

wendy wrote:The second vertex-figure (octagon-ball = oxqxo8ooooo&#tq), is indeed correct. The polar caps are a prismatic layer of x3o8o, of which you see only the eight prisms at the end of a prism. The middle two layers, can be variously read as x4o8o *b3x, where the equatorial plane is both a tiling of x4o8o, and a mirror.

But the octagon ball consists of two cuboctahedra, rotated by 45 degrees around an axis through the square-centres. So there are two intersecting o4o8x *b3x, whose octagons appear as the octagonal layers of the rCO x4o3x, and there is a common dividing symmetry that turns one of the four-fold axies into an eight-fold one.

The lines of longitude are also x4o8o, these are variously the square faces of x4o3x (particularly the set of 12, the set of 8 are the equatorial planes), and the square faces of the triangular prisms.

Yes, so far I'm fully with you. But kind of I have to insist:

Your x4o8o *b3x clearly is the same as I'd put it: x4o3x *b8o. That one surely is a uniform hyperbolic honeycomb (or "polytope" if you'd like to put it). The x4o8o tilings (or "cells") then play the role of a mirror. Also already accepted. Thus the mirror image of the so far obtained laminate (i.e. having only x3o4x and x3o8o cells left) still would be uniform. And its vertex figure so far clearly is xqx8ooo&#q.

But then you want to add some caps onto these clearly curved surfaces x3o8o. In fact there you want to add a full layer of triangular prisms each. Sure, this reflects within the vertex figure as getting the full octagonal ball oxqxo8ooooo&#q. But what about the thereby added vertices "inside" or "atop" the x3o8o cells?

My guess here would be, that these wouldn't any longer be part of a further uniform honeycomb. Rather you probably would obtain a biform one. As there you would have for further vertex figure some ox8oo&#q so far. - Sure, you then could say, that these again are parts of the other ones (the octagon balls), so the laminate there might be somehow continued there the other way round again.

Might be. But I don't see how that could take place. For the x3o8o has a different curvature than x4o8o. The latter shall work as a mirror to x4o3x *8o. Thus I assume that the curvature of the latter is the same as that of x4o8o. (And indeed: the "circumradius" of x4o3x *b8o evaluates as 0.321797126 i and that of x4o8o too. But the one for x3o8o evaluates as 0.594603558 i. Each wrt. unit edges, for sure. Simply put it into your spreadsheet, hehe.)

So: how could the prism layer be used likewise as to mime a 2nd mirror?

Well, that 2nd mirror type (if truely existing) then won't be actually the x3o8o of x4o3x *b8o, rather the midway height section of these prisms, so there is still some space for arguments here ... - But at least I cannot see how that one could be supported with any logics ... - Could you help me out here?

Hope, I made my point better understandable now.

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Re: List of uniform honeycombs

Postby wendy » Sat Jun 03, 2017 12:42 pm

The mirror runs through the half-height of the prism.

The edge that runs from the centre of the vertex-figure to the pole, runs between eight triangular prisms, (tips, if you will), this edge has a mirror bisecting it so the image of the centre is the pole-vertex.

You are looking at a flat x3o8o.

If you take the equator of x3o3o5o, which is o3x5o, you would find that the distances are identical in 3d and 4d. This is because the polytopes are concentric with the centre. If you move the o3x5o so that it's a face of o3o3x5o, then the edges are shorter, but the ratio of chords remain the same. The same applies in hyperbolic space, but the edges get bigger.

The trick is to look at the guage theorm, which is about working the edges of x4oPo and o4oPx. In xPo4o, the van Oss polygon (ie the girthing polygon by going 180 degrees across the vertex), is straight. But the edge of x4oPo is longer, but if you move it a half-edge at the centre, the centres of the vertical edges would exactly match the vertices of xPo4o, and the vertices of x4oPo would then correspond to the projection of the first line onto two curving bolospheres.

Thus you can use the spreadsheet, to verify, that x3o8o has the same edge length as x8o4o. This is in the mirror at the waist of the prisms. Now calculate the x4o8o, which replaces the edges of x8o4o with squares, bisected from edge to opposite edge. This turns the triangles into uniform triangle prisms x2x3o, You will see that this edge is the same as that of E x4o8o = E x8o6o. So indeed, the edges of the 'tip' are the same as the rest.

So you really end up with a slab of prisms, with a mirror running entirely in the interior of the layer. And the prisms are uniform.

What happens is that in hyperbolic space, the x3o8o here is curved like a line of lattitude. On the sphere, the line of lattitude is concave to the equator, the great circles are on the other side of the line. In hyperbolic space, an equidistant to a straight plane, is convex to the plane, the straight line descends closer to the plane. The x3o8o in this figure is clearly convex towards the pole, that is, the angles formed by the two tips is less than the angles of the two rCOs. This means that the figure, without the poles would be concaved, and adding the prism layers makes it convex. But without repeating the middle lattitudes again at the pole, it has different kinds of vertices, and hence not uniform.
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Re: List of uniform honeycombs

Postby username5243 » Wed Feb 20, 2019 9:31 pm

So I went back to my tetracomb regiment list (of which I may have an updated version ready at some point) and managed to stumble upon a new contender for largest tetracomb regiment.

The regiment in question has icot symmetry and a colonel with dynkin symbol x3o4x3o4o3*a4/3*c. I came upon this symmetry when finding conjugates of some other icot-symmetry regiments that I listed. Its cells are giddics, icoes, wavitoths, wavaties, and riches. The verf looks like x x4o || q o4x with lacings of length x(8/3) - in other words some distorted variant of squapip (square antiprism prism). This regiment contains giddics and riches in the colonel, and I think you can facet it to get spics, and possibly afdecs and gacocas as well. In other words, this regiment is pretty freaking huge.
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Re: List of uniform honeycombs

Postby Klitzing » Thu Feb 21, 2019 9:01 am

username5243 wrote:The regiment in question has icot symmetry and a colonel with dynkin symbol x3o4x3o4o3*a4/3*c. I came upon this symmetry when finding conjugates of some other icot-symmetry regiments that I listed. Its cells are giddics, icoes, wavitoths, wavaties, and riches. The verf looks like x x4o || q o4x with lacings of length x(8/3)

Interesting one, indeed. - Do you or Jonathan have already a name for that colonel, and an according OBSA ("official Bowers style accronym")?
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Re: List of uniform honeycombs

Postby wendy » Thu Feb 21, 2019 1:31 pm

Does x4o3o4/3x3*a exist? This is essential.
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Re: List of uniform honeycombs

Postby Klitzing » Thu Feb 21, 2019 7:24 pm

wendy wrote:Does x4o3o4/3x3*a exist? This is essential.

The "cells" of x3o4x3o4o3*a4/3*c obviously are (and individually indeed all do exist):
Code: Select all
.3o4x3o4o3*a4/3*c = o4x3o4o      = rich
x3.4x3o4o3*a4/3*c = x3o4o3x4/3*a = giddic
x3o4.3o4o3*a4/3*c = o3x3o4o      = ico
x3o4x3.4o3*a4/3*c = o3x3o4x4/3*b = wavitoth
x3o4x3o4.3*a4/3*c = o3x4o3x4/3*b = wavaty

Cf. rich, giddic, ico, wavitoth, wavaty.

How would you aim to relate x4o3o4/3x3*a = x3x4o3o4/3*a = 2frico to that figure?

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Re: List of uniform honeycombs

Postby Klitzing » Fri Feb 22, 2019 10:26 am

Here moreover comes the full incidence matrix of that non-convex tetracomb of username5243:
Code: Select all
x3o4x3o4o3*a4/3*c   (N,M → ∞)

o3o4o3o4o3*a4/3*c | 72NM |     8     8 |    4    16     8    4     8 |    8    4    8    8    2    4    2 |  4   4   1  2   1 verf = xo4ox xq&#y, with y=x(8/3)
------------------+------+-------------+-----------------------------+------------------------------------+------------------
x . . . .         |    2 | 288NM     * |    1     2     2    0     0 |    2    2    1    2    1    0    0 |  1   2   1  1   0
. . x . .         |    2 |     * 288NM |    0     2     0    1     2 |    2    0    2    1    0    2    1 |  2   1   0  1   1
------------------+------+-------------+-----------------------------+------------------------------------+------------------
x3o . . .         |    3 |     3     0 | 96NM     *     *    *     * |    2    2    0    0    0    0    0 |  1   2   1  0   0
x . x . . *a4/3*c |    8 |     4     4 |    * 144NM     *    *     * |    1    0    1    1    0    0    0 |  1   1   0  1   0
x . . . o3*a      |    3 |     3     0 |    *     * 192NM    *     * |    0    1    0    1    1    0    0 |  0   1   1  1   0
. o4x . .         |    4 |     0     4 |    *     *     * 72NM     * |    2    0    0    0    0    2    0 |  2   1   0  0   1
. . x3o .         |    3 |     0     3 |    *     *     *    * 192NM |    0    0    1    0    0    1    1 |  1   0   0  1   1
------------------+------+-------------+-----------------------------+------------------------------------+------------------
x3o4x . . *a4/3*c |   24 |    24    24 |    8     6     0    6     0 | 24NM    *    *    *    *    *    * |  1   1   0  0   0 gocco
x3o . . o3*a      |    6 |    12     0 |    4     0     4    0     0 |    * 48NM    *    *    *    *    * |  0   1   1  0   0 oct
x . x3o . *a4/3*c |   24 |    12    24 |    0     6     0    0     8 |    *    * 24NM    *    *    *    * |  1   0   0  1   0 quith
x . x . o3*a4/3*c |   24 |    24    12 |    0     6     8    0     0 |    *    *    * 24NM    *    *    * |  0   1   0  1   0 quith
x . . o4o3*a      |    6 |    12     0 |    0     0     8    0     0 |    *    *    *    * 24NM    *    * |  0   0   1  1   0 oct
. o4x3o .         |   12 |     0    24 |    0     0     0    6     8 |    *    *    *    *    * 24NM    * |  1   0   0  0   1 co
. . x3o4o         |    6 |     0    12 |    0     0     0    0     8 |    *    *    *    *    *    * 24NM |  0   0   0  1   1 oct
------------------+------+-------------+-----------------------------+------------------------------------+------------------
x3o4x3o . *a4/3*c |  288 |   288   576 |   96   144     0  144   192 |   24    0   24    0    0   24    0 | NM   *   *  *   * wavaty
x3o4x . o3*a4/3*c |   96 |   192    96 |   64    48    64   24     0 |    8   16    0    8    0    0    0 |  * 3NM   *  *   * wavitoth
x3o . o4o3*a      |   24 |    96     0 |   32     0    64    0     0 |    0   16    0    0    8    0    0 |  *   * 3NM  *   * ico
x . x3o4o3*a4/3*c |  144 |   288   288 |    0   144   192    0   192 |    0    0   24   24   24    0   24 |  *   *   * NM   * giddic
. o4x3o4o         |   3M |     0   12M |    0     0     0   3M    8M |    0    0    0    0    0    M    M |  *   *   *  * 24N rich

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Re: List of uniform honeycombs

Postby Polyhedron Dude » Sat Feb 23, 2019 7:23 am

username5243 wrote:So I went back to my tetracomb regiment list (of which I may have an updated version ready at some point) and managed to stumble upon a new contender for largest tetracomb regiment.

The regiment in question has icot symmetry and a colonel with dynkin symbol x3o4x3o4o3*a4/3*c. I came upon this symmetry when finding conjugates of some other icot-symmetry regiments that I listed. Its cells are giddics, icoes, wavitoths, wavaties, and riches. The verf looks like x x4o || q o4x with lacings of length x(8/3) - in other words some distorted variant of squapip (square antiprism prism). This regiment contains giddics and riches in the colonel, and I think you can facet it to get spics, and possibly afdecs and gacocas as well. In other words, this regiment is pretty freaking huge.


I looked at it closer, it appears to be in the xo'oxo army. Afdecs and gacocas do show up inside it, so it will be one large regiment. Afdecs will behave like affixthi here since orientation will matter - uh oh! A good prefix for the name could be antiprisamtofrustary due to the shape of the verf.
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Re: List of uniform honeycombs

Postby username5243 » Sat Feb 23, 2019 10:41 pm

Yeah. I remember when I stumbled upon tetracombs that had trigonal versions of this verf (scyropot comes to mind), but the square version is much, much worse. Also keep in mind that, as with most of their other occurrences, the spics/giddics in that regiment will act like it has 32 members because each icoic member can appear in two different orientations.

I'm still working on updating the full list, I've started making notes on the spreadsheet as to which regiments are cojugate, along with a list of the facets of each colonel. As a "sneak preview" I thought I'd show the symmetry groups I've found regiment colonels in. I'm using Klitzing's Dynkin notation here, but on the sheet I'm using shorthands similar to what Hedrondude does on his site.

Code: Select all
Cyclopentachorics:
o3o3o3o3o3*a
(No other groups with colonels. I know that, for instance, there's a cypit faceting with a duhd shaped verf that's Wythoffian, but haven't looked into non-colonels too much.)

Tesseractics:
o4o3o3o4o
o4/3o3o3o4o
o4/3o3o3o4/3o
o4o3o3o4/3o4*c
o4o3o3o4o4/3*c
o4/3o3o3o4/3o4*c
o4/3o3o3o4o4/3*c
o3o3o4o *b4o4/3*c
o3o3o4o *b4/3o4*c
o3o3o4/3o *b4o4/3*c
o3o3o4/3o *b4/3o4*c
o3o3o4o~o4/3*c (~ means infinity)
o4/3o3o4*a4/3o3o4*a
o4/3o3o4*a4o3o4/3*a
o4o3o4/3*a4o3o4/3*a

Icositetrachorics:
o3o3o4o3o
o3o3o4/3o3o
o3o3o4o3o4/3*c
o3o4o3o *b3o4/3*c
o3o4/3o3o *b3o4*c
o3o3o4o3o4/3*b
o3o3o4/3o3o4*b
o3o3o4o3o4*b4/3*d
o3o3o4/3o3o4/3*b4*d
o3o3o4o3o4*b *c4/3*e
o3o3o4/3o3o4/3*b *c4*e
o3o3o4o3o4/3*a
o3o4o3o3o4*a4/3*c
o3o4/3o3o3o4/3*a4*c
o3o4o3o4o3*a4/3*c
o3o4/3o3o4/3o3*a4*c

Demitesseractic:
o3o3o *b3o4o
o3o3o *b3o4/3o
o3o3o4o4/3*b3o
o3o3o4/3o4*b3o
o3o3o4o4/3o3*a
o4o4/3o3o3o3*a3*c
o3o3o4o4o3*b4/3*d
o3o3o4/3o4/3o3*b4*d

Quarter tesseractic:
o3o3o *b3o *b3o
(No other groups with colonels, and all memebrs of this group are in the demitesseractic group. There may be Wythoffians with quarter tesseractic symmetry in other regiments though.)


these are all the symmetry groups I've discovered so far that contain colonels. Other Wythoffian hoeycombs exist in groups not listed above, but those grups don't form colonels.

This is actually the first time I've come back to these in a while - my main project in polytope related studies so far has been trying to list more of the 6D uniforms, I've got a list started with mostly the smaller regiments with pyramidal verfs listed fully along with most Wythoffians. There are some pretty interesting ones in there - including a few symmetry groups I've found that look like a triangle with "tails" attached to all the nodes of the triangular graph. I'll probably post about it (in a new thread) more once I have more of that list done.
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Re: List of uniform honeycombs

Postby Klitzing » Tue Feb 26, 2019 6:16 am

Just to mention, in a private mail PolyhedronDude meanwhile suggested the name "square antitrapezary disicositetratidistesseratiapeiratic tetracomb" with acronym "sazdideta" for that original x3o4x3o4o3*a4/3*c of username5243.
--- rk
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Re: List of uniform honeycombs

Postby username5243 » Tue Feb 26, 2019 11:07 am

I actually have the list all done. I'll attach it here (it's zipped again, because the forum doesn't seem to allow attaching Excel files directly). I'm not sure about all the names (especially some of the later ones among the icoics - I'm not very good at coming up with unique names for things), but I'm far more confident about this list being closer to completeness than the last one I posted. In order, the columns are: Short name - Long name - Symbol (in Hedrondude's form) - Verf - Number of members (if it can be determined from one of Hedrondude's polyteron regiment counts - if not there's question marks in that column) - Facets (of colonel only) - Notes (which includes info about conjugate regiments and a few short names of some prism-type honeycombs used as facets).
Attachments
tetracombs.xlsx.zip
(20.09 KiB) Downloaded 32 times
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Re: List of uniform honeycombs

Postby ubersketch » Tue Mar 05, 2019 1:39 am

Oh Hi UN5243, didn't know you were interested in this stuff. I don't know why people haven't tried to enumerate euclidean honeycombs. think I have my own investigation of this topic somewhere. i don't know where it is and ill try and find it.
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Re: List of uniform honeycombs

Postby polychoronlover » Sun Sep 01, 2019 6:12 am

Here is the cell listing of each member of the rich regiment. I checked each member against the corresponding member (or pair of members) with a topologically equivalent verf from the gacoca regiment, using my post on May 8, 2017, which I had verified to be correct with a Python script, as my reference.

Every cell is listed the same number of times it occurs in asymmetric positions in the figure. For example, if four co verfs appear in a verf faceting with rectangular prism symmetry, co is listed twice for that honeycomb. When I typed up this list of regiment counts it appears I thought there were 12 members with P(4) symmetry, but it turned out that one of them was fissary. (The gacoca counts seem to be right, though.) I typed this list by hand, so please let me know if you find any mistakes.

Code: Select all
R(4) symmetry (o4o3x4o) (7 members):
oct co
oct cho that
oct oho hoha
oct tha
co tha
oho that
cho hoha

S(4) symmetry (o3x4o *b3o) (25 members):
oct co oho sha hoha
oct co cho tha hoha
oct co tha sha
oct oho that
oct cho that tha sha
co thah squat
co thah oho that squat
thah oho cho hoha hoha squat
thah oho cho that that squat
thah cho that squat
thah cho tha hoha squat
thah co tha tha squat
thah co co tha sha
thah co hoha cho sha
thah oho oho that sha hoha
thah oho hoha sha
thah oho that tha sha
thah cho cho that hoha sha
thah sha tha
co oho that tha sha
co cho that
co tha sha
oho cho that hoha sha
oho tha hoha
cho hoha sha

S(4) symmetry (x3o4x *b3x) (6 members):
cho toe
toe hoha
oho toe tha
co toe that
oct oct toe that
oct oct oho toe

P(4) symmetry (o3x3o3x3*a) (11 members):
oho toe that
oct oct co cho toe that
oct oct oho toe hoha tha
co cho toe hoha tha
oct oct co oho toe sha hoha
oct oct co toe that sha hoha
oct oct oho cho toe sha tha
oct oct cho toe that tha sha
co toe that sha hoha
oho cho toe sha tha
cho toe sha hoha

Fissary case:
co oho toe that sha tha


EDIT: I just realized the fissary was actually an inter-regimental compound.
EDIT: "thah cho tha tha squat" is actually "thah co tha tha squat". This was an error in typing my list and does not appear in my notes.
EDIT: "oct oct oho toe" instead of "oct oho toe"
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Re: List of uniform honeycombs

Postby polychoronlover » Sat Sep 14, 2019 5:11 pm

Here is a spreadsheet of all known uniform honeycombs, arranged by regiment and numbered according to my post on May 4, 2017:

https://docs.google.com/spreadsheets/d/1VrA41fAv8amjCrBxH618pWw-HBl03G8QwkVrdzmpXwI/edit?usp=sharing

The acronyms in black are official Bowers-style Acronyms to the best of my knowledge. The acronyms in red were created by username5243, and the names in green were coined by me.
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Re: List of uniform honeycombs

Postby polychoronlover » Sun Sep 22, 2019 4:25 pm

I'm planning to write code to display projections of some of these honeycombs given a list of coordinates. For example, I made this projection of stut cadoca yesterday:

stut_cadoca_published.png
stut_cadoca_published.png (4.4 KiB) Viewed 298 times


On each level, the goccoes take up half of the zones of square symmetry, while the cubes and quiths overlap and take up the other half. Which polyhedra belong to which zones alternates between levels.
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Re: List of uniform honeycombs

Postby Mercurial, the Spectre » Sun Sep 22, 2019 9:05 pm

Sorry to interrupt, but I do like finding uniform snub hyperbolic honeycombs:
o4s4o3o - alternated rectified square tiling honeycomb (verf is a rectified triangular prism).
s4s4o3o - snub square tiling honeycomb (verf is a tridiminished icosahedron with the lateral edges being sqrt(2) instead of 1). This is actually the second member of the family represented by sPs4o3o, with P = 3 being the snub 24-cell.
s4o4o4s - alternated runcinated order-4 square tiling honeycomb (verf is a rectified square antiprism, in general this can be extended to s4oPo4s, with a vertex figure equivalent to a rectified P-antiprism)

I believe these are the only ones that are alternated facetings from honeycombs that are not regular themselves.

The case of x4s4o *b3s (as a partial snub) might actually be non-uniform (tried finding for the vertex figure but failed), but s3s4o4x does indeed exist, compare that to prissi.

Cheers!
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All human beings are born free and equal in dignity and rights. They are endowed with reason and conscience and should act towards one another in a spirit of brotherhood.
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Re: List of uniform honeycombs

Postby polychoronlover » Tue Sep 24, 2019 7:33 pm

You might enjoy this thread from Wendy Krieger:

http://hi.gher.space/forum/viewtopic.php?f=25&t=1795&p=18908

The thread refers to:
  • Alternated x4o4o4x
  • Rectified x4o4o4x
  • Truncated x4o4o4x
  • Gyrated alternated x4o4o4x
  • Diminished alternated x4o4o4x

and others.
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Re: List of uniform honeycombs

Postby polychoronlover » Sun Oct 13, 2019 5:54 am

Here is the edge skeleton of Skivpacoca:

skivpacoca.png
skivpacoca.png (3.49 KiB) Viewed 221 times


Skivpacoca has o4o3o4o symmetry. The goccoes take up one set of zones with cubic symmetry, and the sircoes take up the other set. You can also see a side view of the octagrammic prisms connecting the goccoes.
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