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Inconsistencies

By the definition here, "A polyhedron which has regular faces and is transitive on its edges is said to be quasiregular", all regular polyhedra are quasiregular. By the definition at Polyhedron, "vertex-transitive and edge-transitive (and hence has regular faces) but not face-transitive", no regular polyhedron is quasiregular. Thus these two definitions are inconsistent. Furthermore, the enumeration of convex quasiregular polyhedra given here, which includes the octahedron but excludes the other regular polyhedra, is inconsistent with both definitions.

In fact the two external sites cited disagree as to whether the octahedron should be regarded as quasiregular. But both agree that to be quasiregular, faces with n sides must alternate with faces with m sides at each vertex -- implying an even number of faces at each vertex. Then the octahedron is quasiregular or not, depending on whether you allow n=m or not, but the other regular polyhedra (having 3 or 5 faces at each vertex) are not.

(posted by 24.58.33.52 on 06:06, 3 January 2008).

Hi, you are quite right about the inconsistencies. The literature on quasiregular polyhedra is pretty sparse, but it appears to be one of those many areas where geometers define one thing and then describe something else. One of the difficulties is that there are several interesting quasiregular properties, but nobody has ever methodically figured out which properties are fundamental and which are consequences of these. Feel free to make clarifications as to the nature of the muddle.
As for the octahedron, it can be seen as quasiregular by the definition at Polyhedron, if we colour alternate faces black and white so there are two kinds - a figure sometimes also called the tetratetrahedron.
And as for other web sites, they are for the most part a terrible load of rubbish. Mathworld perpetuates many shameful myths about polyhedra. George Hart is a lot more reliable, though his classification of quite so many star polyhedra as quasiregular is unusual (I happen to agree with him, but that is a long and continuing story).
HTH -- Steelpillow (talk) 12:55, 3 January 2008 (UTC)[reply]

Missing entries?

If non-convex polyhedra with non-convex vertex figures count, there's a number missing, firstly the edge-sharing forms (is there an "official" term?) of the pictured five:

Also, these have only singly symmetrical but nevertheless alternating vertex figures. Dunno if that's an additional problem.

Didn't want to add these directly for two reasons - the section for non-convex quasiregulahedrons is titled with "examples", and I notice the discussion on what the authorities consider quasiregular anyway... --Tropylium (talk) 19:28, 20 February 2008 (UTC)[reply]

You have the right idea. The first list are indeed quasiregular (though the second list are not - they do not have edges all wthin a single symmetry orbit). We can also find quasiregular examples among the apeirohedra - plane tessellations and infinite skew polyhedra. Trouble is, none of this seems to have been published in an adequate reference (It's on George Hart's website but AIUI personal websites are deemed inadequate). A few months ago I submitted a paper on exactly this to the Mathematical Intelligencer, so we shall have to wait and see. -- Steelpillow (talk) 20:58, 22 February 2008 (UTC)[reply]

More missing entries

under the list of convex quasi-regular polyhedra, it says the the cuboctohedron and the icosahedron are the only two. However in the category "Quasiregular Polyhedra", it lists the rhombic dodecahedron and rhombic tricontahedron as well. Shouldn't they be included then? —Preceding unsigned comment added by Timeroot (talk • contribs) 17:39, 4 July 2008 (UTC)[reply]

They're given here under Quasiregular_polyhedron#Quasiregular_duals. Tom Ruen (talk) 05:18, 5 July 2008 (UTC)[reply]
See also the remark in that section, about differences of opinion as to whether these two polyhedra are quasiregular or not. Cromwell regards them as quasiregular, while Coxeter et. al. do not. -- Cheers, Steelpillow (Talk) 11:50, 5 July 2008 (UTC)[reply]

Lead needs to be more accessible

The lead for this uses too much technical jargon. Move the mathematical definition into the article and make the lead into an introduction which is accessible to an intelligent, well educated non-mathematician.

I'm trained as an engineer and I couldn't make head nor tail of this intro. filceolaire (talk) 09:51, 23 November 2009 (UTC)[reply]

A worthy complaint! Hopefully better now? Tom Ruen (talk) 10:22, 23 November 2009 (UTC)[reply]
Much better. Thanks. filceolaire (talk) 12:54, 23 November 2009 (UTC)[reply]

Two errors in second diagram

In the diagram "Regular and quasiregular figures" there are two errors.

In the third column, second row, the Schläfli symbol is missing. It should be {3 3} arranged vertically.

In the fourth column, fourth row, the caption should not be "pentagonal tiling". This thing is quasiregular not regular. Maybe "tetrapentagonal tiling"?

I haven't time to fix these now. Maproom (talk) 13:59, 18 February 2010 (UTC)[reply]

Thanks. I have fixed the tetrapentagonal tiling. The Schläfli symbol poses a problem. These symbols are supposed to be unique to a given figure. Clearly, applies to two of the examples given, thus providing a counter-example to long-held mathematical folk lore. I think that all we can do here is to faithfully reflect published sources, imperfect as they are. -- Cheers, Steelpillow (Talk) 16:38, 18 February 2010 (UTC)[reply]
The 5.4.5.4 tiling does not belong in that table! It ought to show {5,4} as , the rectification of {5,5}. —Tamfang (talk) 22:06, 18 February 2010 (UTC)[reply]
Yes, it belongs in the table above - that will mess up somebody's nice layout. The appropriate "single-height" symbol is {5.4}2, and none other. -- Cheers, Steelpillow (Talk) 20:34, 19 February 2010 (UTC)[reply]

I have moved this from the article because the references given do not include the majority of these figures. Indeed, they do not consistently recognise all the original "parents" as quasiregular - the most reliable sources such as Coxeter do not.

As an example of the problem, the tetrahemihexahedron does not have a Schläfli symbol distinct from that of the cuboctahedron. Coxeter did not regard it as quasiregular, even though he sometimes gave definitions of quasiregularity which are consistent with it. We are once again into the inconsistencies of modern theory, and we must be careful to avoid wp:or. -- Cheers, Steelpillow (Talk) 18:44, 28 February 2010 (UTC)[reply]

here is the bit I have moved

The nonconvex ones are:

  1. The medial rhombic triacontahedron,
  2. the great rhombic triacontahedron,
  3. the duals of the hemipolyhedra,
  4. and the three triambic icosahedra, duals of the three ditrigonal polyhedra.

pictured below.


Medial rhombic triacontahedron
V5.5/2.5.5/2
Great rhombic triacontahedron
V3.5/2.3.5/2
Tetrahemihexacron
V3.4.3/2.4
Octahemioctacron
V3.6.3/2.6

Hexahemioctacron
V4.6.4/3.6
Small icosihemidodecacron
V3.10.3/2.10
Small dodecahemidodecacron
V5.10.5/4.10
Great icosihemidodecacron
V10/3.3.10/7.3

Great dodecahemidodecacron
V5/2.10/3.5/3.10/3
Great dodecahemicosacron
V5.6.5/4.6
Small dodecahemicosacron
5/2.6.5/3.6
Small triambic icosahedron
V5.5/3.5.5/3.5.5/3

Medial triambic icosahedron
V3.5/2.3.5/2.3.5/2
Great triambic icosahedron
V(3.5.3.5.3.5)/2

I'll bite. Why did David Eppstein (talk · contribs) add Chiral polytope to the See also? —Tamfang (talk) 22:46, 30 August 2012 (UTC)[reply]

Because his discussion of abstract chiral polytopes discusses their relation to quasiregularity? — Cheers, Steelpillow (Talk) 13:02, 1 September 2012 (UTC)[reply]

Sixfold vertices

Do any of the sources referenced explicitly state that sixfold vertices can be treated as quasiregular? I have only ever seen quasiregularity discussed in the context of fourfold vertices. — Cheers, Steelpillow (Talk) 11:26, 29 August 2013 (UTC)[reply]

Why wouldn't they be treated as quasiregular? Of course there are no ditrigonal polyhedra, regular or quasiregular; but that's a matter of what can up up to 360 degrees, not of definition. Maproom (talk) 12:16, 29 August 2013 (UTC)[reply]
Actually, there are ditrigonal polyhedra. Even if we ban coplanar faces, using star polygons solves the problem, as they can have internal angles smaller than 60°. Additionally, the faces do not all have to circle forwards, and there is nothing stopping them from wrapping round the vertex more than once. Thus there exist ditrigonal polyhedra: the quasiregular ones are the small ditrigonal icosidodecahedron, ditrigonal dodecadodecahedron, and great ditrigonal icosidodecahedron. Double sharp (talk) 13:56, 14 April 2014 (UTC)[reply]
I was wrong, at least in part. In their classic paper on Uniform polyhedra, Coxeter et. al. give a definition of quasiregularity compatible with sixfold vertices and identify a number of such polyhedra. But I was also right in part. In that same paper they describe nine polyhedra as "semi-regular" even though they appear to meet the given definition of quasiregularity. (These are all "hemi" polyhedra having faces passing through their centres, hence their Schwarz triangles and associated Wythoff symbols do not display the characteristics of quasiregularity even though the polyhedra themselves do. Presumably this is why they were missed.) If we treat these as quasiregular when Coxeter et. al. (or anybody else) did not, we are guilty of WP:OR. If the fact that these polyhedra meet the definition cannot be referenced, then however true and obvious we might think it, saying so is WP:OR. This also applies to tilings of the plane having sixfold vertices, which I have not seen addressed in this context. I admit I have not read widely on such tilings, so good sources might be out there, but are they? That both answers your first question and provides counterexamples to your subsequent assertion. I assume you mean, there are no convex ditrigonal polyhedra in Euclidean space. — Cheers, Steelpillow (Talk) 14:24, 29 August 2013 (UTC)[reply]
George W. Hart lists non-convex ditrigonals as quasi-regular: [1]. — Stannic (talk) 15:51, 29 August 2013 (UTC)[reply]
George's website is self-published and, for the most part if not all, is not peer-reviewed. As such we cannot regard it as an authoritative source. It can at best be used to support material from such sources. — Cheers, Steelpillow (Talk) 18:04, 29 August 2013 (UTC)[reply]


Coxeter's 1954 paper [2] (Section 7), title "The regular and quasiregular polyhedra p | q r" says "p | q 2" are regular, and all the others with this Wythoff symbol are quasiregular, including right triangles "2 | p q" and in general triangles none are regular, all 3 singularly active mirrors make quasiregulars. Tom Ruen (talk) 21:08, 29 August 2013 (UTC)[reply]
The Hemipolyhedron don't qualify as quasiregular because Wythoff symbol, p/(p − q) p/q | r, shows two active mirrors. I notice at Small icosihemidodecahedron it says: It is given a Wythoff symbol, 3/2 3 | 5, but that construction represents a double covering of this model. Tom Ruen (talk) 22:04, 29 August 2013 (UTC)[reply]
Here's a graphic File:Vertex figures.png based on Johnson's names at List of uniform polyhedra by vertex figure (Only online source Mathworld I think, [3]). So it show the quasiregular with a rectangular vertex figure, and (hemi-polyhedra) as versi-regular with crossed-rectangle. Tom Ruen (talk) 23:01, 29 August 2013 (UTC)[reply]
Johnson (2000) classified uniform polyhedra according to the following:
  1. Regular (regular polygonal vertex figures): pq, Wythoff symbol q|p 2
  2. Quasi-regular (rectangular or ditrigonal vertex figures): p.q.p.q 2|p q, or p.q.p.q.p.q, Wythoff symbol 3|p q
  3. Versi-regular (orthodiagonal vertex figures), p.q*.-p.q*, Wythoff symbol q q|p
  4. Truncated regular (isosceles triangular vertex figures): p.p.q, Wythoff symbol q 2|p
  5. Versi-quasi-regular (dipteroidal vertex figures), p.q.p.r Wythoff symbol q r|p
  6. Quasi-quasi-regular (trapezoidal vertex figures): p*.q.p*.-r q.r|p or p.q*.-p.q* p q r|
  7. Truncated quasi-regular (scalene triangular vertex figures), p.q.r Wythoff symbol p q r|
  8. Snub quasi-regular (pentagonal, hexagonal, or octagonal vertex figures), Wythoff symbol p q r|
  9. Prisms (truncated hosohedra),
  10. Antiprisms and crossed antiprisms (snub dihedra)
@Tom, You say that the hemipolyhedra don't qualify because of their Wythoff symbols (on the sphere). I'm sorry, but where in the definition of quasiregularity does it demand a specific form of Wythoff symbol on the sphere? My understanding of quasiregularity is that its definition is based on the arrangement of faces around each vertex. FYI the Mathworld reference to Johnson (2000) is to a draft ms on Uniform Polytopes that is still yet to be published, a classic example of why websites like Mathworld are not reputable sources. Johnson's classification scheme also appears to be used in the List of uniform polyhedra by vertex figure, but there is no proper citation of him (or of anyone else!) on that page. — Cheers, Steelpillow (Talk) 11:11, 30 August 2013 (UTC)[reply]
In fact, with the exception of the orientable octahemioctahedron, these Wythoff symbols do not even represent the hemipolyhedra themselves, but rather degenerate versions of them where all elements are doubled. Double sharp (talk) 13:58, 14 April 2014 (UTC)[reply]

Quasiregular polychora and honeycombs

In Regular Polytopes (book) Coxeter says {4,3,4}, is the only regular honeycomb in Euclidean 3-space, and h{4,3,4}, or is the only quasiregular honeycomb. So I'm not sure of the full definition, but in this case, it has a quasiregular vertex figure. So from that case alone here's a list of qualifying forms, each has alternating of two types of cells, although they are also regular if the two cells are the same type (like a checkerboard). Tom Ruen (talk) 01:23, 13 October 2013 (UTC)[reply]

Quasiregular 4D polytopes:

Quasiregular 3D Euclidean honeycombs

Quasiregular 3D compact hyperbolic honeycombs

Quasiregular 3D Paracompact_uniform_honeycombs

According Norman Johnson (in private communication)

A recursive definition of quasi-regular, which agrees with how the term is used by Coxeter, is as follows:

  • A polygon is quasi-regular if its symmetry group is transitive on the vertices. [regular and truncated ]
  • An n-polytope (n > 2) is quasi-regular if its symmetry group is transitive on the vertices and its vertex figures are quasi-regular.

Rectified polyhedra [1-ring Coxeter diagram] are quasi-regular, but so are the ditrigonary polyhedra, which are not obtained by rectification. For n > 3, rectified regular n-polytopes are generally not quasi-regular.

- Tom Ruen (talk) 03:31, 13 October 2013 (UTC)[reply]

Well, Tom, that's all very fascinating but it appears to be unpublished. As such it is WP:OR and has no place here. — Cheers, Steelpillow (Talk) 09:14, 14 October 2013 (UTC)[reply]
As I said I only have explicit source statement for the Euclidean honeycomb h{4,3,4}, so that is defendable on the article. But for the full list above, that's why I put it here on the talk page. I'll continue looking for what else Coxeter said about Quasiregular honeycombs. Tom Ruen (talk) 09:30, 14 October 2013 (UTC)[reply]

Degenerate case

The common core of a dihedron and hosohedron would be a degenerate figure with n+2 faces (2 n-gons, n digons), 2n edges, and n vertices. It's not currently listed here. Should it be? Does it have a name? Joule36e5 (talk) 22:17, 27 January 2015 (UTC)[reply]

Let's see - = = {n,2} (or = = ). So the regular dihedra are also quasiregular, with 2 colors, one for each hemisphere rather than a single color. So the way the dihedrons are colored now is actually quasiregular. Tom Ruen (talk) 22:28, 27 January 2015 (UTC)[reply]
"Does it have a name?" It's a (pentagonal, whatever) prism, isn't it? Maproom (talk) 22:49, 27 January 2015 (UTC)[reply]
I think Tom's analysis is correct. Like the octahedron, dihedra have both regular and quasiregular colorings. I'd go further and suggest that even-numbered hosohedra do as well (depending on your definition of quasiregularity). But I'll bet you won't find any of that in any reliable source. The hosohedra degenerate to infinite prisms in Euclidean space, but these are not the same structures as the prism polyhedra. — Cheers, Steelpillow (Talk) 07:28, 28 January 2015 (UTC)[reply]
Yes, I see now that Tom is right. A prism isn't quasiregular, it's just a truncated hosohedron. Maproom (talk) 08:05, 28 January 2015 (UTC)[reply]
Hosohedra of the form {2,2q} would indeed be quasiregular: the colouring would be as shown at Hosohedron#Kaleidoscopic symmetry. You can imagine these improper forms on the table by adding a row at the top for {p,2} (the dihedra) and a column to the left for {2,2q} (the hosohedra). (But does {1,2} count as quasiregular? It would be a singular exceptional sense, because I don't see how {1,2q} would work for q > 1.) Double sharp (talk) 10:33, 28 January 2015 (UTC)[reply]
{1,2q} may be constructed on the projective plane, where it is simply q lines through a point. I suppose one could call them projective hosohedra. The dual {2q,1} projective monohedron would be a single 2q-gon wrapped around a line subdivided into q segments. At this point, the wise among us begin to think, "hang on, do I have q or 2q edges here? I need to understand this kind of thing before I decide what is or is not a valid, or even degenerate, polytope". — Cheers, Steelpillow (Talk) 11:02, 28 January 2015 (UTC)[reply]
(All off-topic) I would call them hemishosohedra and hemidihedra. A hemidihedron has q edges, but its face has 2q edges. It is dodgy regarding such things as polyhedra, but they are perfectly respectable regular maps. Maproom (talk) 12:56, 28 January 2015 (UTC)[reply]

To get to a quasiregular even-sided hosohedron, t{p,2}, , I think if you have to look at Norman Johnson's unpublished opinion, i.e. a hosohedron is more like a polygon, (t{p}, ) than a polyhedron. Tom Ruen (talk) 18:47, 28 January 2015 (UTC)[reply]

A recursive definition of quasi-regular, which agrees with how the term is used by Coxeter, is as follows:

  • A polygon is quasi-regular if its symmetry group is transitive on the vertices. [regular and truncated ]
  • An n-polytope (n > 2) is quasi-regular if its symmetry group is transitive on the vertices and its vertex figures are quasi-regular.

Rectified polyhedra [1-ring Coxeter diagram] are quasi-regular, but so are the ditrigonary polyhedra, which are not obtained by rectification. For n > 3, rectified regular n-polytopes are generally not quasi-regular.

p.s. This is the only paper [4] that I've found on "quasiregular polygons", but also "quasiregular prisms"(!?). Tom Ruen (talk) 19:09, 28 January 2015 (UTC)[reply]
Glad to see there's been response here, but I don't think anybody has actually discussed the object I described, which I'll call a "hosodihedron" by analogy with "cuboctahedron" and "icosadodecahedron". It's a "flat prism" -- it has two n-gonal faces, but unlike a dihedron (which connects the two n-gons to each other) or a true prism (which puts n squares between them), they're connected by a ring of digons. Joule36e5 (talk) 03:43, 29 January 2015 (UTC)[reply]
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