Almost-uniform tilings?

[Marek14]

Here's something that I found out recently.

In hyperbolic plane, we can put, say, pentagons and triangles together in (5,5,5,3) tiling. But that tiling is not uniform -- you can't make it so each vertex looks the same.

Surprisingly, it turns out that when you double this tiling (six pentagons and two triangles per vertex, arranged as (5,5,5,3,5,5,5,3)), there IS a way to make it uniform. The key is that two (5,5,5,3) "halves" that form a vertex do not have to be identical. So, does that mean that there exists an "almost-uniform" version of (5,5,5,3) that is still periodic, just has two distinct classes of vertices instead of one class?

[wendy]

The only polytope answering to a vertex-cycle of (5,5,5,3,5,5,5,3) is the archiform: "(1) (2) 3(3,4) 5(5,6) 3(7,8)", which reduces to (s,s,s,3,s,5,s,3), where s are snub-faces, here a pentagon.

The reason that i doubt that this could be dismantled into something less by using an unwrap as proposed, is that the double-cycle above is a snub, and is not symmetric on both halves, unless you draw the line through the pentagon (5,6), and the snub face between the digons at (1) and (2).

I an not sure how significant this is, but (5,5,5,3) can be constructed with horocyclic segments, specifically of orders 3, 7, ... [alternating lucas-numbers], whereas anything that is finite, and constructed by conway's archiform rules necessarily can not have such segments. I don't know how this fares in the wrapping.

Conway has given us a very good test for uniform polytopes. We can suppose a 'quasi-uniform' tiling has vertecies with the same ordering, up to reversed, but not a larger scale uniformity. Thus any of the vc = (5,5,5,3) would be quasi-uniform, but not uniform. This lines up quite nicely with quasi-crystals, which have fragments of a symmetry like {5,10/3}, but breaks down on the larger scale.

[Marek14]

It took some time, but I finally have an almost-uniform (5,5,5,3) tiling!

It uses two kinds of triangles: red (with magenta as their mirror image), and green, and two kinds of pentagons: yellow (with cyan as their mirror image), and blue.

It has three kinds of vertices, but it's described by a simple rule and can be extended indefinitely.

[mr_e_man]

Interesting. After seeing this, I considered "almost-uniform" tilings in the Euclidean plane, and found that no such thing exists; any CRF Euclidean tiling with congruent vertices must be uniform. (You probably already knew this.) What about 3D Euclidean space?

[Marek14]

Interesting. After seeing this, I considered "almost-uniform" tilings in the Euclidean plane, and found that no such thing exists; any CRF Euclidean tiling with congruent vertices must be uniform. (You probably already knew this.) What about 3D Euclidean space?

We still haven't figured out how to describe a general 3D tiling. But we're getting pretty good with general 2D tilings (see here: https://zenorogue.github.io/tes-catalog/).

Yesterday, we spent some time considering (3,4,5,5) before concluding that this particular combination cannot be made into tiling, not even an aperiodic one.

[mr_e_man]

Yesterday, we spent some time considering (3,4,5,5) before concluding that this particular combination cannot be made into tiling, not even an aperiodic one.

I can see that easily: both vertices shared by the triangle and the square must be completed by pairs of pentagons; but then the triangle's third vertex has the form 5.3.5.x, which is not 3.4.5.5, regardless of what x is.

[Marek14]

Yesterday, we spent some time considering (3,4,5,5) before concluding that this particular combination cannot be made into tiling, not even an aperiodic one.

I can see that easily: both vertices shared by the triangle and the square must be completed by pairs of pentagons; but then the triangle's third vertex has the form 5.3.5.x, which is not 3.4.5.5, regardless of what x is.

Oh, we considered allowing (3,4,5,5) as a combination, so (3,5,4,5) could also occur, but it doesn't work even with that relaxed condition.

[mr_e_man]

Okay, that's less obvious. But I can see that also, after a few minutes of sketching; eventually we get a vertex with two triangles or two squares.

[mr_e_man]

With your relaxed condition (each vertex has a certain set of faces, but they can be arranged in any way), there are uncountably many 2D Euclidean tilings. Take the rhombitrihexagonal tiling, and gyrate some of the "hexagonal cupolas"; the vertices are 3.4.4.6 and 3.4.6.4.

Or take the elongated triangular tiling (3.3.3.4.4), and place a "fault line" of 3.3.4.3.4 vertices, then continue with 3.3.3.4.4 for any distance, and place another "fault line", etc. See this picture.

[mr_e_man]

For 3D Euclidean tilings made of Platonic solids, there are just three possible edge types: cube⁴, tet².oct², and tet.oct.tet.oct. From this it follows that there are also only three possible vertex types. One type forms the cubic honeycomb; the vertex figure is an octahedron. The second type forms the octet honeycomb; the vertex figure is a cuboctahedron (a triangular gyrobicupola). The third type forms the gyrated octet honeycomb; the vertex figure is a triangular orthobicupola. If we require congruent vertices, then the result must be one of these three tilings, which are uniform. If we relax this condition, then there are uncountably many possible tilings, but they must have the form given in the link; they're composed of layers of the octet honeycomb.

Generalizing to tilings made of uniform solids, there are uncountably many possible tilings even if we require congruent vertices. Simply insert triangular prisms between octet layers; the vertex figure is a triangular cupola augmented with a hexagonal pyramid. Again the octet layers can have two different orientations.

[mr_e_man]

2D hyperbolic tilings with congruent vertices are also uncountably infinite. Take the uniform 4.4.4.5 tiling, and notice that a column of squares is perpendicular to infinitely many rows of squares. One half of the tiling can slide along any row, while leaving the other rows intact (though the original column is broken). By default, each square is connected by edges to 2 pentagons, or by vertices to 4 pentagons. After sliding, each square in the row is connected by edges to 1 pentagon, and by vertices to 2 pentagons. A default row corresponds to a 0, and a slid row corresponds to a 1, so a tiling corresponds to a real number in binary; these are uncountable.

But what exactly do you mean by "almost-uniform"? Do you want several vertex types, or a single vertex type? And does "vertex type" refer to congruence (local), or symmetry (global), or just the set of faces?

[mr_e_man]

I noticed that a 3.5.5.5 tiling has three conceivable arrangements of triangles around a pentagon. Two of them appear in the tiling you found. The third has 5-fold symmetry; perhaps it appears in a different 3.5.5.5 tiling.

Apparently 3.4.3.4.4 can be made almost-uniform, with two vertex types according to symmetry. The squares come in blocks of three; the middle square is a centre of 2-fold rotation symmetry. There are two triangle types: one is a centre of 3-fold rotation symmetry, and the other is cut (off-centre) by a line of glide-reflection symmetry.

[Marek14]

Actually, you are right -- I have been looking at it today and I found a 3.5.5.5 tiling like that.
It's more complicated: the original one had three types of vertices; this one has five (apparently, there are no solutions for 2 or 4 types).

3.4.3.4.4, you say? I'll have a look.

[Marek14]

First of all, I continued my exploration of 5-uniform (5,5,5,3) tilings and found two more.

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Next, I looked at the (4,4,3,4,3) tiling you mentioned. I found it, but I also found two more:

44343_1_small.png

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44343_2_small.png

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[Marek14]

As for what "almost uniform" means: there is more than one vertex type, but there is still a finite number of the types and the tiling is overall periodic.

[Marek14]

And three more configurations of (4,4,3,4,3), using three kinds of vertices.

[mr_e_man]

I thought that uniform tilings were uniquely determined by their vertex configurations. But it looks like there are three different uniform 3.4.4.4.4 tilings!

[Marek14]

Well, sure. See here: https://bendwavy.org/klitzing/explain/t ... .htm#aaaab

I've known this for a long time now

[mr_e_man]

As for what "almost uniform" means: there is more than one vertex type, but there is still a finite number of the types and the tiling is overall periodic.

Do you allow several vertex configurations?

Do you allow several sets of faces?

Examples: 4.4.3.4.3 and 4.4.4.3.3 are different vertex configurations with the same set of faces. 3⁸ and 3².8³ are different vertex configurations with different sets of faces.

[Marek14]

As for what "almost uniform" means: there is more than one vertex type, but there is still a finite number of the types and the tiling is overall periodic.

Do you allow several vertex configurations?

Do you allow several sets of faces?

Examples: 4.4.3.4.3 and 4.4.4.3.3 are different vertex configurations with the same set of faces. 3⁸ and 3².8³ are different vertex configurations with different sets of faces.

Well, I suppose both would be allowed. For example, this is a tiling that uses two different kinds of hyperbolic octagons: (8,4} and {8,8}.

There are two basic approaches to classification: starting from a set of tiles or starting from a vertex figure (they are duals of each another). In the first group, I made a lot of tessellations based on polyforms (examples can be seen here: https://zenorogue.github.io/tes-catalog/).

[wendy]

John Conway described that while they are described by the vertex figure, the cycle of polygons is not enough to describe it.

What you further need is to describe the outbound and inbound edges, where they fall in the cycle, and if they change parity. This is essentially the orbifold.

The first in Mr_e's diagrams is [1,2] [3,5] [4], where [1,2] is the triangle, and [3,5] is the purple squares and [4] are between the purple triangles.

The second is [1,2] [3,5] (4), where the (4) edge is the centre of an order-2 rotation.

It looks like 3% 2% 2% 2% or (1,2) (3) (4) (5). This is an ordinary snub.

In any case, the first two have orbifold nodes of wanders or miracles, which you tell by the presence of non-consecutive numbers in the edges. The third one is a fairly ordinary snub, with squares (as does Miller's mosnster). A pair of consecutive edges in brackets is a cone, or rotation-polygon.

W

[Marek14]

John Conway described that while they are described by the vertex figure, the cycle of polygons is not enough to describe it.

What you further need is to describe the outbound and inbound edges, where they fall in the cycle, and if they change parity. This is essentially the orbifold.

The first in Mr_e's diagrams is [1,2] [3,5] [4], where [1,2] is the triangle, and [3,5] is the purple squares and [4] are between the purple triangles.

The second is [1,2] [3,5] (4), where the (4) edge is the centre of an order-2 rotation.

It looks like 3% 2% 2% 2% or (1,2) (3) (4) (5). This is an ordinary snub.

In any case, the first two have orbifold nodes of wanders or miracles, which you tell by the presence of non-consecutive numbers in the edges. The third one is a fairly ordinary snub, with squares (as does Miller's mosnster). A pair of consecutive edges in brackets is a cone, or rotation-polygon.

W

Well, yes. That's what I've been doing. With recent advances, I'm able to search solution space for any set of tiles.

However, in some cases, a solution requires a single tile to have multiple edge cycles (each covering one symmetrical part of the edges). The notion of parity is a bit more complicated as well in cases where some tiles have axes of symmetry passing through them and some have not.

[mr_e_man]

While looking through the Wolfram Physics Project, I recognized some causal graphs as 4.4.3.4.3 tilings. The most relevant image is with the words "or after 500 steps". But it's not almost-uniform, and it's not even a true hyperbolic tiling; it has a cusp at the centre.

[Marek14]

Oh, I just realized, looking at your user name, that you're probably the person who replied to me on my recent mathstackexchange question!

So, let me tell you what I've been doing.

Since I found that (4,10,4,10) and (5,6,5,6) share an edge length, I've made an analysis of hybrid tilings that use these four polygons with this edge lengths. There are seven different vertex configurations that are allowed for these tilings ((4,4,10,10), (4,10,4,10), (4,5,6,10), (4,6,5,10), (4,5,10,6), (5,5,6,6), and (5,6,5,6)), and my analysis is currently complete up to 5 vertex types. My program/computer combination is powerful enough to do 6 vertex types, but that will take a while before it's complete.

Smallest solutions have 3 vertex types -- one is attached.

I also took a look at the other combination: (3,10,3,10) and (4,5,4,5) which also share an edge length. This, however, turned to be much harder. The smallest solution requires 5 types of vertices, and it's unique to boot. I have found a few more with 6 vertex types, but they are much scarcer than the previous type.

This was the main impetus of my question, to find other possibilities for hybrid tilings.

I am aware that edge of {3,2n} is generally equal to {2n,n}, and also that edge of (3,4,n,4) is equal to (4,n,2n) (this is the principle behind Johnson solids like diminished rhombicosidodecahedra, I've been able to produce several "diminished" and "gyrated" periodic variants of (3,4,7,4) thanks to that).

My other projects at this point are looking once again at (a,a,a,b) and (a,a,a,a,b) tilings. I've previously enumerated the uniform variants, but now I'm looking at 2-uniform ones as well, which lead to many more interesting tilings with simple structure. (Elongated square gyrobicupola is an example of one of these tilings.)

[mr_e_man]

I thought that uniform tilings were uniquely determined by their vertex configurations. But it looks like there are three different uniform 3.4.4.4.4 tilings!

Well, sure. See here: https://bendwavy.org/klitzing/explain/t ... .htm#aaaab

I've known this for a long time now

In fact, you had already shown those same tilings here viewtopic.php?f=3&t=2051#p26771, which I should have seen before. I didn't need to make new pictures.

[Marek14]

Yeah, there is a lot of possibilities

[Marek14]

Just a note that the tessellation catalog on the link I've posted before is being updated periodically. I have several sub-projects going and I keep adding new examples. One category are tilings based on polyforms made of basic tiles of (6,6,7), (3,7,3,7), or (6,6,8). Another are "polyschwartzes", polyforms made out of basic Schwartz triangles (2,3,7) and (2,4,5). And I started an effort to implement k-uniform Euclidean tilings.