## Almost-uniform tilings?

Higher-dimensional geometry (previously "Polyshapes").

### Almost-uniform tilings?

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?
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

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?
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### Re: Almost-uniform tilings?

mr_e_man wrote: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.
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### Re: Almost-uniform tilings?

Marek14 wrote: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.
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### Re: Almost-uniform tilings?

mr_e_man wrote:
Marek14 wrote: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.
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

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.
Last edited by mr_e_man on Thu Apr 02, 2020 2:34 pm, edited 2 times in total.
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

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?
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### Re: Almost-uniform tilings?

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.
Attachments hyperbolicTiling34344.png (36.46 KiB) Viewed 4828 times
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

First of all, I continued my exploration of 5-uniform (5,5,5,3) tilings and found two more. 5553_3_small.png
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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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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

And three more configurations of (4,4,3,4,3), using three kinds of vertices.
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### Re: Almost-uniform tilings?

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! Attachments hyperbolicTiling34444.png (36.99 KiB) Viewed 4814 times
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### Re: Almost-uniform tilings?

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

I've known this for a long time now Marek14
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### Re: Almost-uniform tilings?

Marek14 wrote: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.
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### Re: Almost-uniform tilings?

mr_e_man wrote:
Marek14 wrote: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/).
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### Re: Almost-uniform tilings?

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] , where [1,2] is the triangle, and [3,5] is the purple squares and  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.

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### Re: Almost-uniform tilings?

wendy wrote: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] , where [1,2] is the triangle, and [3,5] is the purple squares and  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.
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

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.)
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### Re: Almost-uniform tilings?

mr_e_man wrote: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 wrote: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.  ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
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mr_e_man
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### Re: Almost-uniform tilings?

Yeah, there is a lot of possibilities Marek14
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### Re: Almost-uniform tilings?

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.
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### Re: Almost-uniform tilings?

Can anyone explain why, with the edge length of the regular {4,7} tiling, the angle formed by two squares at a vertex is the same as the angle formed by a triangle and a heptagon at a vertex?

2 * 51.428571° = 35.733235° + 67.123908°

I discovered this numerically, and verified it algebraically, but I want to understand it geometrically.

I proved that any CRF hyperbolic tiling with at least one vertex of the form 3.m.n or 4.m.n must be uniform, the only exception being 4.n.2n which can combine with 4.n.4.3 or 4.n.3.4 to make a modified uniform tiling:
Marek14 wrote: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).

In the process, looking at a 4.n.2n vertex next to a 4.4.3.n vertex, I needed to consider fitting other polygons instead of two squares on the other side of the 3.n edge. A single polygon's angle is always smaller than that of two squares. Three triangles, and thus any three or more polygons, is always larger than two squares. A square and anything larger, obviously, is larger than two squares. A triangle and an octagon is larger than two squares. A triangle and a hexagon is smaller than two squares. All that remains is a triangle and a heptagon; hence the above discovery.

(But the edge of {4,7} is too long, longer than that of any 4.n.2n, to be used in such a tiling.)
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mr_e_man
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### Re: Almost-uniform tilings?

mr_e_man wrote:Can anyone explain why, with the edge length of the regular {4,7} tiling, the angle formed by two squares at a vertex is the same as the angle formed by a triangle and a heptagon at a vertex?

2 * 51.428571° = 35.733235° + 67.123908°

I discovered this numerically, and verified it algebraically, but I want to understand it geometrically.

I proved that any CRF hyperbolic tiling with at least one vertex of the form 3.m.n or 4.m.n must be uniform, the only exception being 4.n.2n which can combine with 4.n.4.3 or 4.n.3.4 to make a modified uniform tiling:
Marek14 wrote: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).

In the process, looking at a 4.n.2n vertex next to a 4.4.3.n vertex, I needed to consider fitting other polygons instead of two squares on the other side of the 3.n edge. A single polygon's angle is always smaller than that of two squares. Three triangles, and thus any three or more polygons, is always larger than two squares. A square and anything larger, obviously, is larger than two squares. A triangle and an octagon is larger than two squares. A triangle and a hexagon is smaller than two squares. All that remains is a triangle and a heptagon; hence the above discovery.

(But the edge of {4,7} is too long, longer than that of any 4.n.2n, to be used in such a tiling.)

I've just posted to Tilings Google List with the summary of what I know so far. Reposting here:

I've made a list of Archimedean combinations* that lead to identical or double edge lengths as other combinations. Some of them look... unappealing. For example this one:
edge(13, 27, 27, 30, 30) = edge(18, 18, 23, 25, 29).
*"Archimedean combination" just means that if you fit a particular set of polygon to a vertex, they will have a particular edge length. It doesn't mean that there is actually an Archimedean tiling that would utilize this combination. In addition, each combination except the simplest ones can have its polygons arranged around the vertex in different ways.

There's nothing to suggest why this might be so, and even worse, since polygons on each side are completely different, there is no way to actually build a tiling that utilizes both vertices.

Or take the relationship "edge(5, 5, 10, 10, 13, 17) = edge(5, 6, 6, 12, 17, 18)" -- here you technically *do* have a way to connect both vertices (an edge with pentagon on one side and 17-gon on the other side could connect them), but the numbers are an absolute mess. I ran a search up to 5 vertices (with all possible permutations of numbers in both vertices), and found nothing. My guess is that that could be due to those primes: fitting a 13-gon or a 17-gon into a tiling is a tall order, and this one would have to use both! I think that even if hybrid tiling involving these two vertices existed, it would have more vertex types than my machine can handle.

So far, other interesting combinations I've noticed include the following:

edge(3,3,10,10) = edge(4,4,5,5). Since both combinations include a straight angle, there is also an in-between vertex (3,4,5,10) that allows to connect these two.
Results: So far 1 hybrid solution was constructed, and I'm aware of the existence of several more. The search has stalled, though, because of the computational difficulties.

edge(4,4,10,10) = edge(5,5,6,6). Looks similar to previous case, and also has an in-between vertex, (4,5,6,10). This one is much fertile, I have constructed almost 300 solutions ranging from 3 to 6 vertex types.

edge(4,n,2n) = edge(3,4,4,n). This is a trivial, known example. We can see it in spherical geometry/polyhedra as well: triangular cupola, square cupola and pentagonal cupola are all orbiform and they are based on this relationship if you draw them as spherical tilings instead of polyhedra.

Triangular tilings in general. If you have a triangle in {3,n}, you can remove a vertex from the tiling, blending the n triangles around it into one {n,n/2} polygon. n doesn't have to be even -- if you do this for n=5, for example, you end up with three possible Archimedean combinations that share the same (spherical) edge length, (3,3,3,3,3), (3,3,3,5), and (3,5,5), which all occur on Johnson solids derived from diminishing the icosahedron. If n is even, there will be one combination that has no triangles, if n is odd, there will always be at least one triangle. n=10 is an anomalous case I'll mention later.

I have not researched the last two classes too well, as they seem to be of less interest.

edge(3,5,12,12) = edge(4,5,5,12). There is only one type of edge that can connect these two (between pentagon and dodecagon), but I've tried it and it works. I've built 10 hybrid tilings ranging from 2 to 4 vertex types.

The {5,4} family: {5,4} shares edge length with Archimedean combination (3,3,3,3,4,4). As both can be halved, there's also an in-between vertex (3,3,4,5,5). What's surprising is that there's one other vertex that fits: (3,4,10,20). I have foundsome true hybrids here.

The {18,4} family: {18,4} shares edge length with Archimedean combination (3,3,4,4,6,6), and there's an in-between vertex (3,4,6,18,18) as well. It seems like it could be related to the {5,4} family in some way, but I tried to search ahead, and found no larger example where {n,4} would have the same edge length as (3,3,4,4,m,m).

edge(3,3,3,5,20,20) = edge(3,3,4,5,5,20) = edge(3,4,4,5,5,5). This one might be my favorite of the bunch because all the possible vertices have low symmetry and the search goes through them quickly. On the other hand, they have many combinations and many solutions. I have constructed some true hybrids.

edge(3,4,8,40) = edge(3,5,8,8). My search confirms that there are hybrid solutions.

The {4,5} family: {4,5} shares the edge with (4,4,10,30). True hybrids are confirmed.

The {10,5} family: {10,5} shares the edge with {3,10}, as mentioned before. However, it also shares the edge with {5,6}, making it possible to mix all three at the same vertex. And, doubly also, this edge is twice the edge of {5,4} family, opening the door to new possibilities that use both sizes.

The {8,6} family: {8,6} shares the edge with {4,8}, with in-between vertex (4,4,4,4,8,8,8) possible. Since all polygons involved are even, I think that this one should have an enormous number of solutions.

The {4,7} family: {4,7} shares the edge with three other 7-polygon combinations, obtained by substitution (4,4) -> (3,7): (3,4,4,4,4,4,7), (3,3,4,4,4,7,7), and (3,3,3,4,7,7,7).

{12,4} has twice the edge of (3,3,12,12).

The mystery family: And finally one thing I don't quite understand yet. It turns out that the edge of (n,n,2n,2n,2n,2n) is twice the edge of (4,4,n,n). My search has confirmed this for n=3 to 10, so it's unlikely to be a coincidence. I think this might be a general relationship, but I lack the knowledge to prove it. I should note that the case n=5 is anomalous because (4,4,10,10), as we already know, also shares the edge with (5,5,6,6) and (4,5,6,10), allowing those polygons to also mingle with doubly sized (5,5,10,10,10,10) pentagons and decagons.

As you can see, I've also found the {4,7} connection. Wendy also found that {12,10} and {5,12} share an edge. Right now I'm trying a brute-force approach to compare edge lengths up to {20,6} for small integral ratios.
Are you on the polytope Discord? Lots of things seem to be happening there atm.
Marek14
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### Re: Almost-uniform tilings?

So you didn't find any tilings with those 3,4,7 combinations?

Marek14 wrote:The mystery family: And finally one thing I don't quite understand yet. It turns out that the edge of (n,n,2n,2n,2n,2n) is twice the edge of (4,4,n,n). My search has confirmed this for n=3 to 10, so it's unlikely to be a coincidence. I think this might be a general relationship, but I lack the knowledge to prove it. I should note that the case n=5 is anomalous because (4,4,10,10), as we already know, also shares the edge with (5,5,6,6) and (4,5,6,10), allowing those polygons to also mingle with doubly sized (5,5,10,10,10,10) pentagons and decagons.

Say a tiling with 4.4.2n.2n has edge length s, and n.n.2n.2n.2n.2n has edge length t. Then s is determined by:

2 arcsin(cos(π/4) / cosh(s/2)) + 2 arcsin(cos(π/(2n)) / cosh(s/2)) = π

Dividing by 2 and rearranging:

arcsin(cos(π/4) / cosh(s/2)) = π/2 - arcsin(cos(π/(2n)) / cosh(s/2)) = arccos(cos(π/(2n)) / cosh(s/2))

Taking the cosine, and using cos(θ) = √(1 - sin²(θ)):

√(1 - (cos(π/4) / cosh(s/2))²) = cos(π/(2n)) / cosh(s/2)

Squaring and rearranging, and using some double-angle identities:

1 = (cos²(π/4) + cos²(π/(2n))) / cosh²(s/2)

cosh²(s/2) = (1 + cosh(s))/2 = cos²(π/4) + cos²(π/(2n)) = 1/2 + cos²(π/(2n)) cosh s = 2 cos²(π/(2n)) = 1 + cos(π/n)

And the other length t is determined by:

2 arcsin(cos(π/n) / cosh(t/2)) + 4 arcsin(cos(π/(2n)) / cosh(t/2)) = π

Again dividing by 2 and rearranging:

2 arcsin(cos(π/(2n)) / cosh(t/2)) = π/2 - arcsin(cos(π/n) / cosh(t/2)) = arccos(cos(π/n) / cosh(t/2))

Taking the cosine, and using cos(2θ) = 1 - 2 sin²(θ):

1 - 2 (cos(π/(2n)) / cosh(t/2))² = cos(π/n) / cosh(t/2)

1 - (1 + cos(π/n)) / cosh²(t/2) = cos(π/n) / cosh(t/2)

0 = (1 + cos π/n) (1/cosh(t/2))² + cos(π/n) (1/cosh(t/2)) - 1

Applying the quadratic formula:

1/cosh(t/2) = ( - cos(π/n) ± √(cos²(π/n) + 4 (1 + cos(π/n))) ) / (2 (1 + cos(π/n)))

= ( - cos(π/n) ± (cos(π/n) + 2) ) / (2 (1 + cos(π/n)))

The lower sign gives an impossibility, 1/cosh(t/2) = -1. The upper sign gives:

1/cosh(t/2) = 1/(1 + cos(π/n)) cosh(t/2) = 1 + cos(π/n)

cosh(t/2) = cosh(s)

t = 2s.
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