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Lots of substitution tilings

Posted 5 years ago

I recently updated Substitution Tilings, one of my many items at the Wolfram Demonstrations Project. Some of these were introduced in my blog Shattering the Plane with Twelve New Substitution Tilings Using 2, φ, ψ, χ, ρ. Here are 26 of the 40 tiling currently in Substitution Tilings. Some but not all of these are at the Tilings Encyclopedia.
Wolfram Language code is attached below at the end of this post. SqrtChi tiling SqrtRho tiling SqrtPsi tiling SqrtPhi tiling Quartic Pinwheel tiling SqrtTwo tiling PsiQuad Rho Quad tiling Trib Trap tiling Psi Trap tiling Psi Chord tiling Psi Wedge tiling Trib Chord tiling TwoTriangle tiling RhoQuad tiling Birds and Bees tiling Binary tiling Penrose Rhomb tiling Robinson tiling Kites and Darts tiling Ammann Chair or Scherer Golden Bee Ammann A4 tiling WaltonChair tiling Ammann Phi Chair tiling Triangle Duo tiling Equithirds tiling Tritan tiling limhex tiling Pinwheel tiling

Any corrections, suggestions or additions are welcome.

11 Replies
Posted 1 year ago

I'm not sure what you mean by "algebraic tilings." It doesn't seem to match the "Algebraic Tiling" article by S. K. Stein. I just saw your AlgebraicSubstitutionTiling function. ResourceFunction["AlgebraicSubstitutionTiling"]["KitesandDarts", 0] will give the substitutions used to define the "Kites and Darts" tiling. I wish I could define my own substitutions and see what the resulting tilings would look like. For example: A simple substitution tiling

This uses relatively simple shapes. It's got an inflation factor of sqrt(3). For a more complex example here is the beginnings of substitution tiling with regular 14-gons, 7-gons and other polygons that fit with them. 7-fold girih type tiling

As of now, I haven't computed it's inflation factor. How easy is it to program a new substitution tiling such as one of these? Thanks. John Berglund

POSTED BY: John Berglund

For the algebraic tilings, with just a few weird shapes with weird angles that somehow work together, they are more discovered than made. They always start with an algebraic root that has strange properties. For example, in 3D space, there is a set of 19 points at power distances, based on the plastic constant. There is a classic adage in math and physics: "If it doesn't work, try 1, 0 or the square root." For algebraic geometry, once you've identified an algebraic root, the square root always seems to simplify things geometrically. That's why I wrote the SqrtSpace function. After discovering that made all of my tilings easy, I went through the literature and looked at all other known substitution tilings to see if there were exceptions. So far, there aren't any exceptions.

For non-algebraic with "many" pieces, things are simpler. For example, with multiples of the 5 tetrominoes (Tetris pieces), make larger copies of each of the tetrominoes. This is then a substitution tiling. But there are many ways to do this.

I still haven't found a 3D substitution tiling with the plastic constant, but I'm fairly sure one exists.

Posted 1 year ago

I'm new to the community. The examples are great. How would you go about making up a substitution tiling of your own? It would be nice if you could specify what the substitutions would be for each shape, and have the program show what some patches would look like.

POSTED BY: John Berglund

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POSTED BY: Moderation Team

I've fixed a gap error in AlgebraicSubstitutionTiling that didn't handle complex substitutions well. Now it handles them fine. For example, the new Harvest tiling. Andrew Hudson's Harvest
Here's a few steps of it with my new code. harvest with new code

Here's the Dale Walton quintic from above. DaleWaltonQuintic

But it seems with the changes I've added some minor errors, so I need to work on it more. Then I need to add some complicated substitution tilings that went wonky before due to the gap error.

EDIT: Fixed all errors, I think. Let me know.


Hi Ed, Yes, I like this one-to-five cross tiling, and years ago I spent some time calculating a set of channels that could probably be encoded to force the substitution hierarchy. Here's a printout from one of my notebooks:

Colored Cross Tiling

However, I can't recall my certainty about that proof, and cross hierarchy is relatively difficult compared to the that of Gosper's island:

Gosper Rule Trefoil Tiling

I seem to recall a few other people agreeing about the matching rules, but probably from a subsequent version. It's already been four or five years ago, so I don't know. Anyways, don't forget the Gosper Island!

Later it would probably be worthwhile to try and at least give a second layer to the tiles, which (at least) carries a set of channels sufficient for encoding matching rules.


POSTED BY: Brad Klee

Hi Ed,

Good that you got integer inflation factor tilings in the second post, but there are trivial examples Missing—one square to four, or one equilateral triangle to four.

You might also want to include 3D ABCK tiling by Danzer & Co. I’ve already done some exploration of integer coordinatization, see for example:

which I think you probably published some time earlier. I still can appreciate the result of this demonstration, and think that it suggests more to be done on your program here. If you decompose tiles to edges, how many unique vectors do you get? How are those vectors written out in a canonical basis?

Cheers —Brad

POSTED BY: Brad Klee

Missing—one square to four, or one equilateral triangle to four.

How about one square to five?

AlgebraicSubstitutionTiling[{1,{{-3,-1},{-3,1},{-1,-3},{-1,-1},{-1,1},{-1,3},{1,-3},{1,-1},{1,1},{1,3},{3,-1},{3,1}},  {{1,6,12,7}-> {{1,2,5,4},{5,6,10,9},{8,9,12,11},{3,4,8,7},{4,5,9,8}}} ,
{{1,1,1,1,1}}},5,{"N", "ImageSize"->{600,Automatic}}]

fractal square

Here's another one I was just looking at

AlgebraicSubstitutionTiling[{Root[-1-#1^2+#1^3&,1],{{{4,0,0},{0,0,0}},{{8,-4,-1},{0,-8,7}},{{0,0,-4},{0,0,0}},{{0,0,0},{0,0,0}},{{-2,-3,3},{-6,3,1}},{{2,1,-1},{-6,3,1}},{{4,5,-6},{4,1,-2}},{{0,-3,2},{4,1,-2}}}/4, {{1,2,3}-> {{4,7,3},{1,6,4},{6,5,2},{7,8,4},{4,5,6},{2,8,7},{8,5,4},{5,8,2}}, 
{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}},{1,2,3}-> {{1,2,3}}} ,
{{0,2,3,3,4,4,5,5}+1,{1},{2},{3},{4},{5},{6}}},1,{"N", "ImageSize"->{600,Automatic}}]

not quite psi

It's in the supergolden ratio, psi. But if you scale the area of the big triangle to the component triangles, you get areas psi^{7.7217, 5, 3, 2, 2, 1, 1, 0, 0} ... the big triangle is out of phase for a smooth substitution tiling system with a fixed number of sizes at each step.


I should mention the Demonstrations of Dieter Steemann and the tiling demos of Karl Scherer, particularly Rep-tiles and Irreptiles. I'm still gleaning tilings from these, the Tilings Encyclopedia and IFStile. Hopefully I'll be able to improve my Demonstration Substitution Tilings to be stronger with a lot more tiling systems so that all of them are easily investigated.


So how to make these? Dale Walton sent me a picture of a new tiling. Walton 3017 tiling

"All edges are powers of x=1.2365057033915... (5,6,7) triangle divides into (0,5,6); (3,4,5); (2,4,5)" Where $x^5-x^3-1=0$. This particular root has discriminant 3017. It's an algebraic number field seen giving extremal solutions in Wheels of Powered Triangles and Degenerate Power Simplices.

By the end of the notebook, I get to this image. Walton 3017 tiling

Not quite there. For a full substitution tiling system there should eventually be a fixed number of colors where every color represents congruent triangles. I haven't solved that yet for this tiling.


May as well put in the rest of what I have so far.

Domino Tiling Drafter 3 tiling Drafter 4 tiling half hex tiling kite tiling l tromino tiling p pentomino tiling quarterhex tiling sphinx tiling tritan 4 tiling tromino tiling

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