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.

-- you have earned Featured Contributor Badge Your exceptional post has been selected for our editorial column Staff Picks http://wolfr.am/StaffPicks and Your Profile is now distinguished by a Featured Contributor Badge and is displayed on the Featured Contributor Board. Thank you!

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:

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

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.

--Brad

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}}]

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}}]

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.

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

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.

Attachments:

Walton3017.nb