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Hat tiling: fractal substitution

POSTED BY: Bowen Ping
7 Replies

Merge H7 and H8 to substitute

Since there are at least two ways to separate tiling, one is H8 and H7, and the other is H8 and alternative H7. So here I tried to merge H7 and H8 together to avoid "disagreement" or "difference", and to see if we can get the same result.

From intuition, it is definitely feasible. And the truth is, yes. Here I record something about it.

POSTED BY: Bowen Ping
Posted 2 years ago

Thanks! This is probably exactly what I was asking about, but now I need to take a closer look at the combinatorics how the alternative super-tiles fit together and form larger copies. For reference, we're trying to do something similar to this post on Trilobite and Crab. However, this was done before we started worrying about multi-way graphs, which should improve our organization and reproducibility in this next example.

Thanks again for all your efforts. It's amazing what you've accomplished in such a short time, when these sort of calculations took me years and years to figure out.

POSTED BY: Brad Klee
Posted 2 years ago

I was thinking about if we can find out how these (4,12,12) vertex configurations distribute, it is easy to replace them by corresponding super-tiles, which requires only few information of cluster.

As for the connection between these substitution rule and TralityTree, maybe more likely there is a connection between TralityTree and H-supertile, which are both in a "triangle" shape. We will find out if there is a further connection.

Notice that the $H$ cluster is in 1-to-1 correspondence with reflected tiles only occurring at location $H_1$. The $A+$ edge of $H$ is (from what we've found) the strongest starting place for growing $H_7$ and $H_8$. Counting the two clusters separately, { $H_7$, $H_8$} are in a 2-to-1 correspondence with reflected hats (but the ratio of $H_7$ : $H_8$ is not $1:1$).

Do not be deluded by boundary shape, which is really just an accident of how well the rules have been chosen on edges. My first choice was not so nice, so I kept going through subsequent choices until arriving at something aesthetically pleasing.

I've repeated the code from the WFR site, with extra highlighting to show how two of the symmetric trivalent vertices are selected from the set of eight, and then used to project supertiles to cover the Cartesian space:

Your highlighted $(4,12,12)$ are seen in this image as green and yellow plus one extra gray tile on the outside of the $A+$ edge. Green always falls on an $H_7$ supertile, but in my convention, yellow does not. Yellow tiles are found on both $H_7$ and $H_8$ in an extended position outside of the central $H$ meta-tile.

From what I've heard, Craig made the choice to have both $H_7$ and $H_8$ with complete surrounds of the central tile and thus complete $H$ meta-tiles directly inlaid. This choice is inconsistent with the map from vertex figures, because it would result in overlap of the $H_2$ occurring in every $H_7$. The solution was to parallel-translate the $H_2$ in $H_7$ to the extended location, which gives (imo) a more intrinsically-motivated definition for replacement rules.

Back to my question above: If we suspend belief in the aesthetic choice of completing a surround for every reflected tile, and pursue the tree-motivated definitions more directly, does the resulting substitution system on two-dimensional tiles have any additional nice properties from being defined more naturally?

POSTED BY: Brad Klee

Hi, Brad! This is how to do substitute with alternative H7 and H8. In fact, I think original pair of H7 and H8 is equivalent to the pair of alternative H7 and H8. There are some clues in the following notebook:

POSTED BY: Bowen Ping
POSTED BY: Bowen Ping
Posted 2 years ago

Hi Bowen ~ This is a great article, which needs to be read more closely by relevant experts, congrats!

But I'm still wondering if the original definition of H7/H8 is really the best? Others have agreed with me that these definitions were probably chosen for aesthetic reasons. Unfortunately, the chosen convention is not good for comparing with HatTrialityTree.

If you look under "neat examples" of the triality tree ref. page above, you will find alternative definitions for the H7 / H8 tiles, which are motivated by the structure of the substitution system as realized on a selection of its edge and vertex sets. How does different choice of H7 / H8 affect your control point analysis? Is there a means to prove the tree replacement code from your two-dimensional code or vice versa?

Questions like these are important to the second part of the proof, where we need to show that chosen metatiles only ever form larger and larger supertiles. The proof will be more or less difficult depending on how well we choose meta-tiles and their control points.

POSTED BY: Brad Klee
POSTED BY: EDITORIAL BOARD
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