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Einstein problem solved (aperiodic monotile discovery)

Posted 1 year ago

enter image description here

8 Replies

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POSTED BY: Moderation Team

Can the single tile be generated by some kind of morphing of Penrose's kite and dart?

POSTED BY: Paul Abbott

Can the single tile be generated by some kind of morphing of Penrose's kite and dart?

You could try mapping the "drafter" Triangle into Penrose triangles and hope that the angles / matching rules work out. But I would guess no because how difficult (impossible?) it would be for patterns fitting on a hexagonal grid to ever allow five-fold centers of symmetry.

As far as structural analysis is concerned, Figure 2.12 could be a good place to start. It looks a bit like this space filling tree dragon we discovered at winter school:

enter image description here

In the future we'll probably get a definition of the tree structure itself, separate from definition in tiling context, but who will be the first to find it?

POSTED BY: Brad Klee

But wait! There's more!

At Hat tilings via HTPF equivalence, we've got a lot more code and patterns. For example, each hat tile cuts up a hexagon in the hexagonal grid into 2 or 3 pieces. The division orientation is either Y or upside-down Y. If you color the Y divisions red, you get the pattern below. Code for that and many other items at the link.

enter image description here


It seems from the title image here: that the E. tile can be broken into diamond shapes, each of which could be a kite+dart. So maybe the tile could be broken into kites and darts in a way so that the aperiodicity of the E. tiling would produce aperiodicity in the kites and darts

POSTED BY: Peter Barendse

I didn't get a good svg from the app website, I think: ResourceFunction::usermessage : SVGImport::format: "/Users/pbarendse/Downloads/output(2).svg" does not appear to be a valid SVG file. >>

POSTED BY: Peter Barendse

Hey Peter! Thanks for your interest. We're still actively working on the proof (and now a few others including this one from nks). It turns out after some decently involved combinatorics that the Hat tile is actually a combinatorial hexagon, as is its cluster super tile. Consequently, in addition to the already-awesome HatTrialityTree, we also have HatHexagons. This is fully a new way to build the hat tiling, except that it leaves holes where the reflected tiles are:

hat hexagons

Of course, once you realize (and prove!) that the Hat is actually a hexagon, then it becomes possible to build hat tilings using the SAT solver. That's what we're currently working on for the last phase of the proof. "Oz is ever 浮", to adapt a lyric to the situation.

In the meantime, MoMath is having a contest for Hat-related creations, so if you have anything to contribute, perhaps send it to NYC and see what they say? Even if you don't already have something, you could probably get something quickly using the WFR's... Anyways, have fun and keep puzzling!

POSTED BY: Brad Klee

Many years ago I (and Tom Sibley) proved that any tiling (finite or infinite) with Penrose rhombs is 3-colorable as a map (a question raised by John Conway). And later the same was proved for Kites and Darts. So I wonder if the same is true for hats. Ed: If you send me a pile of hats I can run my 3-coloring program on it. I even made a carpet showing 3-colored rhombs.

POSTED BY: Stan Wagon
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