90 points by todsacerdoti 17 hours ago | 5 comments
blobcode 2 hours ago
I feel like SAT solvers and the like are getting a lot more attention on HN recently (for example https://news.ycombinator.com/item?id=44259476) - justifiably so! I think that they're a great tool that's often criminally underused in industry for a whole subset of problems.
CliffStoll 15 hours ago
Terrific article by Nicolas Hatcher! Aperiodic tilings are fun to make from paper, wood, and ceramics. I've cut tiles from ceramic field tiles.

No surprise that concave cuts in ceramics are a high stress, so Kite and Dart tiles don't work very well (the dart is likely to crack). Same is true for the Turtle, Hat, and Spectre.

Rhombus tiles are everywhere convex, and the P3 Rhombus tiles are easy to cut in a diamond saw (or even a snap-cutter). With a diamond band-saw, it's possible to make Penrose rhombs with curved (parabolic) edges.

But cutting tiles from stock field tiles produces sharp surface edges -- you don't want these as bathroom floor tiles. Also, you waste a lot of the field tile as scrap. To get "friendly" tile shoulders, I'm experimenting with making Penrose tiles directly from high-fired porcelain clay.

kurthr 12 hours ago
Interestingly, when firing your own, you could also make Supertiles from combinatorial collections of Penrose or other aperiodic tiles.
gus_massa 12 hours ago
Do you have some photos? It may be a nice post.
nhatcher 14 hours ago
Hey, thanks! Good luck with your Penrose tiles. I wish I had the time myself :)
akoboldfrying 6 hours ago
Interesting article! I'm stuck on the following claim about tiling Hats though:

> In each center of an hexagon you can have any of the 12 possibilities:

> Any of the 6 rotations of the Hat

> Any of the 6 rotations of the anti-Hat

For this claim to hold, it must be the case that a Hat (or anti-Hat) occupies the same area as a hexagon. But they don't: a hexagon is made of 6 kites, while a Hat is made of 8. So, some hexagons must contain no corresponding (anti-)Hat -- specifically, for every 8 hexagons, there must be 6 (anti-)Hats.

This seems to complicate the SAT formulation. But could the fix be as simple as adding a 13th possibility, "No hat at this hexagon centre occupies more than half of its kites"? Or are additional constraints needed?

nhatcher 1 hour ago
Oof, that is poorly written. I'll update the text.

Notice how every hat has a special "marked" vertex. It is the red dot in this image:

https://www.nhatcher.com/images/hats/hat-marked.png

This is what I mean by "at the center ofthe hexagon you have the hat". What should say is "the center of the hexagon coincides with the marked vertex of a hat". Hope that makes more sense.

17 hours ago
curtisszmania 16 hours ago
[dead]