1

The typical Penrose tiling consists of two deformed quadrangles. But it's there any aperiodic tiling consisting entirely of two or more deformed hexagons? Maybe even one that shares some properties of a hexagonal tiling such as only edge-neighbours?

Tobias Kienzler
  • 6,609
  • 2
    How about the chair tiling? – Dan Rust Aug 09 '14 at 12:39
  • @DanielRust interesting, though it fails on true aperiodicity – Tobias Kienzler Aug 09 '14 at 12:41
  • What's your definition of aperiodicity then? Most take it to mean an FLC non-periodic tiling that is also repetitive (all with respect to translation). The chair tiling satisfies these properties. – Dan Rust Aug 09 '14 at 12:49
  • @DanielRust maybe I missed something up there, is the Penrose tiling also repetitive? – Tobias Kienzler Aug 09 '14 at 12:50
  • Yes. The usual substitution has primitive associated substitution matrix and so it is repetitive (i.e. its associated translational dynamical system is minimal). – Dan Rust Aug 09 '14 at 12:54
  • I should add there are many different definitions of aperiodic tiling, and also an aperiodic set of tiles. So when I asked for your definition it wasn't a rhetorical question, it's perfectly reasonable to consider a different definition. – Dan Rust Aug 09 '14 at 12:56
  • @DanielRust I see. So the chair tiling is definitely a valid answer – Tobias Kienzler Aug 09 '14 at 13:07
  • It would depend on what you're using it for I guess :). With a quick browse of the (admittedly lacking) table of aperiodic sets of tiles, it looks like the Ammann A2 tiles might also be suitable. As with the chair tiles, I don't think this tileset forces aperiodicity without the local matching rules (the decorations on the tiles) but the tiling given by these matching rules is certainly aperiodic. – Dan Rust Aug 09 '14 at 13:17
  • 1
    You may be interested in this: An aperiodic hexagonal tile. It is jut one tile, rather than 2, but they do mirror it, too. Anyway, might provide some good food for thought. – jwd Nov 14 '19 at 23:26
  • @jwd Sounds great, thanks :) – Tobias Kienzler Nov 16 '19 at 19:39

1 Answers1

2

Yes. Use the Penrose Rhombus Tiling as a start. The edges with a round bump, use a straight line. Those with the triangle, use two lines.
enter image description here

Ed Pegg
  • 20,955
  • nice, thanks. I wonder if there something similar can be achieved with purely convex hexagons, and whether edge-only neighbours are possible – Tobias Kienzler Aug 16 '14 at 07:48