4 ms·
It must be the latter. Many trivial shapes admit non-periodic tilings. For example a 2x1 rectangle: you tile the plane with squares, and divide each square in
by enriquto 3y ago
It must be the latter. Many trivial shapes admit non-periodic tilings. For example a 2x1 rectangle: you tile the plane with squares, and divide each square in two in a non-periodic way.
- mabbo 3y ago> and divide each square in two in a non-periodic way. That doesn't actually sound simple. Can you provide an example of how you would do that?
- enriquto 3y agoYou traverse the squares in a spiral pattern starting from a central one. Then each square is divided vertically or horizontally according to the binary digits of pi (or whatever).
- zuminator 3y agoIt doesn't even need to be (pseudo-) random. You can trace a spiral pattern out from the origin like so: 1 vertical up, turn right, 1 horizontal right, turn down, 2 vertical down, turn left, 2 horizontal left, turn up, 3 vertical up, turn right, 3 horizontal right, turn down, etc. The resulting spiral pattern will never repeat.
- BurningFrog 3y agoFor each 2x2 block, decide randomly/arbitrarily to split it horizontally or vertically.
- cvoss 3y agoThis solution won't be accepted without a lot more work. Flipping a coin is not guaranteed to produce an aperiodic tiling, since, in the trivial case where the coin comes up heads indefinitely, you'll be producing a periodic tiling. (You will argue it's not likely to do this, and I'd agree. In fact, I'd conjecture you'll only produce a periodic tiling with measure zero probability.) But what's required is to actually show an aperiodic tiling. If you have a proof that coin flipping will produce such a tiling with non-vanishing probability, then, presumably, embedded in your proof, you will likely find a decision procedure for producing an aperiodic tiling deterministically.
- JadeNB 3y ago> If you have a proof that coin flipping will produce such a tiling with non-vanishing probability, then, presumably, embedded in your proof, you will likely find a decision procedure for producing an aperiodic tiling deterministically. I'm not sure it's obvious to expect a proof that something happens with positive probability to embed a constructive proof of existence (which seems to be what you mean). The difference between these two ideas in combinatorics is exploited with the probabilistic method (https://en.wikipedia.org/wiki/Probabilistic_method https://en.wikipedia.org/wiki/Probabilistic_method).
- IanCal 3y agoThat makes sense but now I'm more confused about Penrose tilings. My understanding was that there were rules about placement to avoid periodic tilings. Thinking out loud, is this down to local Vs global rules? Otherwise I'm a bit confused as to why Penrose tiles were a big deal. To be clear, I'm fully aware the confusion is because I'm missing something.
- penteract 3y agoWith the Penrose tiling, you can add features to the edges of the tiles which enforce the extra placement rules, so it was the first time a set of 2 tiles which tile the plane only aperiodically was known. https://en.wikipedia.org/wiki/Penrose_tiling#Rhombus_tiling_(P3) https://en.wikipedia.org/wiki/Penrose_tiling#Rhombus_tiling_...
- IanCal 3y agoAh I see! Thanks!