5 ms·
Does an aperiodic tiling mean that if you take the entire infinite tiled plane, and while sliding around a copy of the plane, there are zero other places you ca
by cypherpunks01 3y ago
Does an aperiodic tiling mean that if you take the entire infinite tiled plane, and while sliding around a copy of the plane, there are zero other places you can slide it to that matches the original?
I'm just a bit confused by the opt-repeated claim that the tiling "never repeats itself" and I'm not sure if I'm understanding the brief translational symmetry explanation correctly.
- IIAOPSW 3y agoYeah. Its like an irrational number but in 2d.
- mrfox321 3y agoYes, that's the definition of aperiodic tiling. There is no translational symmetry.
- iamgopal 3y agoMy question is, It repeats aperiodically and predictably, like a Fibonacci ? Or aperiodically and unpredictably ? Like a prime numbers ?
- ndsipa_pomu 3y agoIt'll be predictable. Other aperiodic tilings (e.g. Penrose tiles) look to have regular patterns from a quick glance, but those patterns won't have translational symmetry despite looking vaguely symmetrical.
- fanf2 3y agoThe proof that the tiling is aperiodic comes with a recursive algorithm for generating tilings.
- paulddraper 3y agoThe proof doesn't have to be constructive
- gilleain 3y agoYes, exactly. Also this video: https://m.youtube.com/watch?v=IfVwelta1fE https://m.youtube.com/watch?v=IfVwelta1fE Helped me understand an example of a monotile that can tile aperiodically but also has a periodic tiling.
- Glyptodon 3y agoIf that's the case, does it have practical near repeats as the difference in rotation between an selected origin tile and another tile somewhere across infinity becomes bound to approach zero? I'd have guessed that with only 360 degrees of rotation at very large numbers you'd be bound to get a tile that has nearly the same rotation if not the same rotation as another tile. I'd have guessed that for a single tile aperiodic tiling there'd be a 2nd non rotational/translational requirement: that the if rotation is the same, the adjoining tiles of every case of the same rotation would not be the same, but that seems like it'd require infinite positions for two adjacent tiles to interlock, which also seems like at some point would mean two non identical positions are functionally identical for practical purposes unless the shape allows for infinite permutations of adjacency to a single tile that aren't effectively just the same at limit. Very curious about how I'm understanding or not understanding this . Do you get large subsheets as it extends to infinity that differ distinctly from other subsheets by single tiles, with some kind of corellary that subsheets can repeat but there is no subsheet that con cover the plane in totality besides the whole itself?
- gilleain 3y agoUnfortunately I don't have the mathematical skill to answer these questions! It might be worth looking at the original papers - quite readable, I understand - to see how they constructed the proof that their tiles are aperiodic. After all, you cannot just try to construct a large number of tilings and claim aperiodicity. I would imagine (although I could be wrong) that you get an infinite number of copies of each rotation of the tile. My impression would also be that there are only a small finite number of rotations. So it is less about the rotations than the local structure around each tile not repeating, no mater how large a radius you give for your neighbourhood. Like I say, I'm not good enough with this stuff to give a clear explanation here, sadly.