Timeline for Tiling planar integer lattice by finite point sets
Current License: CC BY-SA 4.0
11 events
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| Oct 27, 2022 at 20:32 | history | edited | RavenclawPrefect | CC BY-SA 4.0 |
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| Oct 26, 2022 at 4:59 | comment | added | RavenclawPrefect | @VilleSalo: Ah, you're correct - I missed the use of two tiles in the construction. I've edited the original answer. | |
| Oct 26, 2022 at 4:59 | comment | added | Terry Tao | Our methods are currently only able to demonstrate aperiodicity in high dimensions for tiling with one tile, and undecidability in high dimensions for the problem of tiling with two tiles. It seems reasonable to conjecture though that the problem of tiling with one tile is also undecidable in high dimensions. | |
| Oct 26, 2022 at 4:59 | history | edited | RavenclawPrefect | CC BY-SA 4.0 |
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| Oct 26, 2022 at 4:17 | comment | added | Ville Salo | (Indeed there is no general "monotilification result" in the sense that there are SFTs that are far from being conjugate to the set of tilings with a single tile, e.g. because single tile SFTs have deterministic directions and zero entropy. Of course neither of those things prevents undecidability.) | |
| Oct 26, 2022 at 4:12 | comment | added | Ville Salo | They explicitly say they don't address decidability issues, so hard to say whether they know how to get undecidability, but it does not look like they have a general "monotilification result" that turns anything into a single tile, and aperiodicity doesn't directly imply undecidability. Well, I see that @TerryTao has commented above already, maybe they know more than me. | |
| Oct 26, 2022 at 4:07 | comment | added | Ville Salo | "The structure of tiling problems in $\mathbb{Z}^d$ even in the single-tile case is in general difficult, and undecidable for sufficiently high $d$ as recently shown in Greenfeld and Tao 2022 - see this blog post." is this really claimed by Greenfeld-Tao? I thought they have is a single aperiodic (singleton) tileset. | |
| Oct 26, 2022 at 2:42 | comment | added | Steven Stadnicki | Surely $S=\{\langle 0,0\rangle, \langle 1,0\rangle, \langle 3,0\rangle\}$ can't tile $\mathbb{Z}\times\mathbb{Z}$ by translation? If it could then $\{0, 1, 3\}$ would translate $\mathbb{Z}$ by translation and this is clearly impossible. | |
| Oct 26, 2022 at 1:34 | comment | added | Terry Tao | Even the one-dimensional finite problem of determining all the sets that tile a given cyclic group ${\bf Z}/N{\bf Z}$ is difficult; the main conjecture in this area is the Coven-Meyerowitz conjecture mathscinet.ams.org/mathscinet-getitem?mr=1670646 which remains open in general, though has been some recent progress by Laba and Londner, see e.g., arxiv.org/abs/2207.11809 | |
| Oct 25, 2022 at 23:32 | history | edited | RavenclawPrefect | CC BY-SA 4.0 |
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| Oct 25, 2022 at 19:10 | history | answered | RavenclawPrefect | CC BY-SA 4.0 |