Timeline for Does $\mathbb{Z}\times\mathbb{Z}$ have an aperiodic monotile?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 26, 2023 at 19:31 | vote | accept | Dominic van der Zypen | ||
| May 26, 2023 at 13:38 | comment | added | Ville Salo | If an MO answer doesn't cause any headaches, what are we even doing. I'll add something later | |
| May 26, 2023 at 13:37 | comment | added | Aleksei Kulikov | Ah, I'm stupid and your example actually work, I can't draw pictures apparently (although a little explanation might be better). | |
| May 26, 2023 at 13:36 | comment | added | Ville Salo | If this was math.SE I'd have explained :) | |
| May 26, 2023 at 13:34 | comment | added | Ville Salo | Yes, that's what I mean. The SFT is basically a union of a vertical and a horizontal one-dimensional binary full shift (with a bit of overlap). | |
| May 26, 2023 at 13:31 | comment | added | Aleksei Kulikov | I was about to post the same thing, but I feel that 2-by-2 square actually can not tile $\mathbb{Z}^2$ with only one period -- you can shift by the vector $(2, 0)$ (which you probably think of), but also by $(2, 1)$ (or some variant of it). But now that I think of it, it can be fixed by considering $\{0, 1, 2\}\times \{0, 1, 2\}$ and use some non-repeating sequence to make your argument actually work. | |
| May 26, 2023 at 13:22 | history | answered | Ville Salo | CC BY-SA 4.0 |