3 ms·

Wait a minute. I thought there was a simple non-existence proof something along the lines of: Given the square, you'll need to cut pieces with the outer arc(s)

Wait a minute. I thought there was a simple non-existence proof something along the lines of: Given the square, you'll need to cut pieces with the outer arc(s) of the circle. When you cut any length of convex arc, you also create a piece with and equal length of concave arc which will then require that much more convex arc to fill in the end. In other words, the circle requires a certain amount of arc length, and every time to create a piece to fit that you add a requirement for an equal amount of arc somewhere else. I suppose this could be resolved by some kind of fractal, but that's going to have an infinitely long perimeter.

Sharlin 5y ago
Yes, the pieces here are highly nontrivial in shape, as discussed in the article. But more well-behaved than in earlier results.
jessriedel 5y ago
Yes, as discussed in the article, the shapes are not piecewise smooth curves (i.e., not the sort of shapes you can construct by making a finite number of straight and smoothed cuts). Furthermore, as also mentioned, the areas of the shapes are non-measurable, so the perimeter is probably non-measurable too.
- Sniffnoy 5y ago
  The areas aren't non-measurable, that's part of the point. Doing it with non-measurable pieces was already known.
  - jessriedel 5y ago
    The Quanta article says that doing it non-constructively with non-measurable pieces was already known, and that later results did it partially and then fully constructively (without leaving a measure-zero set). But it does not say in the Quanta article whether or not the pieces in these later results were individually measurable. (The union of non-measurable sets can be measurable.) Are you saying that you know they are?
    - Sniffnoy 5y ago
      Well, when I wrote that I was just inferring it from the article, which while it doesn't explicitly say that they are, is written in a way so as to clearly suggest that they are. But you're right that it doesn't eplicitly say that, now that I look again. But if you click on the link to the actual paper, it says right in the abstract that they're Borel, so yes, they are indeed measurable.
      - jessriedel 5y ago
        Thanks!

Sharlin 5y ago

Yes, the pieces here are highly nontrivial in shape, as discussed in the article. But more well-behaved than in earlier results.

jessriedel 5y ago

Yes, as discussed in the article, the shapes are not piecewise smooth curves (i.e., not the sort of shapes you can construct by making a finite number of straight and smoothed cuts). Furthermore, as also mentioned, the areas of the shapes are non-measurable, so the perimeter is probably non-measurable too.

Sniffnoy 5y ago
The areas aren't non-measurable, that's part of the point. Doing it with non-measurable pieces was already known.
- jessriedel 5y ago
  The Quanta article says that doing it non-constructively with non-measurable pieces was already known, and that later results did it partially and then fully constructively (without leaving a measure-zero set). But it does not say in the Quanta article whether or not the pieces in these later results were individually measurable. (The union of non-measurable sets can be measurable.) Are you saying that you know they are?
  - Sniffnoy 5y ago
    Well, when I wrote that I was just inferring it from the article, which while it doesn't explicitly say that they are, is written in a way so as to clearly suggest that they are. But you're right that it doesn't eplicitly say that, now that I look again. But if you click on the link to the actual paper, it says right in the abstract that they're Borel, so yes, they are indeed measurable.
    - jessriedel 5y ago
      Thanks!