Talk:Kakeya set
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Besicovitch sets <-> needle turning
[edit]I can't understand how it should be able to turn a needle in the depicted Besicovitch sets ... or, if this is not possible, what would be the connection to the Kakeya problem. Could this be added? -- Paul Ebermann (talk) 16:29, 23 January 2008 (UTC)
- I think it would help to better separate the presentation of Besicovitch sets from that of Kakeya needle sets, and add greater information on the latter including the results of Pàl, Van Alphen, Bloom and Schoenberg, and Cunningham (cf. new external links). Perhaps, if I can find the time, I'll attempt a suitable edit. Gtgith (talk) 09:15, 6 June 2009 (UTC)
- Actually, there seems to be confusion between Besicovitch sets and Kakeya (needle) sets. They are defined as the same at the start of the article, but distinguished at the end of the Besicovitch sets section. Gtgith (talk) 09:36, 6 June 2009 (UTC)
- Article edited Gtgith (talk) 10:45, 8 June 2009 (UTC)
Re-Added LInk
[edit]I just reverted the removal of the link in this article. Yes, it's a personal website, however, it's the personal website of Terry Tao who is /was a published author dealing with mathematics. This author appears to be notable and has been published since the 90's. Appears to be okay in this sense. Feel free to revert if you disagree - I won't war. F.U.R hurts Wikipedia 15:15, 2 April 2008 (UTC)
- A published author dealing with mathematics is indeed an accurate description of a Fields medalist. XDDD Scineram (talk) 21:04, 30 December 2008 (UTC)
Misconception
[edit]Hello, Recently with the IP 69.156.205.158. I'm back today to synch the head with the data in the main page. Facts :
- Kekeya needle sets are a sub set of Besicovitch sets
- There are Besicovitch sets with size arbitrarily small
- There is no Kakeya needle set of zero measure
- This alone doesn't mean there is a Kekeya needle sets of size zero or arbitrarily small.
F. Cunningham, our last source, demonstrated how using techniques similar to the Besicovitch sets of infinitesimal size he could make everything outside of an area of arbitrarily small. As far as our sources goes, it's the smallest we can get.
I don't know how far this misconception that a Kekeya needle set could be of null size is going. For what I know, we were quoting F. Cunningham as our source for that claim. Obviously upon reading his paper, he don't make that claim. It might worth a section, but I won't go into OR territories alone on this topic. Of course, if anyone have a good secondary source about this or it contrary, it would be appreciated. Iluvalar (talk) 16:11, 5 November 2015 (UTC)
- The Cunningham result is for Kakeya needle sets that are further constrained to be simply connected and lie within the unit disk. The Alphen result, earlier in the section, already establishes the existence of arbitrarily-small Kakeya needle sets that are not so constrained. Joule36e5 (talk) 04:25, 15 December 2015 (UTC)
- Is that what the issue is? I thought it was the distinction between having measure zero and having measure that is nonzero but arbitrarily small. —David Eppstein (talk) 04:36, 15 December 2015 (UTC)
- The article (correctly) stated that a Besicovich set can have measure zero, and that a Kakeya needle set can have measure epsilon but not measure zero.
However, there was also a mistake where Cunningham's pi/108 constant was omitted. So after fixing the typo, Iluvalar deleted the statement, in the mistaken belief that it had been the misquote of Cunningham's result that had caused it to be added to the article in the first place. Joule36e5 (talk) 05:28, 15 December 2015 (UTC)- Are you 100% sure ? Because I'm highly skeptical that someone could make a significant rotation without any significant area. Could you expand the chapter to explain how that is possible if you can read that source ? Iluvalar (talk) 05:47, 2 January 2016 (UTC)
- Informally: a tiny rotation requires a tiny amount of area, and you need to do a large number of tiny rotations in order to get a full rotation. However, the tiny areas involved can be mostly overlapping, and hence even when we have a large number of them, the total area can still be tiny.
- Are you 100% sure ? Because I'm highly skeptical that someone could make a significant rotation without any significant area. Could you expand the chapter to explain how that is possible if you can read that source ? Iluvalar (talk) 05:47, 2 January 2016 (UTC)
- The article (correctly) stated that a Besicovich set can have measure zero, and that a Kakeya needle set can have measure epsilon but not measure zero.
- Is that what the issue is? I thought it was the distinction between having measure zero and having measure that is nonzero but arbitrarily small. —David Eppstein (talk) 04:36, 15 December 2015 (UTC)
- The Perron tree construction gives you a Besicovitch set of arbitrarily small area, containing a large number of (overlapping) triangles each of which allows you to do a tiny rotation. Those rotations cover all directions. The Pál joins allow you to move the needle from the final orientation of one triangle to a parallel position which is the initial orientation of the next one. All this is already in the article; what part are you skeptical about? Joule36e5 (talk) 05:03, 19 January 2016 (UTC)
- Oops, turns out that the "missing pi/108" was not a mistake after all; Cunningham's Theorem 1 really did establish that the area can be arbitrarily small, even for simply connected sets contained in the radius-1 disk (but no smaller). Theorem 2 deals with an entirely different constraint (star-shaped sets), and establishes pi/180 as a lower bound, not an upper, for that case. Joule36e5 (talk) 10:26, 19 September 2017 (UTC)
Additional unstated conditions?
[edit]Very simply, an annulus with inner radius r > 1 and outer radius (r + 1/r^2) has an arbitrarily small area (~2 pi / r) as r goes to infinity. It fits the needle and allows it to rotate 360 degrees just by going all the way around the annulus.
Does the statement of the problem have additional conditions not listed on the wiki page to exclude this uninteresting solution? — Preceding unsigned comment added by DMPalmer (talk • contribs) 15:49, 2 November 2018 (UTC)
- That doesn't look right to me. For an annulus of inner radius r to contain a unit segment, the outer radius must be at least sqrt(r^2 + 1/4). Then the area of the annulus is exactly pi/4, independent of r. Joule36e5 (talk) 21:09, 2 November 2018 (UTC)