Timeline for The distribution of the number of chord intersections
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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Oct 24, 2017 at 13:48 | vote | accept | Igor Rivin | ||
Oct 24, 2017 at 13:46 | vote | accept | Igor Rivin | ||
Oct 24, 2017 at 13:47 | |||||
Oct 24, 2017 at 9:38 | answer | added | Brendan McKay | timeline score: 2 | |
Oct 23, 2017 at 23:53 | answer | added | Henry | timeline score: 3 | |
Oct 23, 2017 at 1:59 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added experimental wisdom.
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Oct 22, 2017 at 21:21 | comment | added | Igor Rivin | @JamesMartin Yes, the body asked for the distribution, the title for the expectation, I was spreading confusion :) | |
Oct 22, 2017 at 20:25 | comment | added | James Martin | Ah, I see. At the time I made my comment, the title of the question was "The expected number of chord intersections". So it's somewhat natural that both usul and I thought that you wanted the expected number :) | |
Oct 22, 2017 at 18:33 | history | edited | Igor Rivin | CC BY-SA 3.0 |
clarified the statement
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Oct 22, 2017 at 18:32 | comment | added | Igor Rivin | @JamesMartin usul gives the expectation, I ask for the distribution... | |
Oct 22, 2017 at 18:20 | comment | added | James Martin | @IgorRivin Then you should make that clear in the question! There are lots of ways to choose "random chords". In particular, the model referred to in the post you quoted has chords produced by two uniform points in the disc, rather than two uniform points on the perimeter. (Possibly it was edited since you wrote your question?) In any case, for two uniform points on the circle, usul has a nice simple answer below. | |
Oct 22, 2017 at 15:57 | answer | added | usul | timeline score: 4 | |
Oct 22, 2017 at 15:52 | comment | added | Igor Rivin | @GerhardPaseman The chords come from $2n$ uniform points on the circle, so the lengths are what they are... | |
Oct 22, 2017 at 15:33 | comment | added | Gerhard Paseman | It really depends on how chord lengths are distributed. For any collection of n chords (with no length appearing more than twice, and none a diameter), you can get from 0 to n choose 2 intersections by rotating chords individually. However, if they are all small , 0 is more likely, and if they are all near the length of a diameter, n choose 2 is more likely. Gerhard "You Might Use Circular Reasoning" Paseman, 2017.10.22. | |
Oct 22, 2017 at 15:22 | history | asked | Igor Rivin | CC BY-SA 3.0 |