Timeline for Estimating the volume of a union of balls
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 30, 2014 at 20:10 | comment | added | user24451 | @tmyklebu You raised a good point. I will check again my numerics, because it gives me a non-zero volume. But, you are right that I'm working with the intersection of a bunch of balls with the cube. So the balls are not constrained to fit inside the cube. | |
| Aug 30, 2014 at 10:27 | comment | added | tmyklebu | @NicolasEssis-Breton: OK, that's really easy. The volume is, to a few hundred decimal places, zero. The biggest ball that fits inside the unit cube in 500 dimensions has volume $(\pi/4)^{250} / 250! \approxeq 2 \cdot 10^{-519}$, and you don't have many of them. Are you working with the intersection of a bunch of balls with the cube instead? | |
| Aug 29, 2014 at 18:41 | comment | added | user24451 | @tmyklebu It's a 500 dimensional space, with 2,000 balls. ($d=500$, $n=2000$) | |
| Aug 29, 2014 at 18:16 | comment | added | tmyklebu | What dimension are we talking about? How many balls? Naive Monte Carlo will converge, but it won't converge very fast; you get an error of about $1/\sqrt{n}$ in your ratio after $n$ samples. | |
| Aug 27, 2014 at 19:25 | vote | accept | CommunityBot | ||
| Aug 27, 2014 at 15:06 | comment | added | usul | Adding to the question, is random sampling considered better or worse than counting points on a discrete grid? | |
| Aug 27, 2014 at 7:22 | answer | added | Bjørn Kjos-Hanssen | timeline score: 1 | |
| Aug 27, 2014 at 5:41 | review | Close votes | |||
| Aug 27, 2014 at 10:17 | |||||
| Aug 27, 2014 at 3:13 | history | asked | user24451 | CC BY-SA 3.0 |