Timeline for Generating random curves with fixed length and endpoint distance
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Aug 31 at 10:57 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Apr 29, 2017 at 17:21 | comment | added | ARi | 'underdetermined' are you talking about the number of points, you can choose more if required. A nice transformation with two fixed points can be easyly arrived at.. | |
| Apr 29, 2017 at 16:30 | comment | added | Manfred Weis | @ARi the question is for arbitrary types of curves, fractals included; I can't judge, if your proposal would even work in the following simplified setting: given $(0.0)$ and $(\alpha,0)$ as end points as well as $m\lt n-2$ additional points in predefined sequence (so that the interpolation problem is underdetermined) and the task is to find an interpolating polynomial of unit length; in that case it may be quite challenging to get the scaling right. | |
| Apr 29, 2017 at 13:41 | comment | added | ARi | Why not for any chosen degree of the polynomial curve, n, choose n points randomly ( wihin a desired approximation d, within a region of choice-cant be larger than a unit disc), use Lagrangian interpolation to get $C$ which is translated, rotated and scaled so that its end points become (0,0) and ($\alpha$,0). Subject the resulting curve to a non length preserving transform with the abve two as fixed points and a scalar $\lambda$ which is which is varied by a chosen value to get a unit length. As the value of n and d is increased, any continuous curve can be approximated. | |
| Apr 29, 2017 at 13:01 | history | edited | Manfred Weis | CC BY-SA 3.0 |
described an algorithm that generates such curves in full generality
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| Apr 26, 2017 at 19:08 | comment | added | user44143 | @StevenStadnicki, I found it clear enough. We can represent the curves of length 1 by continuous functions from [0,1] to R^n, parameterized by arc length, and we can say that two such curves approximate each other to within epsilon iff the functions f,g satisfy |f(x)-g(x)|<epsilon for all x in [0,1]. That is consistent with everything I've seen from the OP. | |
| Apr 26, 2017 at 18:29 | comment | added | Steven Stadnicki | I think part of the point is that 'curve' can mean a half-dozen things to a half-dozen different people, and you haven't really specified what set you're interested in (though it sounds like you're interested in the broadest possible framing of rectifiable curves). This also plays to the phrase 'approximations of', because arc length is discontinuous with respect to e.g. positional approximation, so you can make the case that 'near' any continuous curve of length $\ell\lt 1$ that goes between your endpoints there are infinitely many curves of length 1 that do the same. | |
| Apr 26, 2017 at 15:01 | answer | added | Jean Duchon | timeline score: 2 | |
| Apr 26, 2017 at 13:09 | answer | added | user44143 | timeline score: 6 | |
| Apr 26, 2017 at 11:51 | answer | added | Joseph O'Rourke | timeline score: 7 | |
| Apr 26, 2017 at 6:38 | comment | added | Manfred Weis | @IzaakMeckler how would you ensure the fixed distance between the endpoints of the curve, when picking points at random and interpolate by "something"; that only works in a trial and error manner. The set of curves with fixed length and endpoint-distance is "static"; therefore asking for its distribution misses something; I agree however that asking for the distribution of the "points" generated by the sampling process, would make sense - that should be taken care of by adjusting parameters during the execution of the algorithm. | |
| Apr 26, 2017 at 5:34 | comment | added | Izaak Meckler | What space of curves are you referring to and with what distribution? It is easy to sample from "reasonable" distributions of curves with two fixed end points. For example, one can pick N points at random in the plane and then smoothly interpolate between them using cubic Hermite splines or something. It really depends on what you need these curves for. | |
| Apr 26, 2017 at 5:05 | history | edited | Manfred Weis | CC BY-SA 3.0 |
clarified questions from the comments
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| Apr 25, 2017 at 17:30 | comment | added | user44143 | Why not let the midpoint of the curve be a random point within (curve length/2) of both the starting point and the end point, and then iterate? | |
| Apr 25, 2017 at 17:08 | comment | added | user21349 | What set of random curves do you have in mind (differentiable, ...?), and what kind of probability measure? | |
| Apr 25, 2017 at 17:08 | comment | added | Henry.L | Do you require any smoothness? Because otherwise you can take a Brownian bridge as pointed out by Bjorn's answer. | |
| Apr 25, 2017 at 17:00 | answer | added | Bjørn Kjos-Hanssen | timeline score: 4 | |
| S Apr 25, 2017 at 10:32 | history | suggested | Jean Duchon | CC BY-SA 3.0 |
small improvement
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| Apr 25, 2017 at 10:26 | review | Suggested edits | |||
| S Apr 25, 2017 at 10:32 | |||||
| Apr 25, 2017 at 4:54 | history | asked | Manfred Weis | CC BY-SA 3.0 |