Timeline for What is the minimum-curvature curve interpolating a given set of points in the plane?
Current License: CC BY-SA 4.0
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| Feb 1, 2023 at 13:56 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| S Dec 20, 2022 at 18:45 | history | bounty ended | Penelope Benenati | ||
| S Dec 20, 2022 at 18:45 | history | notice removed | Penelope Benenati | ||
| Dec 20, 2022 at 18:45 | vote | accept | Penelope Benenati | ||
| Dec 20, 2022 at 13:20 | vote | accept | Penelope Benenati | ||
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| Dec 19, 2022 at 11:36 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 19, 2022 at 4:01 | comment | added | user44143 | @GabeK, if you are asking: “Is there a simple proof that the optimal curvature for three points can always be achieved by a circle?”, then Saul RM’s new answer shows that the optimal curvature can be achieved by at most three circle-arcs and three line segments; from there showing that it can be achieved by a single circle should be straightforward. | |
| Dec 19, 2022 at 1:12 | answer | added | Saúl RM | timeline score: 4 | |
| Dec 18, 2022 at 15:18 | comment | added | Penelope Benenati | @SaúlRM I think it is fine, even if I am pretty sure there should be a method to transform such optimal curve in such a way that in the neighborhood of each point in $x\in X$ the curvature is defined and belongs to $(\kappa',\kappa'')$, where $\kappa'$ and $\kappa''$ are the curvatures of the two arcs connected at point $x$ | |
| Dec 18, 2022 at 13:04 | comment | added | user44143 | @YaakovBaruch, the ease of computing has two things going for it: 1) The optimal union of lines and circle-arcs, which may be optimal overall, can be computed exactly via Tarski-Seidenberg or the like. As a corollary, if the original points are algebraic, the optimal curvature will also be algebraic. 2) The algorithm in my answer gives a starting point, from which one can (repeatedly) identify the arc with maximal curvature and evaluate possible replacements for it. | |
| Dec 18, 2022 at 12:33 | comment | added | Saúl RM | @PenelopeBenenati the thing is, I think the curve with minimal curvature, if it exists, is given by a sequence of arcs of circumference joining the points of $X$. However in the points of $X$, where the arcs of circumference meet, the curvature may not be defined, because the different arcs may have difference curvature. But if that's not a problem I'll try to write it in detail later | |
| Dec 18, 2022 at 12:25 | comment | added | Yaakov Baruch | @MattF. I'm pretty sure that for the points $(0,\pm 1), (\pm a,0)$ for $a>1$ the optimal curve is a pill shape: 2 semicircles or radius $1$ joined by segments of length $2a-2$ (so definitely not an ellipse). If your guess is right in general, the details of the the arcs are going to be determined globally, not locally, so may not be so easy to compute. | |
| Dec 18, 2022 at 12:19 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 18, 2022 at 10:07 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 18, 2022 at 10:06 | comment | added | Penelope Benenati | @YaakovBaruch, initially I formulated the problem with the assumption you mentioned above in your last comment. I am adding it back to the problem text. | |
| Dec 18, 2022 at 10:04 | comment | added | Penelope Benenati | @SaúlRM the curvature is not restricted to belong to a finite portion of the space. Anyway, I think that this does not change the problem. | |
| Dec 18, 2022 at 4:40 | comment | added | Yaakov Baruch | To be clear: the points of $X$ are assumed to be all on the boundary of the convex hull of $X$? | |
| Dec 18, 2022 at 2:30 | comment | added | Saúl RM | @PenelopeBenenati Do you need the curvature to be defined everywhere? Or could it not be defined in a finite set of points? (e.g. the points of $X$) | |
| Dec 16, 2022 at 18:56 | history | edited | user44143 | CC BY-SA 4.0 |
convexity is assumed; the question asks for the curvature
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| Dec 16, 2022 at 18:24 | comment | added | Gabe K | Is there a simple proof that the optimal curve through three points is a circle? It seems like it should be true but I don’t see a quick argument. I suppose that if the triangle is obtuse one can finding many minimizing curves, but it seems like there must be a circular arc through the vertices. | |
| Dec 16, 2022 at 17:38 | answer | added | user44143 | timeline score: 4 | |
| Dec 14, 2022 at 21:55 | comment | added | Penelope Benenati | @MattF. do you think that if each triplet of points in $X$ consists of non-collinear points, then the optimal curve would be just the union of circle arcs solely? | |
| Dec 14, 2022 at 21:46 | comment | added | user44143 | I think an optimal curve will be a union of line segments and circle arcs; if so, it should be easy to compute. | |
| Dec 14, 2022 at 21:38 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
Simplified the whole text
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| Dec 14, 2022 at 9:51 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| S Dec 13, 2022 at 11:12 | history | bounty started | Penelope Benenati | ||
| S Dec 13, 2022 at 11:12 | history | notice added | Penelope Benenati | Draw attention | |
| Dec 11, 2022 at 14:33 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 10, 2022 at 15:56 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 9, 2022 at 21:53 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
edited title
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| Dec 9, 2022 at 21:39 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
I added the constraint that the curve minimizer $\gamma$ is convex.
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| Dec 9, 2022 at 21:31 | answer | added | Saúl RM | timeline score: 9 | |
| Dec 9, 2022 at 21:25 | comment | added | Penelope Benenati | @SaúlRM The curvature can be "arbitrarily low" in some specific points, but how can the maximum curvature at point $\mathbf{p}$ over all points $\mathbf{p}\in\gamma$ be "arbitrarily low" ? | |
| Dec 9, 2022 at 21:19 | comment | added | Saúl RM | @PenelopeBenenati The problem is that, as I said above, there are simple closed curves of arbitrarily low curvature passing through any given finite set of points of ℝ2. So there is no curve minimizing the maximum curvature. I can draw in an answer what such a curve would look like if you want. | |
| Dec 9, 2022 at 21:07 | comment | added | Penelope Benenati | @MattF. as I just wrote in the comment for SaúlRM, informally speaking, the idea is very simple: I want to measure how curve must necessarily be (i.e., I am interested in the maximum curvature over all points of the curve itself) a curve passing through a given sets of points $X$ on a plane $P$, when $X$ belong to a convex polygon. | |
| Dec 9, 2022 at 16:29 | comment | added | Jason Starr | What does "pairwise non-collinear" mean? Every pair of points is contained in a line. | |
| Dec 9, 2022 at 16:13 | comment | added | Saúl RM | I think it would be useful to give some extra conditions, e.g. the curves being contained in $S$. For example given a finite set of points in $\mathbb{R}^2$ it seems there are curves with arbitrarily low curvature passing through all the points | |
| Dec 9, 2022 at 16:02 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 9, 2022 at 13:47 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 9, 2022 at 13:39 | history | edited | Penelope Benenati | CC BY-SA 4.0 |
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| Dec 9, 2022 at 13:29 | history | asked | Penelope Benenati | CC BY-SA 4.0 |