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Re: ListCurvePathPlot
*To*: mathgroup at smc.vnet.net
*Subject*: [mg95942] Re: ListCurvePathPlot
*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
*Date*: Fri, 30 Jan 2009 05:45:17 -0500 (EST)
*Organization*: Uni Leipzig
*References*: <gls253$hq3$1@smc.vnet.net>
*Reply-to*: kuska at informatik.uni-leipzig.de
Hi,
and more over, it does it wrong because I try your example with
data = Table[
RandomReal[{.8, 1.2}] {Cos[t], Sin[t]}, {t, 0, 2 \[Pi] - 2 \[Pi]/6,
2 \[Pi]/32}];
got the data
data-{{0.899074, 0}, {0.931267, 0.185241}, {1.10439, 0.457454}, {0.955484,
0.638434}, {0.795648, 0.795648}, {0.547609, 0.819554}, {0.318697,
0.769402}, {0.232974, 1.17124}, {0, 1.18407}, {-0.207593,
1.04364}, {-0.459194, 1.10859}, {-0.639312, 0.956798}, {-0.664321,
0.664321}, {-0.953214, 0.636917}, {-0.966874, 0.400492}, {-0.893422,
0.177713}, {-1.02206,
0}, {-1.09618, -0.218044}, {-0.995037, -0.412158}, {-0.775199, \
-0.517972}, {-0.722555, -0.722555}, {-0.643561, -0.963157}, \
{-0.373253, -0.901112}, {-0.232702, -1.16987}, {0, -0.93512}, \
{0.189465, -0.952506}, {0.338656, -0.817588}};
and the result is
ListCurvePathPlot[data, InterpolationOrder -> 3, Axes -> False,
InterpolationOrder -> 16]
not curved, not interpolated and broken into *two* pieces.
May be you don't understand the function because it is buggy and useless ...
Regards
Jens
David Park wrote:
> I don't understand the new ListCurvePathPlot, which the Help page says:
> "attempts to reconstruct smooth curves defined by the specified set of
> points."
>
>
>
> This plot routine also has the option: InterpolationOrder. And the word
> "Curve" appears not only in the name but repeatedly in the descriptions.
> But look at the following example:
>
>
>
> data = Table[
>
> RandomReal[{.8, 1.2}] {Cos[t], Sin[t]}, {t, 0, 2 \[Pi] - 2 \[Pi]/6,
>
> 2 \[Pi]/6}];
>
> ListCurvePathPlot[data,
>
> InterpolationOrder -> 3,
>
> Axes -> False]
>
>
>
> I don't see anything "smooth" or "curvy" about the results! So it seems
> that these terms are a misdirection in understanding the use of the routine.
> It appears that what the routine actually does is reorder the points to give
> some simpler line (not curve) than the original set of points. But what is
> the criterion for this? Is this some well know computational geometry
> algorithm? Was InterpolationOrder included as an option by mistake? Did the
> implementation change from the original intention? What is the purpose of
> the routine? What is the relation of this and the spline functions?
>
>
>
> David Park
>
> djmpark at comcast.net
>
> <http://home.comcast.net/~djmpark> http://home.comcast.net/~djmpark/
>
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