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/ >

**Follow-Ups**:**Re: Re: ListCurvePathPlot***From:*DrMajorBob <btreat1@austin.rr.com>