Re: ListCurvePathPlot

*To*: mathgroup at smc.vnet.net*Subject*: [mg96111] Re: ListCurvePathPlot*From*: ucervan at gmail.com*Date*: Thu, 5 Feb 2009 04:40:06 -0500 (EST)*References*: <200901291058.FAA18253@smc.vnet.net> <200902010941.EAA22811@smc.vnet.net>

The problem with InterpolationOrder is going to be fixed in the next release. Thanks for pointing out this oversight and we appreciate reporting this issue to the Tech Support team. With respect to what is the use of this function, consider the problem of having an arbitrary sequence of 2d points that you know are part of a connected curve, and you want to reconstruct the curve. One solution is given by FindShortestTour[], try this: pts = Table[{x = RandomReal[{0, 6 Pi}]; r = x/(2 Pi); r Sin[x], r Cos[x]}, {200}]; FindShortestTour[pts]; // Timing On my system this takes around 7 sec. The shortest tour problem is an NP problem, if you have time try the previous example with 1000 pts. So, in general the problem of reconstructing a curve using a shortest tour approach will be very expensive for large number of samples. If you are in a situation where you can accept an approximation to the curve and allow for possible gaps, then you should consider FindCurvePath[], try: FindCurvePath[pts]; // Timing on my system this takes 0.032s for 200 pts and 0.36sec for 1000 pts. FindCurvePath[] is a fast curve reconstruction, which may or not be connected. In general if you have a sufficiently dense sample of "connected" points, then the solution found by FindCurvePath will be very close to the real shortest tour. If you give a random sequence of points, then the result may seem useless since the assumption for the FindCurvePath algorithm is that the data is part of a curve. This computational geometry problem is know as Curve Reconstruction and is used for hand writing recognition, extraction of features in image processing, raster to vector graphics operations, etc. Going back to ListCurvePathPlot, the implementation of InterpolationOrder is just: pts = Table[{x = RandomReal[{0, 2 Pi}]; x, Sin[x]}, {15}]; plot = ListCurvePathPlot[pts]; plot /. {Line[pp_] -> BSplineCurve[pp, SplineDegree -> 3]} We appreciate any comments and we encourage you to report any problems to tec. support. -Ulises Cervantes Wolfram Research Kernel Developer

**References**:**Re: Re: Re: ListCurvePathPlot***From:*peter <plindsay.0@gmail.com>