Re: Re: Complex Plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg88315] Re: [mg88288] Re: Complex Plot*From*: "Jean-Marc Gulliet" <jeanmarc.gulliet at gmail.com>*Date*: Thu, 1 May 2008 03:22:45 -0400 (EDT)*References*: <fv9a83$mjo$1@smc.vnet.net> <200804301103.HAA04939@smc.vnet.net>

On Wed, Apr 30, 2008 at 2:52 PM, W_Craig Carter <ccarter at mit.edu> wrote: > Hello Jean-Marc, > In 6.0, we can use ContourPlot for implicit equations, but the timing > seems much slower that what you are getting: > > Timing[cp1 = > ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23}];] > (*about 1 second my machine) > > cp1 > > Timing[cp2 = > > ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, > 23},PlotPoints-> 25];] > (*about 3 seconds my machine*) > > cp2 > > Timing[cp3 = > > ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, > 23},PlotPoints-> 50];] > (*about 15 seconds my machine*) > > Timing[cp3 = > > ContourPlot[Sin[x^2*y] == Log[x/y], {x, -12, 35}, {y, -8, 23}, > PlotPoints -> 75];] > (*about 36 seconds my machine*) > > The estimate is about 1/2 hour on my machine (if no swapping)... > > So, unless your machine is about 6 times faster than mine, the > implicit version is slower... > > However, PlotPoints->50, seems to be sufficient to see the lovely structure. Hi Graig, My machine is about 10% slower than yours, indeed. (I have just ran the test you provided.) The "magic" in my original code was to use a high number of points but to turn off the adaptive algorithm by setting the option MaxRecursion to zero. For instance, the fourth plot (75 points) took about 40 seconds with the default settings, while it took only 0.75 second with MaxRecursion->0. That explains the huge discrepancy in timing and why I could claim that it took only few minutes for Mathematica to draw the plot with a sampling of 500 points. (Note that the memory consumption is relatively slow too.) Timing[cp1 = ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, MaxRecursion -> 0, PlotPoints -> 25];] (*about 0.04 seconds my machine*) cp1 Timing[cp2 = ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, MaxRecursion -> 0, PlotPoints -> 25];] (*about 0.07 seconds my machine*) cp2 Timing[cp3 = ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, MaxRecursion -> 0, PlotPoints -> 50];] (*about 0.29 seconds my machine*) cp3 Timing[cp4 = ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, MaxRecursion -> 0, PlotPoints -> 75];] (*about 0.73 seconds my machine*) cp4 > > On Wed, Apr 30, 2008 at 7:03 AM, Jean-Marc Gulliet > <jeanmarc.gulliet at gmail.com> wrote: > > Chris Degnen wrote: > > > > > Can anyone tell me how I can re-express this > > > equation in terms of x for plotting? > > > > > > Sin[x^2*y]==Log[x/y] > > > > > > ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, > > ImageSize -> 500] > > > > ContourPlot[Sin[x^2*y] - Log[x/y], {x, -12, 35}, {y, -8, 23}, > > ImageSize -> 500, ColorFunction -> (White &), ContourStyle -> Black, > > MaxRecursion -> 0, PlotPoints -> 500] > > > > > Hope this helps, > > -- Jean-Marc > > > > > > > > -- > W. Craig Carter > Best regards. -- Jean-Marc