10.4.1 for Microsoft Windows (64-bit) (April 17, 2016)

Mathematica graphics

Crazy result.But the documentation have not any specification for this.And the BezierFunction and BSplineFunction have a same behavior in my PC.

And when I input like this

temp = RandomInteger[100, {3, 2}];
ParametricPlot[BezierFunction[temp][t], {t, 0, 1}]

We'll get the right answer.If anybody can confirm this.I'll add a bug for this post and give a support to wolfram.

  • $\begingroup$ happens also in V9. $\endgroup$ – kglr Apr 24 '16 at 5:27
  • $\begingroup$ You say BSplineFunction[] in the title, but BezierFunction[] in the body. Which one do you actually want? $\endgroup$ – J. M.'s torpor Apr 24 '16 at 5:38
  • $\begingroup$ try ParametricPlot[ Evaluate@BezierFunction[RandomInteger[100, {3, 2}]][t], {t, 0, 1}]? $\endgroup$ – kglr Apr 24 '16 at 5:41
  • $\begingroup$ ... or ParametricPlot[ BezierFunction[RandomInteger[100, {3, 2}]][t], {t, 0, 1}, Evaluated -> True] $\endgroup$ – kglr Apr 24 '16 at 5:42
  • $\begingroup$ @J.M. It's a typo.And thanks for your point out that. $\endgroup$ – yode Apr 24 '16 at 5:45

What is happening: For every value of t that is sampled a new triple of Randominteger pairs is generated and a new BezierFunction is constructed. So every t that is sampled is processed with a different function. This can be seen using Trace on a simpler version of the problem.

Trace[ParametricPlot[BezierFunction[RandomInteger[100, {3, 2}]][t], {t, 0, 1}, 
  PlotPoints ->3, MaxRecursion -> 0,  AspectRatio -> 1], BezierFunction| RandomInteger]

Mathematica graphics

Solution: Evaluateing the first argument of ParametricPlot or using the option Evaluated->True gives the expected output.

ParametricPlot[Evaluate@BezierFunction[RandomInteger[100, {3, 2}]][t], {t, 0, 1}, 
 AspectRatio -> 1]

Mathematica graphics

ParametricPlot[BezierFunction[RandomInteger[100, {3, 2}]][t], {t, 0, 1}, 
 Evaluated -> True, AspectRatio -> 1]

Mathematica graphics


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