Skip to main content

plotting - Where is the other half of my fourth degree Bézier curve?


Bug introduced in 10.0.0 and fixed in 10.2




When I enter the following command, half of the fourth-degree spline is missing from the graph (i.e. it doesn't touch the last point). Is this a bug in Mathematica or am I fundamentally misunderstanding something about Bézier curves?


Manipulate[

Graphics[{BezierCurve[pts, SplineDegree -> 4], Dashed, Green,
Line[pts]}, PlotRange -> {{-.5, 1.5}, {-.5, 1.5}},
Frame -> True], {{pts, {{0, 0}, {.5, 0}, {.5, .5}, {1, .5}, {1,
1}}}, Locator, LocatorAutoCreate -> True}]

Here's a screenshot of the output: missing part of fourth-degree Bézier curve


What makes me suspicious is that the same behavior shows up in Mathematica's own help file on BezierCurve.


screenshot of mathematica's help files on BezierCurve



Answer



Since there is really something wrong with the BezierCurve, I made this work-around:



Clear[bezierCurve];

bezierCurve[pts_] :=
First@ParametricPlot[
BezierFunction[pts, SplineDegree -> Length[pts] - 1][t], {t, 0, 1}]

Manipulate[
Graphics[{bezierCurve[pts], Dashed, Green, Line[pts]},
PlotRange -> {{-.5, 1.5}, {-.5, 1.5}},
Frame -> True], {{pts, {{0, 0}, {.5, 0}, {.5, .5}, {1, .5}, {1,

1}}}, Locator, LocatorAutoCreate -> True}]

Since the spline degree is always one less than the length of the point list, I didn't adhere to the built-in syntax where the degree is specified separately through an option. I just let the function bezierCurve compute the appropriate degree automatically, to reduce the potential for error.


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...