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graphics - Undocumented form for FilledCurve[]


New in Mathematica 8 is FilledCurve (and its cousin JoinedCurve). The docs state that this function can take a list of segments or components, each in the form of a Line, a BezierCurve, or a BSplineCurve. But the examples show alteration of a glyph, with a resulting JoinedCurve that has a different pattern. Here is a simplified example:


ImportString[ExportString[
    Style["\[FilledCircle]", Bold, FontFamily -> "Helvetica", FontSize -> 12],
    "PDF"],"TextMode" -> "Outlines"][[1, 1, 2, 1, 1]]

gives


FilledCurve[{{{1, 4, 3}, {1, 3, 3}, {1, 3, 3}, {1, 3, 3}}}, 
  {{{10.9614, 7.40213}, {10.9614, 10.2686}, {8.51663, 12.3137}, 
   {5.79394, 12.3137}, {3.05663, 12.3137}, {0.641063, 10.2686}, 

   {0.641063, 7.40213}, {0.641063, 4.53319}, {3.05663, 2.48813}, 
   {5.79394, 2.48813}, {8.53125, 2.48813}, {10.9614, 4.51856}, 
   {10.9614, 7.40213}}}]

The second argument is a list of points representing the glyph. The first list is some representation of smoothing the boundary given by the second list. For example, displaying the points alone as connected line segments


Graphics@Line[ImportString[ExportString[
 Style["\[FilledCircle]", Bold, FontFamily -> "Helvetica", 
  FontSize -> 12], "PDF"], "TextMode" -> "Outlines"][[1, 1, 2, 1, 1, 2, 1]]]

gives



Connected line segments for FilledCurve[] glyph


Compare that with the JoinedCurve version:


Graphics[ImportString[
  ExportString[Style["\[FilledCircle]", Bold, FontFamily -> "Helvetica", 
    FontSize -> 12], "PDF"], "TextMode" -> "Outlines"][[1, 1, 2, 1, 1]]
  /. FilledCurve[a__] :> JoinedCurve[a]]

JoinedCurve version of the gylph



  1. What exactly do the triples in the first list represent?


  2. How can I reconstruct the smooth circle using other Graphics primitives, such as BSplineCurve?


My motivation is to be able to extract the curves from 2D glyphs and plot them in Graphics3D[] (e.g., with the replacement {x_Real,y_Real}:>{Cos[theta]x,Sin[theta]x,y}), since FilledCurve[] only works with Graphics[].



Answer



The first element in the triples seems to indicate the type of curve used for the segment where 0 indicates a Line, 1 a BezierCurve, and 3 a BSplineCurve. I haven't figured out yet what 2 does.


Edit: When the first element of the triple is 2, the segment will be a BezierCurve similar to option 1 except that with option 2, an extra control point is added to the list to make sure that the current segment is tangential to the previous segment.


The second digit indicates how many points to use for the segment, and the last digit the SplineDegree. To convert the FilledCurve to a list of Graphics primitives, you could therefore do something like


conversion[curve_] :=
   Module[{ff},


       ff[i_, pts_, deg_] :=
           Switch[i,
               0, Line[Rest[pts]],
               1, BezierCurve[Rest[pts], SplineDegree -> deg],
               2, BezierCurve[
                   Join[{pts[[2]], 2 pts[[2]] - pts[[1]]}, Drop[pts, 2]], 
                   SplineDegree -> deg],
               3, BSplineCurve[Rest[pts], SplineDegree -> deg]
               ];


       Function[{segments, pts},
               MapThread[ff,
                   {
                       segments[[All, 1]],
                       pts[[Range @@ (#1 - {1, 0})]] & /@
                           Partition[Accumulate[segments[[All, 2]]], 2, 1, {-1, -1}, 1],
                       segments[[All, 3]]
                       }
                   ]
               ] @@@ Transpose[List @@ curve]

       ]

Then for the example in the original post,


curve = FilledCurve[{{{1, 4, 3}, {1, 3, 3}, {1, 3, 3}, {1, 3, 3}}}, 
  {{{10.9614, 7.40213}, {10.9614, 10.2686}, {8.51663, 12.3137}, 
   {5.79394, 12.3137}, {3.05663, 12.3137}, {0.641063, 10.2686}, 
   {0.641063, 7.40213}, {0.641063, 4.53319}, {3.05663, 2.48813}, 
   {5.79394, 2.48813}, {8.53125, 2.48813}, {10.9614, 4.51856}, 
   {10.9614, 7.40213}}}];
curve2 = conversion[curve]


gives


{BezierCurve[{{10.9614, 7.40213}, {10.9614, 10.2686}, {8.51663, 
    12.3137}, {5.79394, 12.3137}}, SplineDegree -> 3], 
 BezierCurve[{{5.79394, 12.3137}, {3.05663, 12.3137}, {0.641063, 
    10.2686}, {0.641063, 7.40213}}, SplineDegree -> 3], 
 BezierCurve[{{0.641063, 7.40213}, {0.641063, 4.53319}, {3.05663, 
    2.48813}, {5.79394, 2.48813}}, SplineDegree -> 3], 
 BezierCurve[{{5.79394, 2.48813}, {8.53125, 2.48813}, {10.9614, 
    4.51856}, {10.9614, 7.40213}}, SplineDegree -> 3]}


and Graphics[curve2] produces


converted FilledCurve


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