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graphics - Character edge finding


The following line of code finds the edge of a character:


pic = Binarize[GradientFilter[Rasterize[Style["\[Euro]", FontFamily -> "Times"], 
                              ImageSize -> 200] // Image, 1]]

Mathematica graphics


The coordinates of the edges can be found as follows:


pdata = Position[ImageData[pic], 1];

Test:



Graphics[Point[pdata]]

Mathematica graphics


However, the points are not sorted in an order usable by Line or Polygon:


Graphics[Polygon[pdata]]

Mathematica graphics


This brings me to my question:



  • What would be an efficient method to sort the coordinates so that it would plot properly with Line or Polygon?



Additionally,



  • How to thin and smooth the set of points?

  • How to deal with characters with holes in them, like the ones below?


Mathematica graphics or Mathematica graphics



Answer



I think there is a neat solution. We have curios function ListCurvePathPlot:


pic = Thinning@Binarize[GradientFilter[Rasterize[Style["\[Euro]", 

FontFamily -> "Times"], ImageSize -> 200] // Image, 1]];

pdata = Position[ImageData[pic], 1];

lcp = ListCurvePathPlot[pdata]

enter image description here


Now this is of course Graphics containing Line with set of points


lcp[[1, 1, 3, 2]]


enter image description here


So of course we can do something like


Graphics3D[Table[{Orange, Opacity[.5],Polygon[(#~Join~{10 n})&
/@ lcp[[1, 1, 3, 2, 1]]]}, {n, 10}], Boxed -> False]

enter image description here


I think it works nicely with "8" and Polygon:


pic = Thinning@Binarize[GradientFilter[
Rasterize[Style["8", FontFamily -> "Times"], ImageSize -> 500] //Image, 1]]; 
pdata = Position[ImageData[pic], 1]; lcp = ListCurvePathPlot[pdata]


enter image description here


And you can do polygons 1-by-1 extraction:


Graphics3D[{{Orange, Thick, Polygon[(#~Join~{0}) & /@ lcp[[1, 1, 3, 2, 1]]]},
  {Red, Thick, Polygon[(#~Join~{1}) & /@ lcp[[1, 1, 3, 3, 1]]]},
  {Blue, Thick, Polygon[(#~Join~{200}) & /@ lcp[[1, 1, 3, 4, 1]]]}}]

enter image description here


=> To smooth the curve set ImageSize -> "larger number" in your pic = code.


=> To thin the curve to 1 pixel wide use Thinning:



 Row@{Thinning[#], Identity[#]} &@Binarize[GradientFilter[
 Rasterize[Style["\[Euro]", FontFamily -> "Times"], 
 ImageSize -> 200] // Image, 1]]

enter image description here


You can do curve extraction more efficiently with Mathematica. A simple example would be


text = First[
   First[ImportString[
     ExportString[
      Style["\[Euro] 9 M-8 ", Italic, FontSize -> 24, 

       FontFamily -> "Times"], "PDF"], "PDF", 
     "TextMode" -> "Outlines"]]];

Graphics[{EdgeForm[Black], FaceForm[], text}]

enter image description here


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