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16

A function to subdivide the triangles based on their gray level: h[{v1_, v2_, v3_}] := With[{a = EuclideanDistance[v1, v2], b = EuclideanDistance[v1, v3], c = EuclideanDistance[v2, v3]}, With[{s = (a + b + c)/2}, (2 Sqrt[s (s - a) (s - b) (s - c)])/c ]] shadeTri[tri_, col_, f1_: 1, fc_: 1] := If[col > .8, tri, With[{v = tri[[1, ...


20

(These solutions loosely adhere to OP's request in the question. They were mostly made because of similar art/solutions pointed in the comments.) "Delaunay raster" like Here is a solution related to "Delaunay raster" discussed in the question comments. It is based on the Mathematica documentation page "Create a Mesh Region from Image Data". I changed ...


27

Here's something to get the ball rolling: img = ColorConvert[Import["http://i.stack.imgur.com/FaE06.jpg"], "Grayscale"]; DelaunayMesh[ ImageCorners[img, 1, 70*^-6], MeshCellStyle -> {{2, All} -> White, {1, All} -> GrayLevel[0.5], {0, All} -> Black} ] I'm looking forward to better results :-)


0

LogPolar[x_, y_] := {Log[Sqrt[x^2 + y^2]], ArcTan[x, y]} ImageTransformation[img, LogPolar[#[[1]], #[[2]]] &, DataRange -> {{-Pi, Pi}, {-Pi, Pi}}] Juts for clarification. I am allmost shure that the above doese the inverse Log Polar Transform. To actualy do the transform one would have to use ImageForwardTransform[]. Alternatively ...


1

Share a solution just applicable to 10.4 or later version.The code is very terse,but it seem to have some mysterious bug in it.And I have post it as a discuss.Firstly I make a function name of diskMake diskMake[region_, n_] := Module[{p, rad, dist, temRegion = region}, SeedRandom[1]; Reap[Do[p = RandomPoint[temRegion]; rad = If[(dist = Abs[...



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