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I have a list of 2D points such as in the image.

coord = {{0, 0}, {10, 0}, {20, 0}, {30, 0}, {25, 10}, {0, 10}, {0, 
5}};

enter image description here

I would like to determine the longest "edge" length in a way that if 2 segments are in same line, they would be considered as part of 1 edge. For example, segments 1-2, 2-3, 3-4 are continuous and in the same line, so we consider it as 1 edge connecting Vertex 1 and Vertex 4. In this example, the longest edge length would be the distance from vertex 1 to vertex 4. How can I determine the longest edge length in Mathematica?

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1 Answer 1

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Update: The function in the original answer does not work for arbitrary polygons. The following seems to work

ClearAll[nonCollinearHull]
nonCollinearHull = Module[{coords = #, 
 angles = ArcTan @@@ (Subtract @@@ Partition[#, 2, 1 , {1, 1}]),
 rotation, lengths},
rotation = LengthWhile[Reverse[angles], # == angles[[1]] &];
lengths = Length /@ Split[RotateRight[angles, rotation]];
TakeList[RotateRight[coords, rotation], lengths][[All, 1]]] &;

Examples:

coord = {{0, 0}, {10, 0}, {20, 0}, {30, 0}, {25, 10}, {0, 10}, {0, 5}};
lines = Line /@ Partition[nonCollinearHull[coord], 2, 1];
longest = Last@SortBy[lines, N@ArcLength[#] &];
Graphics[{EdgeForm[Gray], FaceForm[], Polygon@coord, Blue, 
  PointSize[Large], Point@coord, Opacity[.5, Green], 
  AbsolutePointSize[15], Point[nonCollinearHull[coord]], 
  Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest}]

enter image description here

Using

SeedRandom[123]
coord2 = DeleteDuplicates[#[[FindShortestTour[#][[2]]]] & @
   DeleteDuplicates@RandomInteger[10, {50, 2}];

we get

enter image description here

And with

SeedRandom[123]
coord3 = DeleteDuplicates[#[[FindShortestTour[#][[2]]]] &@ 
   DeleteDuplicates@RandomInteger[20, {200, 2}]];

enter image description here

Alternatively, you can use MaximalBy to define longest:

SeedRandom[777777]
coord = MapIndexed[{#2[[1]], #} &, Accumulate[RandomInteger[{-2, 2}, 50]]];
lines = Line /@ Partition[nonCollinearHull[coord], 2, 1];
longest = MaximalBy[lines, N@ArcLength[#] &];
Graphics[{EdgeForm[Gray], FaceForm[], Line@coord, Blue, 
  PointSize[Large], Point@coord, Opacity[.5, Green], 
  AbsolutePointSize[15], Point[nonCollinearHull[coord]], 
  Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest}]

enter image description here

Original answer:

Using the function noncollinearF from this answer:

ClearAll[noncollinearF]
noncollinearF[verts_] := Function[{k}, Nor @@ (RegionMember[ConvexHullMesh[#], k] & /@ 
  Subsets[Complement[verts, {k}], {2}])]

lines = Line /@ Partition[Pick[#, noncollinearF[#] /@ #], 2, 1, {1, 1}]& @ coord;
longest = Last@SortBy[lines, N@ArcLength[#] &]; 
Graphics[{EdgeForm[Gray], FaceForm[], Polygon@coord, 
  Blue, PointSize[Large], Point@coord, 
  Thickness[.03], CapForm["Round"], Opacity[.5], Red, longest}]

enter image description here

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3
  • $\begingroup$ Why ConvexHullMesh and not just Line? $\endgroup$
    – swish
    Jan 4, 2019 at 7:18
  • $\begingroup$ It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example. $\endgroup$
    – swish
    Jan 4, 2019 at 7:45
  • $\begingroup$ @swish, thank you. Updated with an alternative that does not have the issue. $\endgroup$
    – kglr
    Jan 4, 2019 at 19:47

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