How to determine the longest edge in a graph?












9














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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share|improve this question



























    9














    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?






    list-manipulation graphics







    share|improve this question

























      9












      9








      9


      1





      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?






      list-manipulation graphics







      share|improve this question













      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?






      list-manipulation graphics




      list-manipulation graphics






      share|improve this question













      share|improve this question











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      asked yesterday









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          1 Answer
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          9














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



          ClearAll[nonCollinearHull]
          
nonCollinearHull = DeleteCases[#, Alternatives @@ 
          
  (SequenceCases[PadRight[#, 1 + Length@#, "Periodic"],
          
    {a_, Longest[b__], c_} /; (And @@ (RegionMember[ConvexHullMesh[{a, c}]] /@ {b})) :> b, 
          
      Overlaps -> True])] &;


          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



          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






          share|improve this answer























          • Why ConvexHullMesh and not just Line?
            – swish
            3 hours ago










          • It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
            – swish
            3 hours ago











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          9














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



          ClearAll[nonCollinearHull]
          
nonCollinearHull = DeleteCases[#, Alternatives @@ 
          
  (SequenceCases[PadRight[#, 1 + Length@#, "Periodic"],
          
    {a_, Longest[b__], c_} /; (And @@ (RegionMember[ConvexHullMesh[{a, c}]] /@ {b})) :> b, 
          
      Overlaps -> True])] &;


          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



          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






          share|improve this answer























          • Why ConvexHullMesh and not just Line?
            – swish
            3 hours ago










          • It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
            – swish
            3 hours ago
















          9














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



          ClearAll[nonCollinearHull]
          
nonCollinearHull = DeleteCases[#, Alternatives @@ 
          
  (SequenceCases[PadRight[#, 1 + Length@#, "Periodic"],
          
    {a_, Longest[b__], c_} /; (And @@ (RegionMember[ConvexHullMesh[{a, c}]] /@ {b})) :> b, 
          
      Overlaps -> True])] &;


          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



          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






          share|improve this answer























          • Why ConvexHullMesh and not just Line?
            – swish
            3 hours ago










          • It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
            – swish
            3 hours ago














          9












          9








          9






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



          ClearAll[nonCollinearHull]
          
nonCollinearHull = DeleteCases[#, Alternatives @@ 
          
  (SequenceCases[PadRight[#, 1 + Length@#, "Periodic"],
          
    {a_, Longest[b__], c_} /; (And @@ (RegionMember[ConvexHullMesh[{a, c}]] /@ {b})) :> b, 
          
      Overlaps -> True])] &;


          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



          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






          share|improve this answer














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



          ClearAll[nonCollinearHull]
          
nonCollinearHull = DeleteCases[#, Alternatives @@ 
          
  (SequenceCases[PadRight[#, 1 + Length@#, "Periodic"],
          
    {a_, Longest[b__], c_} /; (And @@ (RegionMember[ConvexHullMesh[{a, c}]] /@ {b})) :> b, 
          
      Overlaps -> True])] &;


          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



          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







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 23 hours ago









          kglr

          177k9198406




          177k9198406












          • Why ConvexHullMesh and not just Line?
            – swish
            3 hours ago










          • It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
            – swish
            3 hours ago


















          • Why ConvexHullMesh and not just Line?
            – swish
            3 hours ago










          • It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
            – swish
            3 hours ago
















          Why ConvexHullMesh and not just Line?
          – swish
          3 hours ago




          Why ConvexHullMesh and not just Line?
          – swish
          3 hours ago












          It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
          – swish
          3 hours ago




          It breaks if a coordinate list starts in the middle of the longest edge. Try RotateLeft[coord, 2] for the original example.
          – swish
          3 hours ago


















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