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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    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?










    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?










      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






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      asked yesterday









      N.T.C

      38917




      38917






















          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

          votes








          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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