Draw circles and compute -

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
7
down vote

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The following code draws the polygons of VoronoiMesh[pts]:



SeedRandom[3]; 
pts = RandomReal[-1, 1, 25, 2];
mesh = VoronoiMesh[pts];
vertices = MeshCoordinates[mesh];
Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices]]]


The output is:



enter image description here



The black points are pts, the red ones the vertices of the Voronoi mesh.



My questions:



  • How can I draw circles around each point in pts with a given radius $r$, such as $0.18$?

  • How can I compute <sum of circle areas>-<area of overlaps of the circles>, i.e.
    $$left(sum_cinmathrmCirclesmathrmArea(c)right)-left(sum_oinmathrmOverlapsmathrmArea(o)right)$$



computational-geometry




share|improve this question


























    up vote
    7
    down vote

    favorite
    1












    The following code draws the polygons of VoronoiMesh[pts]:



    SeedRandom[3]; 
    pts = RandomReal[-1, 1, 25, 2];
    mesh = VoronoiMesh[pts];
    vertices = MeshCoordinates[mesh];
    Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices]]]


    The output is:



    enter image description here



    The black points are pts, the red ones the vertices of the Voronoi mesh.



    My questions:



    • How can I draw circles around each point in pts with a given radius $r$, such as $0.18$?

    • How can I compute <sum of circle areas>-<area of overlaps of the circles>, i.e.
      $$left(sum_cinmathrmCirclesmathrmArea(c)right)-left(sum_oinmathrmOverlapsmathrmArea(o)right)$$



    computational-geometry




    share|improve this question
























      up vote
      7
      down vote

      favorite
      1









      up vote
      7
      down vote

      favorite
      1






      1





      The following code draws the polygons of VoronoiMesh[pts]:



      SeedRandom[3]; 
      pts = RandomReal[-1, 1, 25, 2];
      mesh = VoronoiMesh[pts];
      vertices = MeshCoordinates[mesh];
      Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices]]]


      The output is:



      enter image description here



      The black points are pts, the red ones the vertices of the Voronoi mesh.



      My questions:



      • How can I draw circles around each point in pts with a given radius $r$, such as $0.18$?

      • How can I compute <sum of circle areas>-<area of overlaps of the circles>, i.e.
        $$left(sum_cinmathrmCirclesmathrmArea(c)right)-left(sum_oinmathrmOverlapsmathrmArea(o)right)$$



      computational-geometry




      share|improve this question














      The following code draws the polygons of VoronoiMesh[pts]:



      SeedRandom[3]; 
      pts = RandomReal[-1, 1, 25, 2];
      mesh = VoronoiMesh[pts];
      vertices = MeshCoordinates[mesh];
      Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices]]]


      The output is:



      enter image description here



      The black points are pts, the red ones the vertices of the Voronoi mesh.



      My questions:



      • How can I draw circles around each point in pts with a given radius $r$, such as $0.18$?

      • How can I compute <sum of circle areas>-<area of overlaps of the circles>, i.e.
        $$left(sum_cinmathrmCirclesmathrmArea(c)right)-left(sum_oinmathrmOverlapsmathrmArea(o)right)$$




      computational-geometry





      share|improve this question













      share|improve this question




      share|improve this question








      edited Sep 8 at 22:33









      Lukas Lang

      5,2181525




      5,2181525










      asked Sep 8 at 13:11









      Eman

      926




      926




















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          6
          down vote



          accepted










          I am not sure are you interested in the union of the disks or the union sans the intersections. The code below can be used for both cases.



          Implicit regions



          r = 0.18;
          regs = ImplicitRegion[
           Sqrt[(x - #[[1]])^2 + (y - #[[2]])^2] <= r, x, y] & /@ pts;


          Circles drawing



          Show[mesh, 
           Graphics[Cyan, Circle[#, r] & /@ pts, Black, Point[pts], Red, 
           Point[vertices]], Axes -> True]


          enter image description here



          Union



          c = Ceiling[Max[pts] + r, 0.5];

          AbsoluteTiming[
           dregs = DiscretizeRegion[RegionUnion[regs], -2, 2, -2, 2, 
           ImageSize -> Medium]
           ]


          enter image description here



          RegionMeasure[dregs]

          (* 1.99653 *)


          Union Intersection



          ires = DeleteCases[
           Flatten[Table[
           RegionIntersection[regs[[i]], regs[[j]]], i, 1, 
           Length[regs], j, i + 1, Length[regs]]], _EmptyRegion];

          AbsoluteTiming[
           dires = DiscretizeRegion[RegionUnion[ires], -c, c, -c, c, 
           ImageSize -> Medium, Frame -> True, PlotRange -> -c, c, -c, c]
           ]


          enter image description here



          RegionMeasure[dregs] - RegionMeasure[dires]

          (* 1.56288 *)





          share|improve this answer






















          • Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
            – Eman
            Sep 8 at 13:39







          • 1




            @Eman See my edit. It should be close to what you want.
            – Anton Antonov
            Sep 8 at 13:50










          • Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
            – Eman
            Sep 8 at 14:20

















          up vote
          5
          down vote













          radius = .18;
          disks = Disk[#, radius] & /@ pts;
          25 Area[disks[[1]]]



          2.54469




          Area[RegionUnion[disks]]



          2.03381




          Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices], 
           FaceForm[Opacity[.5,LightGreen]],EdgeForm[Thick,Darker@Green], disks]]


          enter image description here






          share|improve this answer






















          • Thanks for your help.
            – Eman
            Sep 8 at 14:18






          • 1




            @eman, my pleasure.
            – kglr
            Sep 8 at 14:24

















          up vote
          3
          down vote













          I think it is much better to use RegionMeasure on the undiscretized regions. For instance:



          disks = Disk[#, .18]& /@ pts;

          ru = RegionMeasure @ RegionUnion[disks]
          RegionMeasure @ DiscretizeRegion @ RegionUnion[disks]



          2.03381



          1.99723




          The region measure of the discretized version is almost 2% off. Similarly:



          ires = DeleteCases[
           Flatten @ Table[
           RegionIntersection[disks[[i]], disks[[j]]],
           i, 1, 25,
           j, i + 1, 25
           ],
           _EmptyRegion
          ];

          int = RegionMeasure @ RegionUnion @ ires
          RegionMeasure @ DiscretizeRegion @ RegionUnion @ ires



          0.44757



          0.439748




          So a more accurate answer would be:



          ru - int



          1.58624




          Finally, it is possible to get this result directly by using BooleanCountingFunction:



          RegionMeasure @ BooleanRegion[
           BooleanCountingFunction[1, 25],
           disks
          ]



          1.58624




          although this version is slower than Antons.






          share|improve this answer




















          • Thanks so much for your help.
            – Eman
            Sep 9 at 14:02










          Your Answer




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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          6
          down vote



          accepted










          I am not sure are you interested in the union of the disks or the union sans the intersections. The code below can be used for both cases.



          Implicit regions



          r = 0.18;
          regs = ImplicitRegion[
           Sqrt[(x - #[[1]])^2 + (y - #[[2]])^2] <= r, x, y] & /@ pts;


          Circles drawing



          Show[mesh, 
           Graphics[Cyan, Circle[#, r] & /@ pts, Black, Point[pts], Red, 
           Point[vertices]], Axes -> True]


          enter image description here



          Union



          c = Ceiling[Max[pts] + r, 0.5];

          AbsoluteTiming[
           dregs = DiscretizeRegion[RegionUnion[regs], -2, 2, -2, 2, 
           ImageSize -> Medium]
           ]


          enter image description here



          RegionMeasure[dregs]

          (* 1.99653 *)


          Union Intersection



          ires = DeleteCases[
           Flatten[Table[
           RegionIntersection[regs[[i]], regs[[j]]], i, 1, 
           Length[regs], j, i + 1, Length[regs]]], _EmptyRegion];

          AbsoluteTiming[
           dires = DiscretizeRegion[RegionUnion[ires], -c, c, -c, c, 
           ImageSize -> Medium, Frame -> True, PlotRange -> -c, c, -c, c]
           ]


          enter image description here



          RegionMeasure[dregs] - RegionMeasure[dires]

          (* 1.56288 *)





          share|improve this answer






















          • Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
            – Eman
            Sep 8 at 13:39







          • 1




            @Eman See my edit. It should be close to what you want.
            – Anton Antonov
            Sep 8 at 13:50










          • Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
            – Eman
            Sep 8 at 14:20














          up vote
          6
          down vote



          accepted










          I am not sure are you interested in the union of the disks or the union sans the intersections. The code below can be used for both cases.



          Implicit regions



          r = 0.18;
          regs = ImplicitRegion[
           Sqrt[(x - #[[1]])^2 + (y - #[[2]])^2] <= r, x, y] & /@ pts;


          Circles drawing



          Show[mesh, 
           Graphics[Cyan, Circle[#, r] & /@ pts, Black, Point[pts], Red, 
           Point[vertices]], Axes -> True]


          enter image description here



          Union



          c = Ceiling[Max[pts] + r, 0.5];

          AbsoluteTiming[
           dregs = DiscretizeRegion[RegionUnion[regs], -2, 2, -2, 2, 
           ImageSize -> Medium]
           ]


          enter image description here



          RegionMeasure[dregs]

          (* 1.99653 *)


          Union Intersection



          ires = DeleteCases[
           Flatten[Table[
           RegionIntersection[regs[[i]], regs[[j]]], i, 1, 
           Length[regs], j, i + 1, Length[regs]]], _EmptyRegion];

          AbsoluteTiming[
           dires = DiscretizeRegion[RegionUnion[ires], -c, c, -c, c, 
           ImageSize -> Medium, Frame -> True, PlotRange -> -c, c, -c, c]
           ]


          enter image description here



          RegionMeasure[dregs] - RegionMeasure[dires]

          (* 1.56288 *)





          share|improve this answer






















          • Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
            – Eman
            Sep 8 at 13:39







          • 1




            @Eman See my edit. It should be close to what you want.
            – Anton Antonov
            Sep 8 at 13:50










          • Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
            – Eman
            Sep 8 at 14:20












          up vote
          6
          down vote



          accepted







          up vote
          6
          down vote



          accepted






          I am not sure are you interested in the union of the disks or the union sans the intersections. The code below can be used for both cases.



          Implicit regions



          r = 0.18;
          regs = ImplicitRegion[
           Sqrt[(x - #[[1]])^2 + (y - #[[2]])^2] <= r, x, y] & /@ pts;


          Circles drawing



          Show[mesh, 
           Graphics[Cyan, Circle[#, r] & /@ pts, Black, Point[pts], Red, 
           Point[vertices]], Axes -> True]


          enter image description here



          Union



          c = Ceiling[Max[pts] + r, 0.5];

          AbsoluteTiming[
           dregs = DiscretizeRegion[RegionUnion[regs], -2, 2, -2, 2, 
           ImageSize -> Medium]
           ]


          enter image description here



          RegionMeasure[dregs]

          (* 1.99653 *)


          Union Intersection



          ires = DeleteCases[
           Flatten[Table[
           RegionIntersection[regs[[i]], regs[[j]]], i, 1, 
           Length[regs], j, i + 1, Length[regs]]], _EmptyRegion];

          AbsoluteTiming[
           dires = DiscretizeRegion[RegionUnion[ires], -c, c, -c, c, 
           ImageSize -> Medium, Frame -> True, PlotRange -> -c, c, -c, c]
           ]


          enter image description here



          RegionMeasure[dregs] - RegionMeasure[dires]

          (* 1.56288 *)





          share|improve this answer














          I am not sure are you interested in the union of the disks or the union sans the intersections. The code below can be used for both cases.



          Implicit regions



          r = 0.18;
          regs = ImplicitRegion[
           Sqrt[(x - #[[1]])^2 + (y - #[[2]])^2] <= r, x, y] & /@ pts;


          Circles drawing



          Show[mesh, 
           Graphics[Cyan, Circle[#, r] & /@ pts, Black, Point[pts], Red, 
           Point[vertices]], Axes -> True]


          enter image description here



          Union



          c = Ceiling[Max[pts] + r, 0.5];

          AbsoluteTiming[
           dregs = DiscretizeRegion[RegionUnion[regs], -2, 2, -2, 2, 
           ImageSize -> Medium]
           ]


          enter image description here



          RegionMeasure[dregs]

          (* 1.99653 *)


          Union Intersection



          ires = DeleteCases[
           Flatten[Table[
           RegionIntersection[regs[[i]], regs[[j]]], i, 1, 
           Length[regs], j, i + 1, Length[regs]]], _EmptyRegion];

          AbsoluteTiming[
           dires = DiscretizeRegion[RegionUnion[ires], -c, c, -c, c, 
           ImageSize -> Medium, Frame -> True, PlotRange -> -c, c, -c, c]
           ]


          enter image description here



          RegionMeasure[dregs] - RegionMeasure[dires]

          (* 1.56288 *)






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Sep 8 at 14:12

























          answered Sep 8 at 13:32









          Anton Antonov

          21.6k164107




          21.6k164107











          • Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
            – Eman
            Sep 8 at 13:39







          • 1




            @Eman See my edit. It should be close to what you want.
            – Anton Antonov
            Sep 8 at 13:50










          • Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
            – Eman
            Sep 8 at 14:20
















          • Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
            – Eman
            Sep 8 at 13:39







          • 1




            @Eman See my edit. It should be close to what you want.
            – Anton Antonov
            Sep 8 at 13:50










          • Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
            – Eman
            Sep 8 at 14:20















          Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
          – Eman
          Sep 8 at 13:39





          Thanks so much for your answer. But, I want to draw circle around each pts point, which is represented by black points and keeps also the voronoi diagram. Then, detect the overlapping areas occurs between these circles. Then calculate the sum of these circles' areas subtracted by the overlapping areas. I think, there is a difference between the circle and the disk.
          – Eman
          Sep 8 at 13:39





          1




          1




          @Eman See my edit. It should be close to what you want.
          – Anton Antonov
          Sep 8 at 13:50




          @Eman See my edit. It should be close to what you want.
          – Anton Antonov
          Sep 8 at 13:50












          Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
          – Eman
          Sep 8 at 14:20




          Thanks so much for your help and your edit. I really appreciate that. Your edit is what I was looking for. Thanks so much.
          – Eman
          Sep 8 at 14:20










          up vote
          5
          down vote













          radius = .18;
          disks = Disk[#, radius] & /@ pts;
          25 Area[disks[[1]]]



          2.54469




          Area[RegionUnion[disks]]



          2.03381




          Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices], 
           FaceForm[Opacity[.5,LightGreen]],EdgeForm[Thick,Darker@Green], disks]]


          enter image description here






          share|improve this answer






















          • Thanks for your help.
            – Eman
            Sep 8 at 14:18






          • 1




            @eman, my pleasure.
            – kglr
            Sep 8 at 14:24














          up vote
          5
          down vote













          radius = .18;
          disks = Disk[#, radius] & /@ pts;
          25 Area[disks[[1]]]



          2.54469




          Area[RegionUnion[disks]]



          2.03381




          Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices], 
           FaceForm[Opacity[.5,LightGreen]],EdgeForm[Thick,Darker@Green], disks]]


          enter image description here






          share|improve this answer






















          • Thanks for your help.
            – Eman
            Sep 8 at 14:18






          • 1




            @eman, my pleasure.
            – kglr
            Sep 8 at 14:24












          up vote
          5
          down vote










          up vote
          5
          down vote









          radius = .18;
          disks = Disk[#, radius] & /@ pts;
          25 Area[disks[[1]]]



          2.54469




          Area[RegionUnion[disks]]



          2.03381




          Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices], 
           FaceForm[Opacity[.5,LightGreen]],EdgeForm[Thick,Darker@Green], disks]]


          enter image description here






          share|improve this answer














          radius = .18;
          disks = Disk[#, radius] & /@ pts;
          25 Area[disks[[1]]]



          2.54469




          Area[RegionUnion[disks]]



          2.03381




          Show[mesh, Graphics[Black, Point[pts], Red, Point[vertices], 
           FaceForm[Opacity[.5,LightGreen]],EdgeForm[Thick,Darker@Green], disks]]


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Sep 9 at 5:40

























          answered Sep 8 at 13:33









          kglr

          159k8184384




          159k8184384











          • Thanks for your help.
            – Eman
            Sep 8 at 14:18






          • 1




            @eman, my pleasure.
            – kglr
            Sep 8 at 14:24
















          • Thanks for your help.
            – Eman
            Sep 8 at 14:18






          • 1




            @eman, my pleasure.
            – kglr
            Sep 8 at 14:24















          Thanks for your help.
          – Eman
          Sep 8 at 14:18




          Thanks for your help.
          – Eman
          Sep 8 at 14:18




          1




          1




          @eman, my pleasure.
          – kglr
          Sep 8 at 14:24




          @eman, my pleasure.
          – kglr
          Sep 8 at 14:24










          up vote
          3
          down vote













          I think it is much better to use RegionMeasure on the undiscretized regions. For instance:



          disks = Disk[#, .18]& /@ pts;

          ru = RegionMeasure @ RegionUnion[disks]
          RegionMeasure @ DiscretizeRegion @ RegionUnion[disks]



          2.03381



          1.99723




          The region measure of the discretized version is almost 2% off. Similarly:



          ires = DeleteCases[
           Flatten @ Table[
           RegionIntersection[disks[[i]], disks[[j]]],
           i, 1, 25,
           j, i + 1, 25
           ],
           _EmptyRegion
          ];

          int = RegionMeasure @ RegionUnion @ ires
          RegionMeasure @ DiscretizeRegion @ RegionUnion @ ires



          0.44757



          0.439748




          So a more accurate answer would be:



          ru - int



          1.58624




          Finally, it is possible to get this result directly by using BooleanCountingFunction:



          RegionMeasure @ BooleanRegion[
           BooleanCountingFunction[1, 25],
           disks
          ]



          1.58624




          although this version is slower than Antons.






          share|improve this answer




















          • Thanks so much for your help.
            – Eman
            Sep 9 at 14:02














          up vote
          3
          down vote













          I think it is much better to use RegionMeasure on the undiscretized regions. For instance:



          disks = Disk[#, .18]& /@ pts;

          ru = RegionMeasure @ RegionUnion[disks]
          RegionMeasure @ DiscretizeRegion @ RegionUnion[disks]



          2.03381



          1.99723




          The region measure of the discretized version is almost 2% off. Similarly:



          ires = DeleteCases[
           Flatten @ Table[
           RegionIntersection[disks[[i]], disks[[j]]],
           i, 1, 25,
           j, i + 1, 25
           ],
           _EmptyRegion
          ];

          int = RegionMeasure @ RegionUnion @ ires
          RegionMeasure @ DiscretizeRegion @ RegionUnion @ ires



          0.44757



          0.439748




          So a more accurate answer would be:



          ru - int



          1.58624




          Finally, it is possible to get this result directly by using BooleanCountingFunction:



          RegionMeasure @ BooleanRegion[
           BooleanCountingFunction[1, 25],
           disks
          ]



          1.58624




          although this version is slower than Antons.






          share|improve this answer




















          • Thanks so much for your help.
            – Eman
            Sep 9 at 14:02












          up vote
          3
          down vote










          up vote
          3
          down vote









          I think it is much better to use RegionMeasure on the undiscretized regions. For instance:



          disks = Disk[#, .18]& /@ pts;

          ru = RegionMeasure @ RegionUnion[disks]
          RegionMeasure @ DiscretizeRegion @ RegionUnion[disks]



          2.03381



          1.99723




          The region measure of the discretized version is almost 2% off. Similarly:



          ires = DeleteCases[
           Flatten @ Table[
           RegionIntersection[disks[[i]], disks[[j]]],
           i, 1, 25,
           j, i + 1, 25
           ],
           _EmptyRegion
          ];

          int = RegionMeasure @ RegionUnion @ ires
          RegionMeasure @ DiscretizeRegion @ RegionUnion @ ires



          0.44757



          0.439748




          So a more accurate answer would be:



          ru - int



          1.58624




          Finally, it is possible to get this result directly by using BooleanCountingFunction:



          RegionMeasure @ BooleanRegion[
           BooleanCountingFunction[1, 25],
           disks
          ]



          1.58624




          although this version is slower than Antons.






          share|improve this answer












          I think it is much better to use RegionMeasure on the undiscretized regions. For instance:



          disks = Disk[#, .18]& /@ pts;

          ru = RegionMeasure @ RegionUnion[disks]
          RegionMeasure @ DiscretizeRegion @ RegionUnion[disks]



          2.03381



          1.99723




          The region measure of the discretized version is almost 2% off. Similarly:



          ires = DeleteCases[
           Flatten @ Table[
           RegionIntersection[disks[[i]], disks[[j]]],
           i, 1, 25,
           j, i + 1, 25
           ],
           _EmptyRegion
          ];

          int = RegionMeasure @ RegionUnion @ ires
          RegionMeasure @ DiscretizeRegion @ RegionUnion @ ires



          0.44757



          0.439748




          So a more accurate answer would be:



          ru - int



          1.58624




          Finally, it is possible to get this result directly by using BooleanCountingFunction:



          RegionMeasure @ BooleanRegion[
           BooleanCountingFunction[1, 25],
           disks
          ]



          1.58624




          although this version is slower than Antons.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Sep 9 at 6:57









          Carl Woll

          56.3k272147




          56.3k272147











          • Thanks so much for your help.
            – Eman
            Sep 9 at 14:02
















          • Thanks so much for your help.
            – Eman
            Sep 9 at 14:02















          Thanks so much for your help.
          – Eman
          Sep 9 at 14:02




          Thanks so much for your help.
          – Eman
          Sep 9 at 14:02

















           

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