how to generate random convex quadrilaterals using circles (actually inscribed within circles)?











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I would like to generate random convex quadrilaterals using circles, my tutor has suggested to using randomreal to generate 4 numbers and scale the sum of them to 2pi, and then use trigonometry properties to do that, I do not see how that works, can somebody gives some hints about how I should go with it then I can try it out.






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  • 1




    "Use trigonometry" means place them on the unit circle. Then connect the dots.
    – Daniel Lichtblau
    1 hour ago















up vote
1
down vote

favorite












I would like to generate random convex quadrilaterals using circles, my tutor has suggested to using randomreal to generate 4 numbers and scale the sum of them to 2pi, and then use trigonometry properties to do that, I do not see how that works, can somebody gives some hints about how I should go with it then I can try it out.






graphics parametricplot







share|improve this question


















  • 1




    "Use trigonometry" means place them on the unit circle. Then connect the dots.
    – Daniel Lichtblau
    1 hour ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I would like to generate random convex quadrilaterals using circles, my tutor has suggested to using randomreal to generate 4 numbers and scale the sum of them to 2pi, and then use trigonometry properties to do that, I do not see how that works, can somebody gives some hints about how I should go with it then I can try it out.






graphics parametricplot







share|improve this question













I would like to generate random convex quadrilaterals using circles, my tutor has suggested to using randomreal to generate 4 numbers and scale the sum of them to 2pi, and then use trigonometry properties to do that, I do not see how that works, can somebody gives some hints about how I should go with it then I can try it out.






graphics parametricplot




graphics parametricplot






share|improve this question













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asked 1 hour ago









Chonglin Zhu

315




315








  • 1




    "Use trigonometry" means place them on the unit circle. Then connect the dots.
    – Daniel Lichtblau
    1 hour ago














  • 1




    "Use trigonometry" means place them on the unit circle. Then connect the dots.
    – Daniel Lichtblau
    1 hour ago








1




1




"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
1 hour ago




"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
1 hour ago










2 Answers
2






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up vote
1
down vote













ClearAll[randomQuad]

randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;



SeedRandom[777]

Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]


enter image description here






share|improve this answer




























    up vote
    1
    down vote













    Maybe this way?



    n = 1000;
    
x = RandomReal[{0, 2 Pi}, {n, 5}];
    
x[[All, 2 ;;]] *= Divide[(2. Pi), (x.{0., 1., 1., 1., 1.})];
    
x = x.UpperTriangularize[ConstantArray[1., {5, 4}]];
    
quads = Transpose[{Cos[x], Sin[x]}, {3, 1, 2}];


    Convexity test:



    And @@ (Graphics`PolygonUtils`PolygonConvexQ /@ quads)



    True







    share|improve this answer























    • maybe we only use random four points on the circle and to construct a quadrilateral?
      – Chonglin Zhu
      47 mins ago










    • Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
      – Henrik Schumacher
      37 mins ago










    • Thanks, this is very helpful!
      – Chonglin Zhu
      36 mins ago










    • You're welcome!
      – Henrik Schumacher
      35 mins ago











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

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    up vote
    1
    down vote













    ClearAll[randomQuad]
    
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
    

    
SeedRandom[777]
    
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]


    enter image description here






    share|improve this answer

























      up vote
      1
      down vote













      ClearAll[randomQuad]
      
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
      

      
SeedRandom[777]
      
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]


      enter image description here






      share|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        ClearAll[randomQuad]
        
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
        

        
SeedRandom[777]
        
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]


        enter image description here






        share|improve this answer












        ClearAll[randomQuad]
        
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
        

        
SeedRandom[777]
        
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        kglr

        175k9197402




        175k9197402






















            up vote
            1
            down vote













            Maybe this way?



            n = 1000;
            
x = RandomReal[{0, 2 Pi}, {n, 5}];
            
x[[All, 2 ;;]] *= Divide[(2. Pi), (x.{0., 1., 1., 1., 1.})];
            
x = x.UpperTriangularize[ConstantArray[1., {5, 4}]];
            
quads = Transpose[{Cos[x], Sin[x]}, {3, 1, 2}];


            Convexity test:



            And @@ (Graphics`PolygonUtils`PolygonConvexQ /@ quads)



            True







            share|improve this answer























            • maybe we only use random four points on the circle and to construct a quadrilateral?
              – Chonglin Zhu
              47 mins ago










            • Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
              – Henrik Schumacher
              37 mins ago










            • Thanks, this is very helpful!
              – Chonglin Zhu
              36 mins ago










            • You're welcome!
              – Henrik Schumacher
              35 mins ago















            up vote
            1
            down vote













            Maybe this way?



            n = 1000;
            
x = RandomReal[{0, 2 Pi}, {n, 5}];
            
x[[All, 2 ;;]] *= Divide[(2. Pi), (x.{0., 1., 1., 1., 1.})];
            
x = x.UpperTriangularize[ConstantArray[1., {5, 4}]];
            
quads = Transpose[{Cos[x], Sin[x]}, {3, 1, 2}];


            Convexity test:



            And @@ (Graphics`PolygonUtils`PolygonConvexQ /@ quads)



            True







            share|improve this answer























            • maybe we only use random four points on the circle and to construct a quadrilateral?
              – Chonglin Zhu
              47 mins ago










            • Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
              – Henrik Schumacher
              37 mins ago










            • Thanks, this is very helpful!
              – Chonglin Zhu
              36 mins ago










            • You're welcome!
              – Henrik Schumacher
              35 mins ago













            up vote
            1
            down vote










            up vote
            1
            down vote









            Maybe this way?



            n = 1000;
            
x = RandomReal[{0, 2 Pi}, {n, 5}];
            
x[[All, 2 ;;]] *= Divide[(2. Pi), (x.{0., 1., 1., 1., 1.})];
            
x = x.UpperTriangularize[ConstantArray[1., {5, 4}]];
            
quads = Transpose[{Cos[x], Sin[x]}, {3, 1, 2}];


            Convexity test:



            And @@ (Graphics`PolygonUtils`PolygonConvexQ /@ quads)



            True







            share|improve this answer














            Maybe this way?



            n = 1000;
            
x = RandomReal[{0, 2 Pi}, {n, 5}];
            
x[[All, 2 ;;]] *= Divide[(2. Pi), (x.{0., 1., 1., 1., 1.})];
            
x = x.UpperTriangularize[ConstantArray[1., {5, 4}]];
            
quads = Transpose[{Cos[x], Sin[x]}, {3, 1, 2}];


            Convexity test:



            And @@ (Graphics`PolygonUtils`PolygonConvexQ /@ quads)



            True








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 40 mins ago

























            answered 1 hour ago









            Henrik Schumacher

            47.5k466134




            47.5k466134












            • maybe we only use random four points on the circle and to construct a quadrilateral?
              – Chonglin Zhu
              47 mins ago










            • Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
              – Henrik Schumacher
              37 mins ago










            • Thanks, this is very helpful!
              – Chonglin Zhu
              36 mins ago










            • You're welcome!
              – Henrik Schumacher
              35 mins ago


















            • maybe we only use random four points on the circle and to construct a quadrilateral?
              – Chonglin Zhu
              47 mins ago










            • Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
              – Henrik Schumacher
              37 mins ago










            • Thanks, this is very helpful!
              – Chonglin Zhu
              36 mins ago










            • You're welcome!
              – Henrik Schumacher
              35 mins ago
















            maybe we only use random four points on the circle and to construct a quadrilateral?
            – Chonglin Zhu
            47 mins ago




            maybe we only use random four points on the circle and to construct a quadrilateral?
            – Chonglin Zhu
            47 mins ago












            Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
            – Henrik Schumacher
            37 mins ago




            Certainly, this is a trivial problem in the sense that it can be tackled in many ways. I just tried to make it fast by using as much vectorized operations and matrix arithmetic as possible. On my machine, the code above generates about 3 million quadrilaterals per second.
            – Henrik Schumacher
            37 mins ago












            Thanks, this is very helpful!
            – Chonglin Zhu
            36 mins ago




            Thanks, this is very helpful!
            – Chonglin Zhu
            36 mins ago












            You're welcome!
            – Henrik Schumacher
            35 mins ago




            You're welcome!
            – Henrik Schumacher
            35 mins ago


















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