How do I create a random point on the surface of a cube? [duplicate]

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This question already has an answer here:



  • How-to cast a set of random points on the surface of a cuboid?

    2 answers



How do I create a random point on the surface of a cube? I mean on one of its six faces at random.






graphics3d random-process







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marked as duplicate by Carl Woll, AccidentalFourierTransform, Michael E2, Community♦ Aug 11 at 6:50


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
    – AccidentalFourierTransform
    Aug 10 at 23:47














up vote
4
down vote

favorite













This question already has an answer here:



  • How-to cast a set of random points on the surface of a cuboid?

    2 answers



How do I create a random point on the surface of a cube? I mean on one of its six faces at random.






graphics3d random-process







share|improve this question















marked as duplicate by Carl Woll, AccidentalFourierTransform, Michael E2, Community♦ Aug 11 at 6:50


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
    – AccidentalFourierTransform
    Aug 10 at 23:47












up vote
4
down vote

favorite









up vote
4
down vote

favorite












This question already has an answer here:



  • How-to cast a set of random points on the surface of a cuboid?

    2 answers



How do I create a random point on the surface of a cube? I mean on one of its six faces at random.






graphics3d random-process







share|improve this question
















This question already has an answer here:



  • How-to cast a set of random points on the surface of a cuboid?

    2 answers



How do I create a random point on the surface of a cube? I mean on one of its six faces at random.





This question already has an answer here:



  • How-to cast a set of random points on the surface of a cuboid?

    2 answers






graphics3d random-process




graphics3d random-process






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Aug 10 at 18:05

























asked Aug 10 at 14:56









mattiav27

2,00411428




2,00411428




marked as duplicate by Carl Woll, AccidentalFourierTransform, Michael E2, Community♦ Aug 11 at 6:50


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Carl Woll, AccidentalFourierTransform, Michael E2, Community♦ Aug 11 at 6:50


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
    – AccidentalFourierTransform
    Aug 10 at 23:47
















  • ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
    – AccidentalFourierTransform
    Aug 10 at 23:47















ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
– AccidentalFourierTransform
Aug 10 at 23:47




ListPointPlot3D[RandomSample /@ Transpose[RandomReal[0, 1, 2, 1000]~Join~RandomInteger[0, 1, 1000]], BoxRatios -> 1, 1, 1]
– AccidentalFourierTransform
Aug 10 at 23:47










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










ReplacePart[Table[RandomReal, 3], 
 RandomChoice[Range[3]] -> RandomInteger[0, 1]]


If you want points on the surface of a $d$-dimensional unit hypercube:



d = 5;
ReplacePart[Table[RandomReal, d], 
RandomChoice[Range[d]] -> RandomInteger[0, 1]]





share|improve this answer






















  • That's a very clever approach!
    – Henrik Schumacher
    Aug 10 at 20:10










  • I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
    – Carl Woll
    Aug 10 at 20:36










  • @David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
    – Henrik Schumacher
    Aug 10 at 20:41










  • You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
    – Carl Woll
    Aug 10 at 21:46











  • Ah... you are correct. I'll remove my timing claims.
    – David G. Stork
    Aug 10 at 22:03

















up vote
5
down vote













We simply need a MeshRegion that represents the boundary of the unit cube. Then we apply RandomPoint to it. The following shows one of probably many ways to do it.



R = DiscretizeRegion@RegionBoundary[Cuboid];
RandomPoint[R]





share|improve this answer






















  • Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
    – Kuba♦
    Aug 10 at 15:19










  • @Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
    – Henrik Schumacher
    Aug 10 at 15:26






  • 1




    @kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
    – Henrik Schumacher
    Aug 10 at 20:04






  • 2




    I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
    – Carl Woll
    Aug 10 at 21:53






  • 1




    @CarlWoll Okay. I've just sent a report.
    – Henrik Schumacher
    Aug 10 at 22:02

















up vote
2
down vote













Here is how to build it from scratch in case someone finds that interesting:



coords = 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1;
pts = 4, 3, 2, 1, 1, 2, 6, 5, 2, 3, 7, 6, 3, 4, 8, 7, 4, 1, 5, 8, 5, 6, 7, 8;
polygons = Flatten@Normal@GraphicsComplex[coords, Polygon[pts]];
reg = RegionUnion[DiscretizeRegion /@ polygons];
RandomPoint[reg]


I grabbed the coordinates from the documentation for Hexahedron.






share|improve this answer




















  • Is it possible to find the coordinates of the point?
    – mattiav27
    Aug 10 at 15:56










  • @mattiav27 What do you mean? RandomPoint returns coordinates.
    – C. E.
    Aug 10 at 17:57










  • I didn't know it thanks!
    – mattiav27
    Aug 10 at 18:04

















3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










ReplacePart[Table[RandomReal, 3], 
 RandomChoice[Range[3]] -> RandomInteger[0, 1]]


If you want points on the surface of a $d$-dimensional unit hypercube:



d = 5;
ReplacePart[Table[RandomReal, d], 
RandomChoice[Range[d]] -> RandomInteger[0, 1]]





share|improve this answer






















  • That's a very clever approach!
    – Henrik Schumacher
    Aug 10 at 20:10










  • I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
    – Carl Woll
    Aug 10 at 20:36










  • @David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
    – Henrik Schumacher
    Aug 10 at 20:41










  • You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
    – Carl Woll
    Aug 10 at 21:46











  • Ah... you are correct. I'll remove my timing claims.
    – David G. Stork
    Aug 10 at 22:03














up vote
3
down vote



accepted










ReplacePart[Table[RandomReal, 3], 
 RandomChoice[Range[3]] -> RandomInteger[0, 1]]


If you want points on the surface of a $d$-dimensional unit hypercube:



d = 5;
ReplacePart[Table[RandomReal, d], 
RandomChoice[Range[d]] -> RandomInteger[0, 1]]





share|improve this answer






















  • That's a very clever approach!
    – Henrik Schumacher
    Aug 10 at 20:10










  • I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
    – Carl Woll
    Aug 10 at 20:36










  • @David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
    – Henrik Schumacher
    Aug 10 at 20:41










  • You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
    – Carl Woll
    Aug 10 at 21:46











  • Ah... you are correct. I'll remove my timing claims.
    – David G. Stork
    Aug 10 at 22:03












up vote
3
down vote



accepted







up vote
3
down vote



accepted






ReplacePart[Table[RandomReal, 3], 
 RandomChoice[Range[3]] -> RandomInteger[0, 1]]


If you want points on the surface of a $d$-dimensional unit hypercube:



d = 5;
ReplacePart[Table[RandomReal, d], 
RandomChoice[Range[d]] -> RandomInteger[0, 1]]





share|improve this answer














ReplacePart[Table[RandomReal, 3], 
 RandomChoice[Range[3]] -> RandomInteger[0, 1]]


If you want points on the surface of a $d$-dimensional unit hypercube:



d = 5;
ReplacePart[Table[RandomReal, d], 
RandomChoice[Range[d]] -> RandomInteger[0, 1]]






share|improve this answer














share|improve this answer



share|improve this answer








edited Aug 10 at 22:03

























answered Aug 10 at 18:15









David G. Stork

21.5k11646




21.5k11646











  • That's a very clever approach!
    – Henrik Schumacher
    Aug 10 at 20:10










  • I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
    – Carl Woll
    Aug 10 at 20:36










  • @David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
    – Henrik Schumacher
    Aug 10 at 20:41










  • You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
    – Carl Woll
    Aug 10 at 21:46











  • Ah... you are correct. I'll remove my timing claims.
    – David G. Stork
    Aug 10 at 22:03
















  • That's a very clever approach!
    – Henrik Schumacher
    Aug 10 at 20:10










  • I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
    – Carl Woll
    Aug 10 at 20:36










  • @David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
    – Henrik Schumacher
    Aug 10 at 20:41










  • You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
    – Carl Woll
    Aug 10 at 21:46











  • Ah... you are correct. I'll remove my timing claims.
    – David G. Stork
    Aug 10 at 22:03















That's a very clever approach!
– Henrik Schumacher
Aug 10 at 20:10




That's a very clever approach!
– Henrik Schumacher
Aug 10 at 20:10












I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
– Carl Woll
Aug 10 at 20:36




I don't know about faster. Compared to RandomPoint[R, 1000], how long does your method take to generate 1000 random points?
– Carl Woll
Aug 10 at 20:36












@David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
– Henrik Schumacher
Aug 10 at 20:41




@David G. Stork. That depends. If you count in the timing for discretizing the region: Yes, this takes a lot if time. But you have to do that only once.
– Henrik Schumacher
Aug 10 at 20:41












You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
– Carl Woll
Aug 10 at 21:46





You used Table[RandomPoint[R], 10^6] instead of RandomPoint[R, 10^6]. The latter is about 3 orders of magnitude faster.
– Carl Woll
Aug 10 at 21:46













Ah... you are correct. I'll remove my timing claims.
– David G. Stork
Aug 10 at 22:03




Ah... you are correct. I'll remove my timing claims.
– David G. Stork
Aug 10 at 22:03










up vote
5
down vote













We simply need a MeshRegion that represents the boundary of the unit cube. Then we apply RandomPoint to it. The following shows one of probably many ways to do it.



R = DiscretizeRegion@RegionBoundary[Cuboid];
RandomPoint[R]





share|improve this answer






















  • Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
    – Kuba♦
    Aug 10 at 15:19










  • @Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
    – Henrik Schumacher
    Aug 10 at 15:26






  • 1




    @kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
    – Henrik Schumacher
    Aug 10 at 20:04






  • 2




    I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
    – Carl Woll
    Aug 10 at 21:53






  • 1




    @CarlWoll Okay. I've just sent a report.
    – Henrik Schumacher
    Aug 10 at 22:02














up vote
5
down vote













We simply need a MeshRegion that represents the boundary of the unit cube. Then we apply RandomPoint to it. The following shows one of probably many ways to do it.



R = DiscretizeRegion@RegionBoundary[Cuboid];
RandomPoint[R]





share|improve this answer






















  • Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
    – Kuba♦
    Aug 10 at 15:19










  • @Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
    – Henrik Schumacher
    Aug 10 at 15:26






  • 1




    @kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
    – Henrik Schumacher
    Aug 10 at 20:04






  • 2




    I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
    – Carl Woll
    Aug 10 at 21:53






  • 1




    @CarlWoll Okay. I've just sent a report.
    – Henrik Schumacher
    Aug 10 at 22:02












up vote
5
down vote










up vote
5
down vote









We simply need a MeshRegion that represents the boundary of the unit cube. Then we apply RandomPoint to it. The following shows one of probably many ways to do it.



R = DiscretizeRegion@RegionBoundary[Cuboid];
RandomPoint[R]





share|improve this answer














We simply need a MeshRegion that represents the boundary of the unit cube. Then we apply RandomPoint to it. The following shows one of probably many ways to do it.



R = DiscretizeRegion@RegionBoundary[Cuboid];
RandomPoint[R]






share|improve this answer














share|improve this answer



share|improve this answer








edited Aug 10 at 15:25

























answered Aug 10 at 15:13









Henrik Schumacher

38.2k250110




38.2k250110











  • Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
    – Kuba♦
    Aug 10 at 15:19










  • @Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
    – Henrik Schumacher
    Aug 10 at 15:26






  • 1




    @kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
    – Henrik Schumacher
    Aug 10 at 20:04






  • 2




    I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
    – Carl Woll
    Aug 10 at 21:53






  • 1




    @CarlWoll Okay. I've just sent a report.
    – Henrik Schumacher
    Aug 10 at 22:02
















  • Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
    – Kuba♦
    Aug 10 at 15:19










  • @Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
    – Henrik Schumacher
    Aug 10 at 15:26






  • 1




    @kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
    – Henrik Schumacher
    Aug 10 at 20:04






  • 2




    I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
    – Carl Woll
    Aug 10 at 21:53






  • 1




    @CarlWoll Okay. I've just sent a report.
    – Henrik Schumacher
    Aug 10 at 22:02















Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
– Kuba♦
Aug 10 at 15:19




Sorry, not into regions lately, will RegionBoundary be ok too?RandomPoint[DiscretizeRegion@RegionBoundary@Cuboid, 500]
– Kuba♦
Aug 10 at 15:19












@Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
– Henrik Schumacher
Aug 10 at 15:26




@Kuba Ah right! I tried all combinations of Mesh and Boundary but forget about RegionBoundary! Thank's for the hint.
– Henrik Schumacher
Aug 10 at 15:26




1




1




@kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
– Henrik Schumacher
Aug 10 at 20:04




@kirma Yes. What I meant" RegionBoundary@Cuboid returns a list of quadrilaterals wrapped in Polygon. Mathematica would have to run planarity checks when calling RandomPoint. While this would be possible, I can understand that it would not be worthwhile to implement that, in particular since there is already an implementation for simplicial MeshRegions. And an automatica conversion would have also its pitfalls. But I agree that I was also puzzled first when I realized that RandomPoint@RegionBoundary@Cuboid does not evaluate.
– Henrik Schumacher
Aug 10 at 20:04




2




2




I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
– Carl Woll
Aug 10 at 21:53




I believe the need for DiscretizeRegion is a bug, and should be reported. It used to work in 11.1. Also, it works in higher dimensions, e.g., RandomPoint @ RegionBoundary @ Cuboid[0,0,0,0, 1,1,1,1].
– Carl Woll
Aug 10 at 21:53




1




1




@CarlWoll Okay. I've just sent a report.
– Henrik Schumacher
Aug 10 at 22:02




@CarlWoll Okay. I've just sent a report.
– Henrik Schumacher
Aug 10 at 22:02










up vote
2
down vote













Here is how to build it from scratch in case someone finds that interesting:



coords = 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1;
pts = 4, 3, 2, 1, 1, 2, 6, 5, 2, 3, 7, 6, 3, 4, 8, 7, 4, 1, 5, 8, 5, 6, 7, 8;
polygons = Flatten@Normal@GraphicsComplex[coords, Polygon[pts]];
reg = RegionUnion[DiscretizeRegion /@ polygons];
RandomPoint[reg]


I grabbed the coordinates from the documentation for Hexahedron.






share|improve this answer




















  • Is it possible to find the coordinates of the point?
    – mattiav27
    Aug 10 at 15:56










  • @mattiav27 What do you mean? RandomPoint returns coordinates.
    – C. E.
    Aug 10 at 17:57










  • I didn't know it thanks!
    – mattiav27
    Aug 10 at 18:04














up vote
2
down vote













Here is how to build it from scratch in case someone finds that interesting:



coords = 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1;
pts = 4, 3, 2, 1, 1, 2, 6, 5, 2, 3, 7, 6, 3, 4, 8, 7, 4, 1, 5, 8, 5, 6, 7, 8;
polygons = Flatten@Normal@GraphicsComplex[coords, Polygon[pts]];
reg = RegionUnion[DiscretizeRegion /@ polygons];
RandomPoint[reg]


I grabbed the coordinates from the documentation for Hexahedron.






share|improve this answer




















  • Is it possible to find the coordinates of the point?
    – mattiav27
    Aug 10 at 15:56










  • @mattiav27 What do you mean? RandomPoint returns coordinates.
    – C. E.
    Aug 10 at 17:57










  • I didn't know it thanks!
    – mattiav27
    Aug 10 at 18:04












up vote
2
down vote










up vote
2
down vote









Here is how to build it from scratch in case someone finds that interesting:



coords = 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1;
pts = 4, 3, 2, 1, 1, 2, 6, 5, 2, 3, 7, 6, 3, 4, 8, 7, 4, 1, 5, 8, 5, 6, 7, 8;
polygons = Flatten@Normal@GraphicsComplex[coords, Polygon[pts]];
reg = RegionUnion[DiscretizeRegion /@ polygons];
RandomPoint[reg]


I grabbed the coordinates from the documentation for Hexahedron.






share|improve this answer












Here is how to build it from scratch in case someone finds that interesting:



coords = 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1;
pts = 4, 3, 2, 1, 1, 2, 6, 5, 2, 3, 7, 6, 3, 4, 8, 7, 4, 1, 5, 8, 5, 6, 7, 8;
polygons = Flatten@Normal@GraphicsComplex[coords, Polygon[pts]];
reg = RegionUnion[DiscretizeRegion /@ polygons];
RandomPoint[reg]


I grabbed the coordinates from the documentation for Hexahedron.







share|improve this answer












share|improve this answer



share|improve this answer










answered Aug 10 at 15:28









C. E.

47.7k391193




47.7k391193











  • Is it possible to find the coordinates of the point?
    – mattiav27
    Aug 10 at 15:56










  • @mattiav27 What do you mean? RandomPoint returns coordinates.
    – C. E.
    Aug 10 at 17:57










  • I didn't know it thanks!
    – mattiav27
    Aug 10 at 18:04
















  • Is it possible to find the coordinates of the point?
    – mattiav27
    Aug 10 at 15:56










  • @mattiav27 What do you mean? RandomPoint returns coordinates.
    – C. E.
    Aug 10 at 17:57










  • I didn't know it thanks!
    – mattiav27
    Aug 10 at 18:04















Is it possible to find the coordinates of the point?
– mattiav27
Aug 10 at 15:56




Is it possible to find the coordinates of the point?
– mattiav27
Aug 10 at 15:56












@mattiav27 What do you mean? RandomPoint returns coordinates.
– C. E.
Aug 10 at 17:57




@mattiav27 What do you mean? RandomPoint returns coordinates.
– C. E.
Aug 10 at 17:57












I didn't know it thanks!
– mattiav27
Aug 10 at 18:04




I didn't know it thanks!
– mattiav27
Aug 10 at 18:04


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