how to generate random convex quadrilaterals using circles (actually inscribed within circles)?
up vote
1
down vote
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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.
graphics parametricplot
asked 1 hour ago
Chonglin Zhu
315
add a comment |
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
asked 1 hour ago
Chonglin Zhu
315
1
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
1 hour ago
add a comment |
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
asked 1 hour ago
Chonglin Zhu
315
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
asked 1 hour ago
Chonglin Zhu
315
asked 1 hour ago
Chonglin Zhu
315
asked 1 hour ago
Chonglin Zhu
315
asked 1 hour ago
Chonglin Zhu
315
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
add a comment |
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
add a comment |
2 Answers
2
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oldest
votes
up vote
1
down vote
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]
answered 1 hour ago
kglr
175k9197402
add a comment |
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
answered 1 hour ago
Henrik Schumacher
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]
answered 1 hour ago
kglr
175k9197402
add a comment |
up vote
1
down vote
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]
answered 1 hour ago
kglr
175k9197402
add a comment |
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]]
answered 1 hour ago
kglr
175k9197402
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]]
answered 1 hour ago
kglr
175k9197402
answered 1 hour ago
kglr
175k9197402
answered 1 hour ago
kglr
175k9197402
answered 1 hour ago
kglr
175k9197402
175k9197402
add a comment |
add a comment |
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
answered 1 hour ago
Henrik Schumacher
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
add a comment |
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
answered 1 hour ago
Henrik Schumacher
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
add a comment |
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
answered 1 hour ago
Henrik Schumacher
47.5k466134
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
answered 1 hour ago
Henrik Schumacher
47.5k466134
edited 40 mins ago
answered 1 hour ago
Henrik Schumacher
47.5k466134
answered 1 hour ago
Henrik Schumacher
47.5k466134
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
add a comment |
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
add a comment |
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1
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
1 hour ago