how to generate random convex quadrilaterals using circles (actually inscribed within circles)?
6
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 Dec 16 at 15:05
Chonglin Zhu
666
add a comment |
6
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 Dec 16 at 15:05
Chonglin Zhu
666
3
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18
add a comment |
6
6
6
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 Dec 16 at 15:05
Chonglin Zhu
666
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 Dec 16 at 15:05
Chonglin Zhu
666
asked Dec 16 at 15:05
Chonglin Zhu
666
asked Dec 16 at 15:05
Chonglin Zhu
666
asked Dec 16 at 15:05
Chonglin Zhu
666
asked Dec 16 at 15:05
Chonglin Zhu
666
666
3
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18
add a comment |
3
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18
3
3
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18
add a comment |
3 Answers
3
active
oldest
votes
5
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 Dec 16 at 15:16
Henrik Schumacher
48.6k467139
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
add a comment |
3
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Circle,
{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]}]
answered Dec 16 at 15:31
kglr
176k9198404
add a comment |
2
Here's a fun way to use Henrik's stuff (and some other nicely vectorized operations):
n = 550;
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}];
disks = {Range[2, 3], Range[7, 8] , Range[12, 15]};
shiftQuads = quads +
RandomChoice[Flatten@disks, n]*
Transpose[
ConstantArray[{Cos[Subdivide[0., 2. [Pi], n - 1]],
Sin[Subdivide[0., 2. [Pi], n - 1]]}, 4], {2, 3, 1}];
{
Opacity[.15],
Annulus[{0, 0}, {-1, 1} + MinMax@#] & /@ disks,
Thread@{Opacity[.5], RandomColor[n], Thread[Polygon@shiftQuads]}
} // Graphics
answered Dec 17 at 2:29
b3m2a1
26.7k257154
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
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 Dec 16 at 15:16
Henrik Schumacher
48.6k467139
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
add a comment |
5
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 Dec 16 at 15:16
Henrik Schumacher
48.6k467139
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
add a comment |
5
5
5
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 Dec 16 at 15:16
Henrik Schumacher
48.6k467139
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 Dec 16 at 15:16
Henrik Schumacher
48.6k467139
edited Dec 16 at 15:55
answered Dec 16 at 15:16
Henrik Schumacher
48.6k467139
answered Dec 16 at 15:16
Henrik Schumacher
48.6k467139
answered Dec 16 at 15:16
Henrik Schumacher
48.6k467139
48.6k467139
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
add a comment |
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
maybe we only use random four points on the circle and to construct a quadrilateral?
– Chonglin Zhu
Dec 16 at 15:48
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
Dec 16 at 15:59
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
Dec 16 at 15:59
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
Thanks, this is very helpful!
– Chonglin Zhu
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
You're welcome!
– Henrik Schumacher
Dec 16 at 16:00
add a comment |
3
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Circle,
{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]}]
answered Dec 16 at 15:31
kglr
176k9198404
add a comment |
3
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Circle,
{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]}]
answered Dec 16 at 15:31
kglr
176k9198404
add a comment |
3
3
3
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Circle,
{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]}]
answered Dec 16 at 15:31
kglr
176k9198404
ClearAll[randomQuad]
randomQuad = SortBy[#, ArcTan @@ # &] & /@ RandomPoint[Circle, {#, 4}] &;
SeedRandom[777]
Graphics[{Circle,
{Opacity[.5], EdgeForm[Gray], RandomColor, Polygon@#} & /@ randomQuad[5]}]
answered Dec 16 at 15:31
kglr
176k9198404
edited Dec 17 at 0:15
answered Dec 16 at 15:31
kglr
176k9198404
answered Dec 16 at 15:31
kglr
176k9198404
answered Dec 16 at 15:31
kglr
176k9198404
176k9198404
add a comment |
add a comment |
2
Here's a fun way to use Henrik's stuff (and some other nicely vectorized operations):
n = 550;
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}];
disks = {Range[2, 3], Range[7, 8] , Range[12, 15]};
shiftQuads = quads +
RandomChoice[Flatten@disks, n]*
Transpose[
ConstantArray[{Cos[Subdivide[0., 2. [Pi], n - 1]],
Sin[Subdivide[0., 2. [Pi], n - 1]]}, 4], {2, 3, 1}];
{
Opacity[.15],
Annulus[{0, 0}, {-1, 1} + MinMax@#] & /@ disks,
Thread@{Opacity[.5], RandomColor[n], Thread[Polygon@shiftQuads]}
} // Graphics
answered Dec 17 at 2:29
b3m2a1
26.7k257154
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
add a comment |
2
Here's a fun way to use Henrik's stuff (and some other nicely vectorized operations):
n = 550;
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}];
disks = {Range[2, 3], Range[7, 8] , Range[12, 15]};
shiftQuads = quads +
RandomChoice[Flatten@disks, n]*
Transpose[
ConstantArray[{Cos[Subdivide[0., 2. [Pi], n - 1]],
Sin[Subdivide[0., 2. [Pi], n - 1]]}, 4], {2, 3, 1}];
{
Opacity[.15],
Annulus[{0, 0}, {-1, 1} + MinMax@#] & /@ disks,
Thread@{Opacity[.5], RandomColor[n], Thread[Polygon@shiftQuads]}
} // Graphics
answered Dec 17 at 2:29
b3m2a1
26.7k257154
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
add a comment |
2
2
2
Here's a fun way to use Henrik's stuff (and some other nicely vectorized operations):
n = 550;
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}];
disks = {Range[2, 3], Range[7, 8] , Range[12, 15]};
shiftQuads = quads +
RandomChoice[Flatten@disks, n]*
Transpose[
ConstantArray[{Cos[Subdivide[0., 2. [Pi], n - 1]],
Sin[Subdivide[0., 2. [Pi], n - 1]]}, 4], {2, 3, 1}];
{
Opacity[.15],
Annulus[{0, 0}, {-1, 1} + MinMax@#] & /@ disks,
Thread@{Opacity[.5], RandomColor[n], Thread[Polygon@shiftQuads]}
} // Graphics
answered Dec 17 at 2:29
b3m2a1
26.7k257154
Here's a fun way to use Henrik's stuff (and some other nicely vectorized operations):
n = 550;
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}];
disks = {Range[2, 3], Range[7, 8] , Range[12, 15]};
shiftQuads = quads +
RandomChoice[Flatten@disks, n]*
Transpose[
ConstantArray[{Cos[Subdivide[0., 2. [Pi], n - 1]],
Sin[Subdivide[0., 2. [Pi], n - 1]]}, 4], {2, 3, 1}];
{
Opacity[.15],
Annulus[{0, 0}, {-1, 1} + MinMax@#] & /@ disks,
Thread@{Opacity[.5], RandomColor[n], Thread[Polygon@shiftQuads]}
} // Graphics
answered Dec 17 at 2:29
b3m2a1
26.7k257154
answered Dec 17 at 2:29
b3m2a1
26.7k257154
answered Dec 17 at 2:29
b3m2a1
26.7k257154
answered Dec 17 at 2:29
b3m2a1
26.7k257154
26.7k257154
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
add a comment |
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
Great! I will have a try, so many thanks!
– Chonglin Zhu
Dec 18 at 2:30
add a comment |
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3
"Use trigonometry" means place them on the unit circle. Then connect the dots.
– Daniel Lichtblau
Dec 16 at 15:18