Random numbers within a sphere?

[Stavros Christoforou]

Hello everyone,

I was wondering if someone could help me with an issue I have in C++. I
want to select random points within the volume of a sphere. I know how
to get random numbers using srand() and rand(), but have no idea how to
do that within a more complicated geometry. Any help would be greatly
appreciated..

Regards
Stavros

[Karl Heinz Buchegger]

Shezan Baig wrote:

Stavros Christoforou wrote:

No, the probability of being close to the center is higher than the
probability of being around the edges.

May I ask why?

It might be a little hard to explain without graphical aid :)

But basically, to be around the center, you need to have a small
radius. The angles don't really matter here - as long as the radius is
small, then the point will be near the center.

To make the point closer to a particular edge, you need a bigger radius
*and* the correct angle (or direction, or whatever). So the
probability of this happening is less.

What 'edges' are you talking about?

--
Karl Heinz Buchegger
kb******@gascad .at

[Shezan Baig]

Karl Heinz Buchegger wrote:

Shezan Baig wrote:

Stavros Christoforou wrote:

> No, the probability of being close to the center is higher than the > probability of being around the edges.
>

May I ask why?

It might be a little hard to explain without graphical aid :)

But basically, to be around the center, you need to have a small
radius. The angles don't really matter here - as long as the radius is small, then the point will be near the center.

To make the point closer to a particular edge, you need a bigger radius *and* the correct angle (or direction, or whatever). So the
probability of this happening is less.

What 'edges' are you talking about?

--
Karl Heinz Buchegger
kb******@gascad .at

Sorry, I meant a point near the circumference.

[Karl Heinz Buchegger]

Shezan Baig wrote:

Karl Heinz Buchegger wrote:

Shezan Baig wrote:

Stavros Christoforou wrote:
> > No, the probability of being close to the center is higher than the > > probability of being around the edges.
> >
>
> May I ask why?
>

It might be a little hard to explain without graphical aid :)

But basically, to be around the center, you need to have a small
radius. The angles don't really matter here - as long as the radius is small, then the point will be near the center.

To make the point closer to a particular edge, you need a bigger radius *and* the correct angle (or direction, or whatever). So the
probability of this happening is less.

What 'edges' are you talking about?

Sorry, I meant a point near the circumference.

In this case your argumentation doesn't apply. If a sampled
radius is 80% of the speheres radius, then it will be 20% from
the surface, no matter which direction.
--
Karl Heinz Buchegger
kb******@gascad .at

[Joseph Seigh]

On Tue, 08 Feb 2005 14:18:56 GMT, Phlip <ph*******@yaho o.com> wrote:

Fabio Rossi wrote:

Stavros Christoforou wrote:

> Hello everyone,
>
> I was wondering if someone could help me with an issue I have in C++. I
> want to select random points within the volume of a sphere. I know how
> to get random numbers using srand() and rand(), but have no idea how to
> do that within a more complicated geometry. Any help would be greatly
> appreciated..

You could represent any point of the sphere in spherical coordinates: a
radius and two angles. A coordinate is a uniform variable (using rand())

in

its [min,max] interval.

Stavros might need evenly distributed numbers. A polar coordinate scheme
would bunch numbers up around the center, at least.

The fix is to bound the sphere with a cube, tangent on all faces, and find
random points in the cube by finding random xyz points, each >= -r and < r,
where r is the radius of the sphere. Then filter out all points laying
outside the sphere, by comparing x^2 + y^2 + z^2 to r^2.

That's not likely to be an even distribution since you're discarding samples.
The OP need to define what the set of "points" is in the fist place, which
would depend on the coordinate system chosen. The OP would also need to
define the function f(x, y), where x and y are "points", against which the
random distribution is being applied. It could be metric distance between
the points or something else entirely. The OP really needs to repost their
question in comp.programmin g where they deal with these kind of questions
rather than there where it's a little off topic.

--
Joe Seigh

http://atomic-ptr-plus.sourceforge.net/

[Stavros Christoforou]

Joseph Seigh wrote:
The OP really needs to repost their

question in comp.programmin g where they deal with these kind of questions
rather than there where it's a little off topic.

I see. Thank you all for your replies, I'll give it a shot there!

Stavros

[Pete Becker]

Joseph Seigh wrote:

That's not likely to be an even distribution since you're discarding
samples.

It's a very common technique: generate values that are evenly
distributed and discard ones that aren't in the desired range. The
remaining values are as evenly distributed in the target range as the
original ones were in their range.

--

Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)

[phlip2005]

Pete Becker wrote:

Joseph Seigh wrote:

That's not likely to be an even distribution since you're discarding samples.

It's a very common technique: generate values that are evenly
distributed and discard ones that aren't in the desired range. The
remaining values are as evenly distributed in the target range as the

original ones were in their range.

For example, if Pete and me post to a newsgroup for a decade, and your
filter in all the posts where he backs up one of my wild opinions, the
posts will have the same distribution. Except over a much smaller set.
;-)

--
Phlip

[Michael]

Off the top of my head, I think you'd need to look at the partial deriative
of the volume with respect to radius, ie, at a particular distance see what
a delta R did to delta V...... but I've got to go to lectures , ( and a bit
OT!)

Nice probelm tho!!

Mike

"Stavros Christoforou" <S.************ **********@iri. tudelft.nl> wrote in
message news:1f******** *************** ****@news1.tude lft.nl...

No, the probability of being close to the center is higher than the
probability of being around the edges.

May I ask why?

Stavros might need evenly distributed numbers. A polar coordinate
scheme would bunch numbers up around the center, at least.

How so? If the [0, sphere-radius] random number generator is uniform, so would the density with respect to the center.

The density would be selected from a pdf, eg

s = - (log ((double) (rand()+ 1.0)/RAND_MAX))/sigmat

(just a random function)

Therefore my problem focuses more on how to select the points within the
sphere, and my idea so far was similar to Philip's. However, I am sure
something faster and more "code-correct" exists.

Also, sorry if I posted this on the wrong group, but as I am creating
the program on C++ I thought that this would be the appropriate place to
ask questions.
Stavros

[phlip2005]

Ron Natalie wrote:

Stavros might need evenly distributed numbers. A polar coordinate scheme would bunch numbers up around the center, at least.

How so? If the [0, sphere-radius] random number generator is uniform, so would the density with respect to the center.
Instead of random numbers, imagine evenly spaced numbers on a line. Put
the line from the center to the rim of a circle. Add a point to the
circle's plane for each number on the line. Now rotate the line so it
reaches from the center to another point on the rim. Stamp out more
points onto the circle.

Keep going until you have covered the rim in an even number of steps.
Notice you drew N dotted circles, one for each number on the original
line. Now notice how the dots near the center are closer together.

Arbitrarily plugging evenly spaced pseudorandom numbers into polar
coordinates, with no other adjustments, will bunch the numbers up
around the center.
The fix is to bound the sphere with a cube, tangent on all faces, and find random points in the cube by finding random xyz points, each >= -r and < r, where r is the radius of the sphere. Then filter out all points laying outside the sphere, by comparing x^2 + y^2 + z^2 to r^2.

That wouldn't be any better and a lot slower as a lot of samples

would have to be generated, tested, and discarded.

Only slower by the volume of a cube minus the volume of a tangent
sphere. Most points will be inside the sphere.

The performance would still only be dominated by the speed of
generating pseudo-random numbers, and the speed of comparing their
distances. (And note I left the Square Root calculation out of the
distance formula.)

--
Phlip

[Andrey Tarasevich]

Ron Natalie wrote:

Stavros might need evenly distributed numbers. A polar coordinate scheme
would bunch numbers up around the center, at least.

How so? If the [0, sphere-radius] random number generator is uniform, so would
the density with respect to the center.

As always with geometrical probability, what constitutes "even
distribution" is not as obvious as it might seem at the first sight. It
depends on the choice of metric. That's exactly the essence of the
misunderstandin g that takes place in this discussion. For an
illustration look up "bertrand paradox" (or "bertrand's paradox") on
Google, it is directly relevant to the ongoing discussion (see, for
example, here http://www.cut-the-knot.org/bertrand.shtml).

--
Best regards,
Andrey Tarasevich