I am unsure about whether this is true. The ratio of a ball’s volume to its enclosing hypercube’s volume should decrease to 0 as dimensionality increases. Thus, the approach should actually generalize very poorly.
Edit: see @srean's excellent explanation why that won't do.
One way to fix the problem is to sample uniformly not on the latitude x longitude rectangle but the sin (latitude) x longitude rectangle.
The reason this works is because the area of a infinitesimal lat long patch on the sphere is dlong x lat x cosine (lat). Now, if we sample on the long x sin(lat) rectangle, an infinitesimal rectangle also has area dlong x dlat x d/dlat sin(lat) = dlong x dlat cos (lat).
Unfortunately, these simple fixes do not generalize to arbitrary dimensions. For that those that exploit rotational symmetry of L2 norm works best.
To appreciate why, consider strips along two constant latitudes. One along the Equator and the other very close to the pole. The uniformly random polar coordinates method will assign roughly the same number points to both. However the equatorial strip is spread over a large area but the polar strip over a tiny area. So the points will not be uniformly distributed over the surface.
What one needs to keep track of is the ratio between the infinitesimal volume in polar coordinates dphi * dtheta to the infinitesimal of the surface area. In other words the amount of dilation or contraction. Then one has apply the reciprocal to even it out.
This tracking is done by the determinant of the Jacobian.
This gives an algorithm for sampling from a sphere: choose randomly from a cylinder and then project onto a sphere. In polar coordinates:
sample theta uniformly in (0,2pi)
sample y uniformly in (-1,1)
project phi = arcsin(y) in (-pi,pi)
polar coordinates (theta, phi) define describe random point on sphere
EDIT: Plotting it out as a point cloud seems to confirm your suspicion.
Without scaling: https://editor.p5js.org/spyrja/sketches/7IK_RssLI
Scaling fixed: https://editor.p5js.org/spyrja/sketches/kMxQMG0dj
https://observablehq.com/@jrus/stereorandom
At least when trying to end up with stereographically projected coordinates, in general it seems to be faster to uniformly generate a point in the disk by rejection sampling and then transform it by a radially symmetric function to lie on the sphere, rather than uniformly generating a point in the ball and then projecting outward. For one thing, fewer of the points get rejected because the disk fills more of the square than the ball fills of the cube.
[1] https://docs.scipy.org/doc/scipy/reference/generated/scipy.s...
I’d also point out that the usual way Gaussians are generated (sample uniformly from the unit interval and remap via the Gaussian percentile) can be expressed as sampling from a d-dimensional cube and then post-processing as well, with the advantage that it works in one shot. (Edit: typo)
The Box-Muller transform is slow because it requires log, sqrt, sin, and cos. Depending on your needs, you can approximate all of these.
log2 can be easily approximated using fast inverse square root tricks:
constexpr float fast_approx_log2(float x) {
x = std::bit_cast<int, float>(x);
constexpr float a = 1.0f / (1 << 23);
x *= a;
x -= 127.0f;
return x;
}
sqrt is pretty fast; turn `-ffast-math` on. (this is already the default on GPUs) (remember that you're normalizing the resultant vector, so add this to the mag_sqr before square rooting it)
The slow part of sin/cos is precise range reduction. We don't need that. The input to sin/cos Box-Muller is by construction in the range [0,2pi]. Range reduction is a no-op.
For my particular niche, these approximations and the resulting biases are justified. YMMV. When I last looked at it, the fast log2 gave a bunch of linearities where you wanted it to be smooth, however across multiple dimensions these linearities seemed to cancel out.
If you want fast sqrt (or more generally, if you care at all about not getting garbage), I would recommend using an explicit approx sqrt function in your programming language rather than turning on fastmath.
And what's the mathematical basis? (that is, is this technique formalized anywhere?)
It seems insane to me that you run Newton's algorithm straight on the IEEE 754 format bits and it works, what with the exponent in excess coding and so on
A lot of old-school algorithms like CORDIC went the same way.
There's a related technique to compute exponentials with FMA that's somewhat more useful in ML (e.g. softmax), but it has similar precision issues and activation functions are so fast relative to matmul that it's not usually worth it.
It turns out you do have an ultra fast way to compute log_2: you bitcast a float to an integer, and then twiddle some bits. The first 8 bits (after the sign bit, which is obviously zero because we're assuming our input is positive) or whatever are the exponent, and the trailing 23 bits are a linear interpolation between 2^n and 2^(n+1) or whatever. exp_2 is the same but in reverse.
You can simply convert the integer to floating point, multiply by -.5, then convert back to integer. But multiplying -.5 by x can be applied to a floating point operating directly on its bits, but it's more complicated. You'll need to do some arithmetic, and some magic numbers.
So you're bitcasting to an integer, twiddling some bits, twiddling some bits, twiddling some bits, twiddling some bits, and bitcasting to a float. It turns out that all the bit twiddling simplifies if you do all the legwork, but that's beyond the scope of this post.
So there you go. You've computed exp_2(-.5 log_2 x). You're done. Now you need to figure out how to apply that knowledge to the inverse square root.
It just so happens that 1/sqrt(x) and exp(-.5 log x) are the same function. exp(-.5 log x) = exp(log(x^-.5)) = x^-.5 = 1/sqrt(x).
Any function where the hard parts are computing log_2 or exp_2 can be accelerated this way. For instance, x^y is just exp_2(y log_2 x).
Note that in fast inverse square root, you're not doing Newton's method on the integer part, you're doing it on the floating point part. Newton's method doesn't need to be done at all, it just makes the final result more accurate.
Here's a blog here that gets into the nitty gritty of how and why it works, and a formula to compute the magic numbers: https://h14s.p5r.org/2012/09/0x5f3759df.html
You can't do the XOR in floating point representation, but you can if you do the entire algorithm in fixed point or if you retain the random bits before converting to a floating point value.
This decreases the number of random numbers that need to be generated from ~5.72 to ~3.88.
If you convert (non mathematician here!) your sphere into an n-agon with an arbitrarily fine mesh of triangular faces, is the method described by OP still valid. ie generate ...
... now I come to think of it, you now have finite faces which are triangular and that leads to a mad fan of tetrahedrons and I am trying to use a cubic lattice to "simplify" finding a series of random faces. Well that's bollocks!
Number the faces algorithmically. Now you have a linear model of the "sphere". Generating random points is trivial. For a simple example take a D20 and roll another D20! Now, without toddling off to the limit, surely the expensive part becomes mapping the faces to points on a screen. However, the easy bit is now the random points on the model of a sphere.
When does a triangular mesh of faces as a model of a sphere become unworkable and treating a sphere instead as a series of points at a distance from a single point - with an arbitrary accuracy - become more or less useful?
I don't think that will be an issue for IT - its triangles all the way and the more the merrier. For the rest of the world I suspect you normally stick with geometry and hope your slide rule can cope.
pavel_lishin•4d ago
Maybe; my first instinct is that there'll be some bias somewhere.
Maybe I'll have some time tonight to play with this in p5js.
pavel_lishin•4d ago
It looks reasonably random to my eye: https://editor.p5js.org/fancybone/sketches/DUFhlJvOZ
guccihat•3d ago
NoahZuniga•3d ago
pavel_lishin•3d ago
rkomorn•5h ago
cluckindan•6h ago
pavel_lishin•3d ago
spyrja•6h ago
sparky_z•3d ago
-First of all, it's intuitive to me that the "candidate" points generated in the cube are randomly distributed without bias throughout the volume of the cube. That's almost by definition.
-Once you discard all of the points outside the sphere, you're left with points that are randomly distributed throughout the volume of the sphere. I think that would be true for any shape that you cut out of the cube. So this "discard" method can be used to create randomly distributed points in any 3d volume of arbitrary shape (other than maybe one of those weird pathological topologies.)
-Once the evenly distributed points are projected to the surface of the sphere, you're essentially collapsing each radial line of points down to a single point on the sphere. And since each radial line has complete rotational symmetry with every other radial line, each point on the surface of the sphere is equally likely to be chosen via this process.
That's not a rigorous proof by any means, but I've satisfied myself that it's true and would be surprised if it turned out not to be.
pavel_lishin•3d ago
sparky_z•3d ago
You're correctly pointing out that the values of r won't be uniformly distributed. There will be many more points where the value of r is close to 1 then there will be where the value of r is close to 0. This is a natural consequence of the fact that the points are uniformly distributed throughout the volume, but there's more volume near the surface than there is near the center. That's all true.
But now look at the final step. By projecting every point to the surface of the sphere, you've just overwritten every single point's r-coordinate with r=1. Any bias in the distribution of r has been discarded. This step is essentially saying "ignore r, all we care about are the values of theta and phi."
[0]https://en.wikipedia.org/wiki/Spherical_coordinate_system
layer8•7h ago
Or in other words, if you take the “dotted” sphere out of the cube afterwards, you won’t be able to tell by the dots which way it was originally oriented within the cube.
bob1029•7h ago
What would be biased is if you inscribed a cube in the unit sphere. This would require additional transformations to create a uniform distribution. If you simply "throw away" the extra corner bits that aren't used, it won't affect the distribution.