3 ms·
>Let me know if I am wrong. You're wrong on multiple levels. 1. Generating random values for x^2, y^2, z^2 and taking their square root will only give you val
by machiavelli1024 8y ago
>Let me know if I am wrong.
You're wrong on multiple levels.
1. Generating random values for x^2, y^2, z^2 and taking their square root will only give you values in the x,y,z > 0 octant.
(But let's say you "fix" this by randomly multiplying them with -1.)
2. Taking the square root of a uniformly distributed random variable is no longer uniformly distributed.
3. Randomly reordering the coordinates won't fix your bias.
Here's a demonstration in 2D: https://jsfiddle.net/tz85wnxy/59/ https://jsfiddle.net/tz85wnxy/59/
Uncomment line 17, 18, 20 to see how it's still not uniform even if you randomly multiply the coordinates by -1 and reorder them.
- goldenkey 8y agoWorks fine as long as you generate x,y,z randomly and not their squares. Same process but minor tweak. https://jsfiddle.net/w8zLvsy2/ https://jsfiddle.net/w8zLvsy2/
- machiavelli1024 8y agoNope. You still have clustering on the sides. A proper uniform distribution would produce the same grey shade all over the circle, like this: https://jsfiddle.net/tz85wnxy/86/ https://jsfiddle.net/tz85wnxy/86/ Yours have darker areas on the sides.
- cvoss 8y agoThis approach is still incorrect, and the bias toward the set of 4 points S = {(1,0),(-1,0),(0,1),(0,-1)} can be seen in the diagram. There is no way to break out of this bias by flipping or reordering coordinates because those two operations preserve the symmetries of S: - if you flip the x-coord of a point near S, e.g. (-.9,0), you still get a point near S, e.g. (.9,0) - if you swap the x-coord and y-coord of a point near S you still get a point near S. - etc.