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| ▲ | creata a day ago | parent | next [-] |
| https://en.wikipedia.org/wiki/Probability_density_function#V... Affine transformations just scale the probability density by a constant, so a uniform distribution is still uniform. |
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| ▲ | aaronblohowiak a day ago | parent [-] | | ah, excellent point. thanks. this works if they are transforming the triangle to be a right triangle to make a rectangle but if they are making it half a square, that could not be accomplished solely with affine transformations (if I figure correctly...) | | |
| ▲ | cluckindan 18 hours ago | parent [-] | | But you can just map one right triangle to the other without affecting the distribution. |
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| ▲ | o11c a day ago | parent | prev [-] |
| There's no distortion for the purposes of randomness (there would be if you cared about distance between specific points before/after the transformation), but the blog article fails to actually explain the method. Clicking through to SO explains it (but assumes you can read numpy). The `s * u` and `t * v` (where `u` and `v` are vectors) are the transformation from right-triangle (half of square) to triangle (half of parallelgram). |
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| ▲ | saltcured a day ago | parent [-] | | Hmm, is this "car(ing) about distance between specific points" something that would matter if we were going to operate in reverse and splat something at each point to render the triangle? I think that sort of splatting is how I intuitively think about this problem, and likely where I went wrong in thinking there would be a distortion of the density function. I'm not confident whether this actually matters though. Would it still approximate the same distribution regardless of what splat shape is used...? |
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