2 ms·

> It's always going to be faster to skip step 2. What are we gaining by not skipping it? Having any number be possible is nice. Having the entire space be rea

by mlyle 1y ago

> It's always going to be faster to skip step 2. What are we gaining by not skipping it? Having any number be possible is nice. Having the entire space be reachable when you're sampling a solution space and the solution may be smaller than 1/8. Or having small vector quantities not have their angles ridiculously constrained by quantization. Most applications don't need it-- indeed, a double has a lot of precision compared to what most people care about, and throwing away some of it at the low end of the randomness range is no big deal. But sometimes, a double is barely (or not even) enough.

thaumasiotes 1y ago
> Having the entire space be reachable when you're sampling a solution space and the solution may be smaller than 1/8. You're out of luck there; no matter what happens you can only generate floating point numbers. If you're worried that the solution lies within the space of floating point numbers, but not within the part that's reachable by generating uniform samples with 1/2^53 resolution, you need to be much, much, much more worried that the solution isn't a floating point number at all. By contrast, if you're unconcerned with the numeric properties of the numbers you're generating, and what you want is to sample "floating point values" to examine what kind of implementation-specific behavior they might have, you would never want this scheme - you'd want to generate random floating-point values, which you can do by just filling in random bits. Those will be concentrated around zero, which doesn't matter because you're not interpreting them as numbers.
- mlyle 1y ago
  > You're out of luck there; no matter what happens you can only generate floating point numbers. If you're worried that the solution lies within the space of floating point numbers, but not within the part that's reachable by generating uniform samples with 1/2^53 resolution, you need to be much, much, much more worried that the solution isn't a floating point number at all. 2^53 is 15 sigfigs. My answer might be around 10 pico-something and I might need a few digits of precision. Then I've thrown away 11 of my sigfigs and there may not be a numerically stable solution with what's left. As you point out, double precision often isn't enough. But it's much more likely to not be enough if we throw large parts of the precision away.

by mlyle 1y ago

thaumasiotes 1y ago

> Having the entire space be reachable when you're sampling a solution space and the solution may be smaller than 1/8. You're out of luck there; no matter what happens you can only generate floating point numbers. If you're worried that the solution lies within the space of floating point numbers, but not within the part that's reachable by generating uniform samples with 1/2^53 resolution, you need to be much, much, much more worried that the solution isn't a floating point number at all. By contrast, if you're unconcerned with the numeric properties of the numbers you're generating, and what you want is to sample "floating point values" to examine what kind of implementation-specific behavior they might have, you would never want this scheme - you'd want to generate random floating-point values, which you can do by just filling in random bits. Those will be concentrated around zero, which doesn't matter because you're not interpreting them as numbers.

mlyle 1y ago
> You're out of luck there; no matter what happens you can only generate floating point numbers. If you're worried that the solution lies within the space of floating point numbers, but not within the part that's reachable by generating uniform samples with 1/2^53 resolution, you need to be much, much, much more worried that the solution isn't a floating point number at all. 2^53 is 15 sigfigs. My answer might be around 10 pico-something and I might need a few digits of precision. Then I've thrown away 11 of my sigfigs and there may not be a numerically stable solution with what's left. As you point out, double precision often isn't enough. But it's much more likely to not be enough if we throw large parts of the precision away.