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Re:Higher Quality Random Floats by NeutralForest on Wednesday October 18, 2023 @07:13AM (#37927108)
what does subnormal mean?
Re:Higher Quality Random Floats by camel-cdr on Wednesday October 18, 2023 @07:22AM (#37927175)
Usually floating point numbers are defined as `1.fraction * 2^(exponent-bias)`, but if the exponent is zero `0.fraction * 2^(1-bias)` is used, to allow representing very small numbers.

IIRC I also explained it here: https://m.youtube.com/watch?v=VHJUlRiRDCY&t=2774s

Re:Higher Quality Random Floats by NeutralForest on Wednesday October 18, 2023 @07:29AM (#37927224)
thanks!
Re:Higher Quality Random Floats by Aardwolf on Wednesday October 18, 2023 @08:30AM (#37927668)
And the reason subnormals exist at all is that normally there's an implicit 1-bit (without the implicit 1-bit, there would be multiple representations of the same value, e.g. all values with mantissa 0 would represent 0 regardless of exponent), but you could never represent the value 0 itself if the implicit 1-bit is always there. So for the lowest exponent value, the implicit bit is set to 0, to allow representing 0, but also those other very tiny values since there's still an entire mantissa to give meaning to in addition to value 0.
Re:Higher Quality Random Floats by magicalhippo on Wednesday October 18, 2023 @10:21AM (#37929104)
Subnormals or denormalized floats can lead to some unexpected results. IIRC dividing a subnormal by a subnormal returns infinity.

I experienced this where a colleague had tried to reduce the number of divisions by rewriting (a/x)*(b/y)*(c/z) to (a*b*c)/(x*y*z).

The problem was that in certain circumstances the terms were all small, but of comparable magnitude. Thus the latter lead to subnormal divided by subnormal. The fix was to undo the optimization and do the divisions first, which lead to the multiplication of three numbers of order 1.

Re:Higher Quality Random Floats by bee_rider on Wednesday October 18, 2023 @11:19AM (#37929908)
Yeah, I think a good general guideline is that if you see a bunch of divisions like that, the author did it intentionally (to avoid overflow or subnormals).

Anyway, I wonder if the optimization even helps at all. Either way the critical path is a division and two multiplications. That seems like the sort of thing that, between SIMD an OoO execution, a modern CPU ought to be able to work out, haha.

No promises but I’d want to benchmark it before caring.

Re:Higher Quality Random Floats by Aardwolf on Wednesday October 18, 2023 @11:55AM (#37930527)
Maybe depends on the platform? I tried 5e-323 / 6e-323 in javascript console and it returns 0.8333333333333334 (and e-323 should be subnormal for double precision)
Re:Higher Quality Random Floats by leiroigh on Wednesday October 18, 2023 @01:08PM (#37931587)
Depends on control registers like e.g. MXCSR. It's an utter mess. Consider e.g. https://news.ycombinator.com/item?id=32738206
Re:Higher Quality Random Floats by thethirdone on Wednesday October 18, 2023 @03:40PM (#37933715)
> IIRC dividing a subnormal by a subnormal returns infinity.

This should not be true in a conformant implementation of IEEE 754-2008. You can get infinity by dividing by a subnormal (due to overflow), but that should only be possible with sufficiently large numerators.

> The problem was that in certain circumstances the terms were all small, but of comparable magnitude. Thus the latter lead to subnormal divided by subnormal. The fix was to undo the optimization and do the divisions first, which lead to the multiplication of three numbers of order 1.

My guess is that this problem was actually the numerator being subnormal and the denominator underflowing all the way to zero and therefore the result going to infinity.

Re:Higher Quality Random Floats by magicalhippo on Wednesday October 18, 2023 @05:22PM (#37934955)
It's been over a decade, so I struggle to recall the details.

> My guess is that this problem was actually the numerator being subnormal and the denominator underflowing all the way to zero and therefore the result going to infinity.

Yes thinking about you're probably correct.