Juraj Sukop added the comment:
Mark, thank you very much for the thorough reply!
I believe it is not for me to say whether the complication is worth it or not.
What really huge numbers goes, if I'm not mistaken, the original `isqrt` is
already "fast" and only constant speed
Juraj Sukop added the comment:
The rounding down of `l` might compute more than half of the bits so that the
final Heron' step in `isqrt_2` might correct the uncertain low bit if `a - (a*a
> n)` is missing from `isqrt`.
As it currently stands, `a - (a*a > n)` is computed both in
Juraj Sukop added the comment:
What the proof goes, you did most of the work already. Consider the following:
l = (n.bit_length() - 1)//4
a = isqrt(n >> 2*l)
a = ((a << l) + n//(a << l))//2
return a - (a*a > n)
This computes the square root of the (possib
New submission from Juraj Sukop :
This is a follow up to https://bugs.python.org/issue36887 and
https://bugs.python.org/issue36957 .
The new `isqrt` is remarkably simple but it does not split the number at hand
optimally. Ideally one would want to have 2n/n division everywhere but since
the
Juraj Sukop added the comment:
FWIW, there is a new implementation of FMA [1] which is licensed very
permissively [2]. Perhaps it could be used here as well..?
[1] https://github.com/smasher164/fma
[2]
https://github.com/smasher164/fma/commit/4e85d2388c7d4d850be12df918f9431ca687f57a
Juraj Sukop added the comment:
I would say because it has wide applicability, especially considering the
amount of code it adds. It is similar in spirit to `copysign` or `fsum` which
are already there and C99 includes it anyway. For simpler things like dot
product or polynomial evaluation
Changes by Juraj Sukop :
Added file: http://bugs.python.org/file46300/setup.py
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Changes by Juraj Sukop :
Added file: http://bugs.python.org/file46298/xmathmodule.c
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Changes by Juraj Sukop :
Added file: http://bugs.python.org/file46299/example.py
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New submission from Juraj Sukop:
Fused multiply-add (henceforth FMA) is an operation which calculates the
product of two numbers and then the sum of the product and a third number with
just one floating-point rounding. More concretely:
r = x*y + z
The value of `r` is the same as if the
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