paul zimmermann writes:
> if you need to save \beta with respect to the proof of [4], yes maybe you need
> to repeat that proof to explain how you save the extra +1.
I think we can make it work. We have the reciprocal v, and a
corresponding "remainder"
K = \beta^3 - {d_1, d_0} (\beta + v)
in
Dear Niels,
it works with r = (u_0 - q d_1 - p_1 - 1) \bmod \beta line 6 in all
cases, assuming it works with -1 replaced by - [p_0 > 0].
We only need to check the case p_0 = 0.
p_0 = 0 means that q d_0 is divisible by \beta, i.e., R' is multiple of \beta.
Let still be the two low words
Dear Niels,
> > page 1: the division instruction is now much faster than before on modern
> > processors
>
> According to https://gmplib.org/~tege/x86-timing.pdf, they're still an
> order of magnitute slower than multiplication. E.g 86 vs 3 cycles on
> Intel skylake. And in addition,
