29 Jan
2020
29 Jan
'20
4:21 p.m.
nisse@lysator.liu.se (Niels Möller) writes:
If we want to compute v = z^-1 (mod p), but in redc form with v' = vB and z' = zB, then we have
v z = 1 (mod p)
but
v' z' = B^2 (mod p)
So for redc curves we need to compute v' as
v' = (z' / B^2)^-1 (mod p)
I've pushed changes to do this to a new branch invert-with-redc. I think it makes things a bit simpler. I haven't yet looked at any benchmarks, but I'd hope to either see no change or an epsilon improvement.
Regards, /Niels
--
Niels Möller. PGP-encrypted email is preferred. Keyid 368C6677.
Internet email is subject to wholesale government surveillance.