Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
MrFreeDragon
on 01/05/2020, 19:04:29 UTC
-snip-
We do have fast inversions. Inversion in additive group means: -P, and -P is pratically for free.
-snip-

For the whole group with range [0, order] yes it is free to make -P from P. But for the subgroup within a pecified range you need to make scalar operations.