Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
arulbero
on 01/05/2020, 20:06:41 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.

If you shift the range [a,b] -> [-(b-a)/2, (b-a)/2], each point P in this interval has the property that -P is in interval too. No other operations are needed.