Post
Topic
Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
Etar
on 16/06/2020, 12:40:55 UTC

second interval[/u]  = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,....,27,28,29,30,31]  (5 bit)
you want to find P = 11*G, you know only that the P lies in this interval

instead of searching P in the second interval [1*G, 2*G, ...., 31*G]

you search P'=inv(4)*P (a different point) in the

third interval [1*G', 2*G', ......, 31*G']  (that contains the first interval but not at the beginning + P')

The old DPs are now: 8*G' and 20*G' (same points, different private keys). All the points of the first interval are now in the third one too.

The private key of P' = 11, in fact 11*G' = 11*inv(4)*G --> P = 4*P' = 4*11*inv(4)*G = 11*G

Perfect explanation. Thanks!