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Board Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
Jean_Luc
on 01/05/2020, 14:19:34 UTC
I coded the stuff as it is described in the paper using this:

Code:
T={{a,−a}:a∈[−N/2,N/2]},
W={{n+a,−(n+a)}:a∈[−N/4,N/4]

And a of translation of -N/2.G of the public key to solve in order to have as specified:

Quote
Each experiment involved choosing uniformly at random−N/2≤n≤N/2  and  solving the  DLP  for Q=  [n]P.

It found as expected from time to time the symmetric point.
It is slower than the classic version (~2^21.2 on 40bit search) and far from 1.46sqrt(n) Sad

But, they are not very clear on the jumps and especially the average distance of the jumps.
As the wild are on a shorter range, may be steps have to be different than tame's one (not yet tested).
I also didn't coded a stop and retry after a certain number of jump (sqrt(n) ?), i let the algorithm continue outside the range, so it may loose the symmetry...

To be continued...