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Development & Technical Discussion
Re: Pollard's kangaroo ECDLP solver
by
a.a
on
04/10/2021, 15:50:46 UTC
I have a conjecture:
The longest amount of consecutive 1 or 0 is sqrt(bits). So sqrt(120) = 10,95. So probably we can skip all numbes which have more than 11 consecutive 1 and 0.