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    Author Topic: Pollard's kangaroo ECDLP solver  (Read 62387 times)
    WanderingPhilospher
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    May 29, 2020, 06:28:44 PM
     #521

    -snip-
    PS. Actually instead of solving pk110 within the range [2^109 ... 2^110-1] the program solves the key pk110s within the range [0 ... 2^109-1]
    Thanks for the detailed answer. This is very clear now.
    Can you help make it clear to me Smiley

    So the tames work from range 0...meaning 0x0 to 0x3FFFFFFFFFFFFFFFFFFFFFFFFFFFF  (for pubkey 110)?

    The range is shifted, not expanded. Shift means the move of start and end values both by the SAME value.
    Instead of range [2000000000000000000000000000 ... 3fffffffffffffffffffffffffff] the program works with range [0 ... 1fffffffffffffffffffffffffff]

    PS. Do you really like these hex values posted here? They are difficult to understand by a human (as difficult to count "f" and "0" signs). So i represented them earlier by the powers of 2 (2^109 and 2^110) which is much easier to understand.
    I'll have to digest this Smiley the SAME value is throwing me off a little. It makes sense by what you said; you subtracted the 20000....from both values to get 0 and the 1ffff...
    What is the advantage or purpose of shifting?

    As far as ranges, I'm good with either, hex or powers, just wanted to keep it simple (for me, maybe for others) since we input a range start and range end on the input file versus a -bits flag.
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