Re: Is it a hard problem to solve

laicko wrote:
>I know how it deduce. if A(P,abP,aP) = bP, then A(aP, ba^{-1}aP,
>a^{-1}aP)=abP while a^{-1} = a^{phi(o(P))-1} (mod o(P)) .
>my problem is at least as hard as CDH
>
>and if CDH can be solved by algorithm B such that B(P,aP,bP) =abP,
>then we could use it to compute B(aP, baP,a^{-1}aP) =ba^{-1}aP=bP,
>CDH is at least as hard as my problem.
>
>is that right?

Looks right to me!
.

Relevant Pages

  • Re: Is it a hard problem to solve
    ... Suppose we have a magic algorithm A such that A= bP. ... >problem is at least as hard as solving the CDH problem. ... and if CDH can be solved by algorithm B such that B=abP, ...
    (sci.crypt)
  • Re: Is it a hard problem to solve
    ... > Suppose we have a magic algorithm A that on input P, abP, aP produces ... Then A= abP. ... I think you're saying the problem can be solved until the CDH is solved ... indistinguishability maybe think as more strict result of my problem. ...
    (sci.crypt)