Re: Is it a hard problem to solve

>Actually, that's not quite what I meant. Let me restate what I was trying
>to say. Suppose we have a magic algorithm A such that A(P,abP,aP) = bP.
>Then it follows that A(aP,bP,P) = abP (do you see why?), so it follows that
>algorithm A can be used to solve the CDH problem. Corollary: Solving your
>problem is at least as hard as solving the CDH problem.

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?

.

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