Re: enhanced diffie-helman

From: Kristian Gjøsteen (kristiag+news_at_item.ntnu.no)
Date: 08/21/05 

Date: Sun, 21 Aug 2005 20:02:14 +0000 (UTC)

Anna Jansen <ajansen77_spamfree@gmx.de> wrote:
>again a question, hopefully you can help me again:

I can try, but it's late at night now.

>Given: g^a, g^b, g^(ac), g^(abc).

Note that g^c does not appear.

Provided a and c are uniformly and indepently distributed among the
non-zeros, then ac is uniformly distributed and independent of a,
and you may as well rephrase this as: g^a, g^b, g^c, g^{bc}.

>Is it hard to compute: g^{ab}?
>(call this EDH)
>
>Is this equal to the DH-Assumption? Sure it is easy to show, that:
>EDH => DH

Yep.

>But what do you think of DH => EDH? Can I show this, too?

Yep, with the above reformulation. 

-- 
Kristian Gjøsteen