# Is the $\ell$-Diffie Hellman Inversion easy when g is known?

From here they define the $\ell$-Diffie Hellman inversion problem as:

Given $g^{a},g^{a^2}\ldots,g^{a^{\ell}} \in G$, compute $g^{a^{-1}}$

Would this problem become easy if the generator $g$ is also known?

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No.

The stated $l$-DHI problem is believed to remain hard when $g$ is known.

Actually, in the quoted page, it is assumed that $g$ is known throughout. This is obvious in particular in the sections on DLP and CDH.

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I missed that bit at the top of the page. I got confused because some of the other assumptions list g to be explicitly known. –  blz Mar 1 '12 at 6:04
@blz: You did not miss that bit - it was missing, I added it. –  fgrieu Mar 1 '12 at 9:08

Do you know about the paper Variations of Diffie-Hellman Problem (PDF)? The problem you stated is the generalization of the inversion problem stated in that paper. You can use their technique to prove the relation.

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I fail to locate the paper. –  fgrieu Feb 29 '12 at 16:30
Thanks for pointing. I corrected the link. –  Jalaj Feb 29 '12 at 16:43
the link is now perfect, thanks. –  fgrieu Mar 1 '12 at 9:08

No. The generator $g$ is a public parameter of the group $G$. You cannot perform a Diffie-Hellman handshake unless both parties agree on the generator (as well as any other parameter that defines $G$), so naturally any variation of the Diffie-Hellman problem must, by definition, assume the same thing.

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