# Which assumption is stronger (p-BDHI) and DL?

(p-BDHI): p-Bilinear Diffie-Hellman inversion problem

Given $P,sP,s^2P,...,s^pP$. Find $e(P,P)^{\frac{1}{s}}$.

(DL): discrete logarithm problem

Given $P,sP$. Find $s$.

To break (p-BDHI), attacker needs to break DL.

I think the two assumptions have same-level hard problem.

No one is stronger than another. Is it right?

p-BDHI is clearly at least stronger than DL: if you can break the DL problem, you can recover $s$, and then compute $e(P,P)^{1/s}$ (if you are in a group where inversion can be computed efficiently, which I think is always the case in known pairing groups).