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(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?

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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).

On the other end, it is not clear that breaking p-BDHI could allow breaking the discrete logarithm problem, as, intuitively, breaking it only involve computing a group element and does not directly give a way to recover an exponent (this is informal as it could still be that p-BDHI allows too break the discrete logarithm problem, it does just seem far from obvious).

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  • $\begingroup$ I agree that they do not seem equivalent. But I do not buy your argument. Considering 1-BDHI, the obstacle is that you get an element in a different group. If you got s^-1 P instead and your oracle works for any P, you could break CDH (I think) which often allows you to break DL. $\endgroup$ – K.G. Jan 25 '17 at 21:14
  • $\begingroup$ @Geoffroy Couteau. I didn't get your idea why p-BDHI is stronger than DL. $\endgroup$ – myat Jan 26 '17 at 6:26

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