Diffie Hellman Problem (DHP) and Discrete Logarithmic Problem (DLP) I am novice here. I am following a lecture series of Prof. Christof Paar and in his lecture of DLP, he states a fact which I did not understand at all. Can somebody please explain it better to me.I understand DLP and DHP independently.
The statement goes as follow:

"If the only way of solving the DHP requires the DLP, one would say that, the DHP is equivalent to the DLP" However this is not proven yet.

Also he says that if somebody proves it, it would be ground breaking research.
DH for reference:
  
  
  Alice takes a random $a$ and sends $g^a$ to Bob.
  Bob picks a random $b$ and sends $g^b$ to Alice.
  Both compute $g^{ab}$ Alice as $(g^b)^a$, Bob as $(g^a)^b$.
  


As you already explained breaking DLOG will break DH in the following way:

An attacker can e.g.


take $g^a$ and compute $a$ by solving the DLOG problem.
use $g^b$ to compute $(g^b)^a = g^{ab}$.


At this point the attacker knows the shared secret $g^{ab}$ and DH is broken.

However what about the other way around? Does breaking DH mean that we can compute DLOGs? The situation is not so clear...

Breaking DH means that the attacker is given $g, g^a, g^b$ and is able to find the shared secret $g^{ab}$. 

The question is can we somehow leverage this knowledge of $g^{ab}$ to find $a$ or $b$.


If yes, DH and DLOG would be equivalent because breaking one always means breaking the other.
If no, DLOG would be strictly harder than DH because we can break DH without breaking DLOG but not the other way around.