# Weak Decisional Diffie-Hellman Problem

Is this problem still hard?

Given $$(g,g^a,g^b,c)$$ decide if $$c=a\cdot b$$?

If there is an adversary that solves the standard Decisional Diffie-Hellman Problem then it can solve my new problem. But I can't understand that my new problem still hard or not.

Did anyone see this problem or similar to my problem? Can anyone help me?

• Feb 18 at 8:10
• @kelaka This doesn't answer the question. Indeed, as the question states - it is clear that if DDH is easy then so is this. However, this is not DL or CDH or DDH since the actual value $c=a\cdot b$ is given, and not $g^c$. Feb 18 at 9:08
• @YehudaLindell Uh, I read incorrectly, Feb 18 at 9:56
• As an aside, I would note that the similar problem "given $(g, g^a, g^b, c)$ is $c = a / b$" turns out to be easy... Feb 19 at 20:43
Yes it is. It can be formally reduced to the hardness of the decisional square Diffie-Hellman assumption, which states that distinguishing $$(g,g^a,g^{a^2})$$ from random is hard (this is a well established assumption).
• Thank you very much for your answer. I think that my problem reduced to the decisional inverse Diffie-Hellman assumption, which states that distinguishing $(g,g^a,g^{a^{-1}})$ from random. Feb 20 at 3:09
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