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Say we are able to decrypt a Elgamal ciphertext $c$ using only the public key. Apparantly it is now possible to solve the Diffie-Hellman problem (given $g^a, g^b$ calculate $g^{ab}$). How?

I know how it works the otherway. When I'm able to solve DH, I found a way to decrypt Elgamal ciphertext. But I can't come up with a good idea for this problem.

I figured since $c=(c_1,c_2)=(v,w)$ maybe $c_1=g^a, c_2=g^b$ such that the plaintext $p=g^{ab}$, but that doesn't work out.

I'm thankful for any tips.

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  • $\begingroup$ AFAIK ElGamal only reduces to the decisional DH problem, not the the computational DH problem. So this proof shouldn't exist $\endgroup$ Commented Jul 4, 2014 at 15:05
  • $\begingroup$ What's the difference between the decisional DH problem and the computational one? $\endgroup$ Commented Jul 4, 2014 at 15:33
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    $\begingroup$ Decisional is: given $g, g^a, g^b, g^c$, decide whether $g^c = g^{ab}$. Computational is: given $g, g^a, g^b$, compute $g^{ab}$. $\endgroup$
    – fkraiem
    Commented Jul 4, 2014 at 18:28
  • $\begingroup$ Well, apparantly CodesInChaos was wrong (see answer below). $\endgroup$ Commented Jul 4, 2014 at 19:09

1 Answer 1

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Since you're learning, I won't go ahead and give you the answer; I will try to point you in the right direction.

Consider what Elgamal decryption is: if the private key is $a$ (and hence the public key is $g^a$), and you're given a ciphertext $(b, c)$, the ElGamal decryption is $b^{-a} \cdot c$

Now, assume that we have an Oracle that, given an arbitrary public key $g^x$, and a ciphertext $(g^y, z)$, would return the ElGamal decryption. Can you see how you could use that by passing it a public key and ciphertext of your choosing (hint: write out what the decryption would look like in that case)?

(Further hint: given $g^x$, it is easy to compute $g^{-x}$, as that's just the multiplicative inverse)

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  • $\begingroup$ Decryption: $b^{-x} \cdot z = (g^{xy})^{-1} \cdot z = (g^{xy})^{-1} \cdot h^r \cdot p$ with $h$ is public key, $r$ random number, $p$ plaintext. Is that correct? This looks good, but don't I just want $g^{xy}$ (or $(g^{xy})^{-1}$ for that matter)? How can I eleminate the other factors without knowing $r$ and $p$? $\endgroup$ Commented Jul 4, 2014 at 15:53
  • $\begingroup$ @CGFoX: stop at $(g^{xy})^{-1} \cdot z$; do you know the value $z$? $\endgroup$
    – poncho
    Commented Jul 4, 2014 at 16:07
  • $\begingroup$ Oh, of course. Since I know $c$ I also know $z$ and can easily calculate $z^{-1}$. Thank you! $\endgroup$ Commented Jul 4, 2014 at 18:32
  • $\begingroup$ Could I just set $z$ to 1? $\endgroup$ Commented Jul 5, 2014 at 16:33
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    $\begingroup$ @CGFoX: well, $z$ is from the ciphertext, and you get to pick that, so of course you can... (I almost mentioned that myself; I'm glad I didn't; it's better that you thought of it) $\endgroup$
    – poncho
    Commented Jul 5, 2014 at 17:11

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