Would either $\alpha$ or $r$ suffice as challenge?

I am aware that the signature and verification were to adapt. However, what is the motivation behind using two challenges?

DSA Identification Scheme According to "Introduction to Modern Cryptography" by Katz & Lindell


1 Answer 1


Assume the verifier is honest so that the adversary wants to fool the verifier using a previous transcript.

$\alpha$ is necessary. If not, then let $\alpha=0$. The adversary gets $g^k, r, s = k^{-1}xr$ and computes $k^{-1}x$. Then, for some random $t$ it computes $I' = g^{tk}, u' = (tk)^{-1}x$ which can fool the verifier for any random challenge $r'$ by sending $I'$ and $s' = u'r'$.

I am unsure if $r$ is necessary. Please post the full context.

  • $\begingroup$ Without r, if Eve can get the prover to use the same k to sign two different values of alpha then she can recover k, and hence x, the private key. If r is included, this doesn't work. I am trying to find a solution for how alpha lets the private key be recovered from a single signature, or similar, but this is potentially also a good reason. $\endgroup$
    – whatf0xx
    Commented Aug 4, 2023 at 11:42
  • $\begingroup$ I agree with that the attack works when "use the same k" twice. However, this is unnatural since the prover is partly compromised and $k$ is the only randomization of the prover. $\endgroup$
    – xacid
    Commented Aug 17, 2023 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.