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Add solution for 5f

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Tobias Eidelpes 2022-06-21 12:59:54 +02:00
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exam/ex.tex
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@ -276,7 +276,14 @@
$G=\phi_{ch}^{-1}(G_{ch})$ and therefore $G'=\psi(\phi_{ch}^{-1}(G_{ch}))$ $G=\phi_{ch}^{-1}(G_{ch})$ and therefore $G'=\psi(\phi_{ch}^{-1}(G_{ch}))$
so the verifier will always accept. so the verifier will always accept.
\item \TODO \item Suppose $G_{ch}$ is not isomorphic to $G$. $\mathcal{P}$ prepares in
advance for a challenge $ch^*$ and so
$G'=\psi(\phi_{ch^*}^{-1}(G_{ch^*}))$. $\mathcal{P}$ commits to $G'$. If
the challenge by $V$ is $ch^*$ (so $ch=ch^*$), $\mathcal{V}$ accepts,
otherwise it rejects. Because $ch\in\{0,\dots,2^{130}-1\}$, the
probability that $\mathcal{P}$ convinces $\mathcal{V}$ is
$\frac{1}{2^{130}}$ (soundness error).
\item \TODO \item \TODO