From a340ef443e47d7c01bacdee7ca0d0138da4886f0 Mon Sep 17 00:00:00 2001 From: Tobias Eidelpes Date: Tue, 21 Jun 2022 19:14:04 +0200 Subject: [PATCH] Add solution for 5a --- exam/ex.tex | 14 +++++++++++++- 1 file changed, 13 insertions(+), 1 deletion(-) diff --git a/exam/ex.tex b/exam/ex.tex index d479167..91a422f 100644 --- a/exam/ex.tex +++ b/exam/ex.tex @@ -264,7 +264,19 @@ \item \textbf{(33 points)} \begin{enumerate} - \item \TODO + \item Let there be an adversary $\mathcal{A}$ which breaks CGI. We can then + construct an adversary $\mathcal{B}$ which breaks CGI2. + + Suppose $\mathcal{B}$ is given a CGI2 instance + $(\mathcal{G}_a,\mathcal{G}_b)$ where $a\neq b$ and $\mathcal{G}_a$ and + $\mathcal{G}_b$ are in the set of $2^{130}$ graphs isomorphic to + $\mathcal{G}$. The goal of $\mathcal{B}$ is to find an isomorphism $\phi$ + with non-negligible advantage such that $\mathcal{G}_a = + \phi(\mathcal{G}_b)$. $\mathcal{B}$ will give + $(\mathcal{G}_a,\mathcal{G}_b)$ to $\mathcal{A}$ and $\mathcal{A}$ will + output an isomorphism $\phi$ which satisfies $\mathcal{G}_a = + \phi(\mathcal{G}_b)$. $\mathcal{B}$ can then take this isomorphism and + apply it to its own problem to obtain the solution. \item \TODO