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fixed typo

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This commit is contained in:
Aaron Kaiser
2023-03-02 10:46:40 +01:00
parent 69a34c2608
commit 91ec89c040
1 changed files with 2 additions and 2 deletions
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thesis/Abschlussarbeit.tex
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@@ -349,7 +349,7 @@ The adversary has to call the \ioracle oracle with a commitment $\groupelement{R
\label{eq:Adlog} \label{eq:Adlog}
\end{align} \end{align}
Assuming that $r_2 + 2^c c$ is invertable in $\field{L}$ (not equal to $0$) we can use both equations to calculate the discrete logarithm of $\groupelement{A}$. To ensure that $r_2 + 2^c c$ is invertable the reduction has to abort if $-r_2$ equals $2^c c$ with $c$ being randomly choosen in the \ioracle oracle. Assuming that $r_2 + 2^c c$ is invertible in $\field{L}$ (not equal to $0$) we can use both equations to calculate the discrete logarithm of $\groupelement{A}$. To ensure that $r_2 + 2^c c$ is invertible the reduction has to abort if $-r_2$ equals $2^c c$ with $c$ being randomly chosen in the \ioracle oracle.
\begin{figure} \begin{figure}
\hrule \hrule
@@ -452,7 +452,7 @@ Game $G_0$ is defined in Figure \ref{fig:igamewithabort} by ignoring all boxes.
To prove (\ref{eq:advbsdlog}), we define an adversary $\adversary{B}$ attacking \sdlog that simulates $\adversary{A}$'s view on $G_2$. Adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybsdlog} is run in the \sdlog game and has access to \ioracle. \ioracle is perfectly simulated. To prove (\ref{eq:advbsdlog}), we define an adversary $\adversary{B}$ attacking \sdlog that simulates $\adversary{A}$'s view on $G_2$. Adversary $\adversary{B}$ formally defined in figure \ref{fig:adversarybsdlog} is run in the \sdlog game and has access to \ioracle. \ioracle is perfectly simulated.
Finally, consider $\adversary{A}$ output $s^*$. If bad is not set $r_2 + 2^c c^*$ is invertable in $\field{L}$ and $(2^c s^* - r_1)(r_2 + 2^c c^*)^{-1}$ is the discrete logarithm of $\adversary{A}$ to the basis of $\adversary{B}$ as shown in (\ref{eq:Adlog}). Finally, consider $\adversary{A}$ output $s^*$. If bad is not set $r_2 + 2^c c^*$ is invertible in $\field{L}$ and $(2^c s^* - r_1)(r_2 + 2^c c^*)^{-1}$ is the discrete logarithm of $\adversary{A}$ to the basis of $\adversary{B}$ as shown in (\ref{eq:Adlog}).
\item This proves Theorem \ref{theorem:advgamez}. \item This proves Theorem \ref{theorem:advgamez}.
\end{proof} \end{proof}