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Dlog' => GmaeZ: used min entropy

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Aaron Kaiser
2023-03-02 10:35:55 +01:00
parent 851033a324
commit 881bb0c128
1 changed files with 3 additions and 6 deletions
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9
thesis/Abschlussarbeit.tex
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@@ -352,7 +352,6 @@ The adversary has to call the \ioracle oracle with a commitment $\groupelement{R
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 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.
\begin{figure} \begin{figure}
% TODO: set caption
\hrule \hrule
\begin{multicols}{2} \begin{multicols}{2}
\large \large
@@ -398,11 +397,9 @@ Game $G_0$ is defined in Figure \ref{fig:igamewithabort} by ignoring all boxes.
% TODO: hard bezüglich ggen % TODO: hard bezüglich ggen
% TODO: min entropy von {0,1}^{2b} mod L? % TODO: min entropy von {0,1}^{2b} mod L?
\[ \advantage{\igame}{\adversary{A}} \leq \advantage{\sdlog}{\adversary{B}} - \frac{\oraclequeries}{L} \] \[ \advantage{\igame}{\adversary{A}} \leq \advantage{\sdlog}{\adversary{B}} - \frac{\oraclequeries}{2^{\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}} \]
\end{theorem} \end{theorem}
TODO: vielleicht doch eher $\oraclequeries$ durch min entropy von $\{0,1\}^{2b} \pmod L$?
\begin{proof} \begin{proof}
\item \paragraph{\underline{$G_0$:}} Let $G_0 \assign \igame$ be \igame. By definition, \item \paragraph{\underline{$G_0$:}} Let $G_0 \assign \igame$ be \igame. By definition,
@@ -413,9 +410,9 @@ TODO: vielleicht doch eher $\oraclequeries$ durch min entropy von $\{0,1\}^{2b}
\[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1] \] \[ \Pr[G_0^{\adversary{A}} \Rightarrow 1] = \Pr[G_1^{\adversary{A}} \Rightarrow 1] \]
% TODO: wählen von % TODO: wählen von
\item \paragraph{\underline{$G_2:$}} Game $G_2$ aborts if the flag bad is set. For each individual \ioracle query the bad flag is set with probability at most $\frac{1}{L}$, since $c$ is chosen from $\field{L}$ uniformly at random. By the Union bound over all $\oraclequeries$ queries we obtain $\Pr[bad] = \frac{\oraclequeries}{L}$. Since $G_1$ and $G_2$ are identical-until-bad games, we have \item \paragraph{\underline{$G_2:$}} Game $G_2$ aborts if the flag bad is set. For each individual \ioracle query the bad flag is set with probability at most $\frac{1}{2^{\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$, since $c$ is chosen from $\{0,1\}^{2b}$ uniformly at random and then reduced modulo $L$. By the Union bound over all $\oraclequeries$ queries we obtain $\Pr[bad] = \frac{\oraclequeries}{2^{\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}}$. Since $G_1$ and $G_2$ are identical-until-bad games, we have
\[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries}{L} \] \[ |\Pr[G_1^{\adversary{A}} \Rightarrow 1] - \Pr[G_2^{\adversary{A}} \Rightarrow 1]| \leq \Pr[bad] \leq \frac{\oraclequeries}{2^{\log_2(\lceil \frac{2^{2b} - 1}{L} \rceil 2^{-2b})}} \]
\item Finally, Game $G_2$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying \item Finally, Game $G_2$ is well prepared to show that there exists an adversary $\adversary{B}$ satisfying