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Thomas Johnson 11 months ago
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b1841c6e73
1 changed files with 2 additions and 2 deletions
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      presentation.tex

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@@ -42,7 +42,7 @@ Given $n = pq$, where $p$ and $q$ are prime:
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 \begin{itemize}
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 \item<2-> Pick a random element $a \in \mathbb{Z}/n\mathbb{Z}$.
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-\item<3-> Compute the order $r$ of $a$ using the oracle, so that $a^r \equiv 0\ (\mathrm{mod}\ n)$.
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+\item<3-> Compute the order $r$ of $a$ using the oracle, so that $a^r \equiv 1\ (\mathrm{mod}\ n)$.
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 \item<4-> If $r$ is odd or $a^{\frac{r}{2}} \equiv -1\ (\mathrm{mod}\ n)$, restart the procedure.
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 \item<5-> Let $s \equiv a^{\frac{r}{2}}\ (\mathrm{mod}\ n)$. Compute $s + 1$ and $s - 1$. One of these will be a factor of $n$.
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 \end{itemize}
@@ -121,7 +121,7 @@ which is true when the string $b$ is lexicographically after $w$. This can be im
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 \frametitle{Proof (continued...)}
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 At last, consider, given some pivot string $w$ in the $2^k$ space of strings, the formula:
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 $$
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-EncryptsToC(X) && LexicographicallyAfterW(X)
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+EncryptsToC(X) \&\& LexicographicallyAfterW(X)
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 $$
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 Applying the SAT oracle to this formula will tell us if there is a bitstring in the upper half of the search space that is the desired plaintext string. We can run binary search using this, and acquire the desired plaintext in linear time.
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 \end{frame}

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