Prime Algorithms
Författare:
Nadya
Last Updated:
för 9 år sedan
Licens:
Creative Commons CC BY 4.0
Sammanfattning:
Primality Tests and Factoring Algorithms
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Primality Tests and Factoring Algorithms
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% TITLE PAGE
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\title[Prime Algorithms]{Primality Tests and Factoring Algorithms} % The short title appears at the bottom of every slide, the full title is only on the title page
\author{Joel Barnett and Nadya DeBeers} % Your name
\institute[SRJC] % Your institution as it will appear on the bottom of every slide, may be shorthand to save space
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Santa Rosa Junior College \\ % Your institution for the title page
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%\textit{john@smith.com} % Your email address
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\date{\today} % Date, can be changed to a custom date
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% PRESENTATION SLIDES
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\section{Fermat's Primality Test} % Sections can be created in order to organize your presentation into discrete blocks, all sections and subsections are automatically printed in the table of contents as an overview of the talk
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%\subsection{Subsection Example} % A subsection can be created just before a set of slides with a common theme to further break down your presentation into chunks
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\section{Application}
\begin{frame}{Application}
\frametitle{Application}
\begin{itemize}
\item It's fun, right?
\end{itemize}
\medskip
\medskip
The "real" mathematics of the "real" mathematicians, the mathematics of Fermat and Euler and Gauss and Abel and Riemann, is almost wholly "useless." ...It is not possible to justify the life of any genuine professional mathematician on the ground of the "utility" of his work. -— G. H. Hardy, \textit{A Mathematician’s Apology}, 1941
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\begin{frame}{Application}
\begin{itemize}
\item Encryption systems rely on these "useless" number theories developed by Fermat and Euler.
\item Primes are the basis of encryption security
\end{itemize}
\end{frame}
%--------------------------------------
\begin{frame}{Primality tests}
\begin{itemize}
\item Simple Primality Tests
\item Fermat's Primality Algorithm
\begin{itemize}
\item Modulo Arithmetic
\item Fermat's Little Theorem
\item The algorithm
\item Flaws
\end{itemize}
\item Rabin-Miller Primality Test
\end{itemize}
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\begin{frame}{Simple Primality Test}
Prime or Composite?
\begin{enumerate}
\medskip
\item 511
\item 73
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\begin{frame}{Simple Primality Test}
Prime or Composite?
\begin{enumerate}
\medskip
\item 511 Composite
\\
$\frac{511}{2}=2$ \hfill $\frac{511}{3}=170.33$\hfill $\frac{511}{5}=102.2$\hfill . . .\hfill $\frac{511}{7}=73$
\item 73
\end{enumerate}
\end{frame}
%--------------------------------------
\begin{frame}{Simple Primality Test}
Prime or Composite?
\begin{enumerate}
\medskip
\item 511 Composite
\item 73
\\
$\frac{73}{2}=36.5$ \hfill $\frac{73}{3}=24.33$\hfill $\frac{73}{5}=18.25$\hfill $\frac{73}{7}=10.43$\hfill $\frac{73}{8}=9.13$\hfill$\frac{73}{9}=8.11$
\\
\medskip Note: We stop at $\lceil{}\sqrt[]{73}\rceil{}=\lceil{} 8.544 \rceil{}=9$
\end{enumerate}
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\begin{frame}{Simple Primality Test}
This is a long process!
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\begin{frame}{Simple Primality Test}
This is a long process!
\medskip
\\Testing a 400 digit number ($10^{400}$) requires checking approximately ($\sqrt[]{10^{400}}10^{200}$) factors!
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%--------------------------------------
\begin{frame}{Simple Primality Test}
This is a long process!
\medskip
\\Testing a 400 digit number ($10^{400}$) requires checking approximately ($\sqrt[]{10^{400}}10^{200}$) factors!
\medskip
\\Scientists estimate the life of the universe to be only $10^{18}$ seconds, rendering this process impractical.
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\begin{frame}{Fermat's Method}
New Strategy!
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\begin{frame}
\frametitle{Modulo}
Background\\
\begin{block}{Modulo}
\underline{Definition}: Let \textit{m} and \textit{n} be integers and let \textit{d} be a positive integer. We say that \textit{m} is congruent to \textit{n} modulo \textit{d} if, and only if, \textit{d} divides \textit{m-n} and write:\\
\medskip
$m\equiv{n}\mod{d}\iff d |(m-n)$\\
\medskip
Sometimes this is written in the form:\\
$m\equiv n \mod d \iff m \mod{d}=n\mod{d}$
\medskip
\end{block}
Example:\\
$15\equiv 8\mod 7$ because $7|(15-8)$\\
Or $15\equiv 8\mod 7$ because $15\mod{7}=1$ and $8\mod{7}=1$
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\begin{frame}{Fermat's Method}
\begin{block}{Fermat's Little Theorem}
If \textit{p} is a prime number, then, $a^p\equiv a\mod{p}, \forall a\in \mathbb{Z}$.\\
\medskip
This theorem is also often rearranged to the form:\\
\medskip $a^{p-1}\equiv 1\mod{p}$
\medskip
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\begin{frame}{Fermat's Method}
Shortened Proof:
\begin{block}{Fermat's Little Theorem}
If \textit{p} is a prime number, then, $a^p\equiv a\mod{p}, \forall a\in \mathbb{Z}$.\\
\medskip
\end{block}
\underline{Proof} (by induction): Let \textit{p} be any prime number.\\
\smallskip
Base Case: $1^{p} \equiv 1 \mod{p}\iff p|(1^p-1) $, which is true.\\
\smallskip
Inductive Assumption: Assume $k^p\equiv k\mod{p}$ for some integer \textit{k}.\\
That is, assume $p|(k^p-k)$ is true.
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%--------------------------------------------------
\begin{frame}{Fermat's Method}
That is, assume $p|(k^p-k)$ is true.\\\textit{[We must show the (k+1) case follows]}.\\
That is, show $(k+1)^p\equiv (k+1)\mod{p}$\\
$\iff p|(k+1)^p-(k+1)$ \\
$\iff(k+1)^p-(k+1)=p*q, \exists q \in \mathbb{Z}$\\
Using the binomial theorem, the left hand side of above equation becomes:\\$k^p+\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k+1-(k+1)=k^p-k+[\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k]$\\Now,$ \binom{p}{j}=\frac{p!}{j!(p-j)!}$ where \textit{j} is some integer. Since \textit{p} is prime and $j<p$, there is a factor of \textit{p} in the numerator, and no factors of \textit{p} in the denominator.\\
Thus, $\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k=p*m, \exists m \in \mathbb{Z}$
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\begin{frame}{Fermat's Method}
Returning to what we were trying to show: $(k+1)^p-(k+1)=p*q$\\
$\iff k^p-k+[\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k]=p*q$\\
But, $\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k=p*m$, so $k^p-k+[\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k]=k^p-k+p*m$\\
Thus, $k^p-k+[\binom{p}{1}k^{p-1}+\binom{p}{2}k^{p-2}+ . . . +\binom{p}{p-1}k]=p*q$
$\iff k^p-k +p*m=p*q$\\$\iff k^p-k=p(q-m)$\\$p|(k^p-k)$, which is true by our inductive assumption.
Thus $a^p\equiv a\mod{p}, \forall a>1$. To prove this for all integers, the inductive direction
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\section{Second Section}
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\textbf{Treatments} & \textbf{Response 1} & \textbf{Response 2}\\
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Treatment 1 & 0.0003262 & 0.562 \\
Treatment 2 & 0.0015681 & 0.910 \\
Treatment 3 & 0.0009271 & 0.296 \\
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\frametitle{Theorem}
\begin{theorem}[Mass--energy equivalence]
$E = mc^2$
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\begin{theorem}[Mass--energy equivalence]
$E = mc^2$
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\bibitem[Smith, 2012]{p1} John Smith (2012)
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\newblock \emph{Journal Name} 12(3), 45 -- 678.
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\Huge{\centerline{The End}}
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