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Co-authored-by: Mark <mark@betalupi.com>
Co-committed-by: Mark <mark@betalupi.com>

src/Advanced/Cryptography/parts/0 euclidean.tex

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+\section{The Euclidean Algorithm}
+
+\definition{}
+The \textit{greatest common divisor} of $a$ and $b$ is the greatest integer that divides both $a$ and $b$. \par
+We denote this number with $\gcd(a, b)$. For example, $\gcd(45, 60) = 15$.
+
+\problem{}
+Find $\gcd(20, 14)$ by hand.
+
+\begin{solution}
+	$\gcd(20, 14) = 2$
+\end{solution}
+
+\vfill
+
+\theorem{The Division Algorithm}<divalgo>
+Given two integers $a, b$, we can find two integers $q, r$, where $0 \leq r < b$ and $a = qb + r$. \par
+In other words, we can divide $a$ by $b$ to get $q$ remainder $r$.
+
+\theorem{}<gcd_abc>
+For any integers $a, b, c$, \par
+$\gcd(ac + b, a) = \gcd(a, b)$
+
+\problem{The Euclidean Algorithm}<euclid>
+Using the two theorems above, detail an algorithm for finding $\gcd(a, b)$. \par
+Then, compute $\gcd(1610, 207)$ by hand. \par
+
+\begin{solution}
+	Using \ref{gcd_abc} and the division algorithm,
+
+	% Minipage prevents column breaks inside body
+	\begin{multicols}{2}
+		\begin{minipage}{\columnwidth}
+			$\gcd(1610, 207)$ \par
+			$= \gcd(207, 161)$ \par
+			$= \gcd(161, 46)$ \par
+			$= \gcd(46, 23)$ \par
+			$= \gcd(23, 0) = 23$ \par
+		\end{minipage}
+
+		\columnbreak
+
+		\begin{minipage}{\columnwidth}
+			$1610 = 207 \times 7 + 161$ \par
+			$207 = 161 \times 1 + 46$ \par
+			$161 = 46 \times 3 + 23$ \par
+			$46 = 23 \times 2 + 0$ \par
+		\end{minipage}
+	\end{multicols}
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+
+
+\problem{}<extendedeuclid>
+Using the output of the Euclidean algorithm,
+
+\begin{itemize}
+	\item[-] find a pair $(u, v)$ that satisfies $20u + 14v = \gcd(20, 14)$
+	\item[-] find a pair $(u, v)$ that satisfies $541u + 34v = \gcd(541, 34)$
+	% gcd = 1
+	% u = 11; v = -175
+\end{itemize}
+This is called the \textit{extended Euclidean algorithm}. \par
+\hint{
+	You don't need to fully solve the last part of this question. \\
+	Understand how you \textit{would} do it, then move on.
+	Don't spend too much time on arithmetic.
+}
+
+%For which numbers $c$ can we find a $(u, v)$ so that $541u + 34v = c$? \\
+%For every such $c$, what are $u$ and $v$?
+
+\vspace{2mm}
+
+\textbf{Hint:}
+
+After running the Euclidean algorithm, you have a table similar to the one shown below. \par
+You can use a bit of algebra to rearrange these statements to get what you need. \par
+
+\vspace{5mm}
+
+\newdimen\mywidth
+\setbox0=\hbox{Using the Euclidean Algorithm to find that $\gcd(20, 14) = 2$:}
+\mywidth=\wd0
+\begin{minipage}{\mywidth}
+	\begin{center}
+	Using the Euclidean Algorithm to find that $\gcd(20, 14) = 2$: \par
+		$20 = 14 \times 1 + 6$ \par
+		$14 = 6 \times 2 + 2$ \par
+		$6 = 2 \times 3 + 0$ \par
+	\end{center}
+\end{minipage}\par
+\vspace{2mm}
+We now want to write the 2 in the last equation in terms of 20 and 14.
+
+
+
+\begin{solution}
+	Using the output of the Euclidean Algorithm, we can use substitution and a bit of algebra to solve such problems. Consider the following example:
+
+	\begin{multicols}{2}
+		\begin{minipage}{\columnwidth}
+			\textit{Euclidean Algorithm:} \par
+			$20 = 14 \times 1 + 6$ \par
+			$14 = 6 \times 2 + 2$ \par
+			$6 = 2 \times 3 + 0$ \par
+		\end{minipage}
+
+		\columnbreak
+
+		\begin{minipage}{\columnwidth}
+			\textit{Rearranged:} \par
+			$6 = 20 - 14 \times 1$ \par
+			$2 = 14 - 6 \times 2 = \gcd(20, 14)$ \par
+		\end{minipage}
+	\end{multicols}
+
+	Using the right table, we can replace $6$ in $2 = 14 - 6 \times 2$ to get
+	$2 = 14 - (20 - 14) \times 2$, \par
+	which gives us $2 = \gcd(20, 14) = (3)14 + (-2)20$. \par
+
+	\linehack{}
+
+	$\gcd(20, 14) = 20(-2) + 14(3)$ \par
+	$\gcd(541, 34) = 541(11) + 34(-175)$
+\end{solution}
+
+\begin{solution}
+	\huge
+	This problem is too hard. Break it into many.
+\end{solution}
+
+\vfill
+\pagebreak
+