handouts/Advanced/Cryptography/parts/part 1.tex

\definition{}
The \textit{greatest common divisor} of $a$ and $b$ is the greatest integer that divides both $a$ and $b$. \\
We denote this number with $\gcd(a, b)$. For example, $\gcd(45, 60) = 15$.


\theorem{The Division Algorithm}
Given two integers $a, b$, we can find two integers $q, r$, where $0 \leq r < b$ and $a = qb + r$. \\
In other words, we can divide $a$ by $b$ to get $q$ remainder $r$.

\theorem{}<gcd_abc>
For any integers $a, b, c$, \\
$\gcd(ac + b, a) = \gcd(a, b)$

\problem{}
Find $\gcd(20, 14)$ by hand.

\begin{solution}
	$\gcd(20, 14) = 2$
\end{solution}

\vfill

\problem{The Euclidean Algorithm}<euclid_algorithm>
Using the theorems above, detail an algorithm for finding $\gcd(a, b)$.\\
Then, compute $\gcd(1610, 207)$ by hand. \\

\begin{solution}
	Using \ref{gcd_abc} and the division algorthm,

	% 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{Divide and Conquer}
If we are given $a, b, c$, when can we find $u, v$ that satisfy $au + bv = c$?

\problempart{Divide}
Show that if we find a solution $(u, v)$ to $au + bv = \gcd(a, b)$, we can easily find a $(u, v)$ for any other value of $c$. \\
\textcolor{gray}{\textit{Note: } We are not looking for \textit{all} $(u, v)$ that solve $au + bv = c$, we are looking for an easy way to find \textit{any} $(u, v)$.}

\begin{solution}
	Note that $\gcd(a, b)$ divides both a and b. \\
	Therefore, any $c$ must be divisible by $\gcd(a, b)$.
	The smallest such $c$ is $\gcd(a, b)$ itself, and we can get all other tuples $(u, v, c)$ by scaling.
\end{solution}

\vfill

\problempart{Conquer}<extend_e_algorithm>
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}
For which numbers $c$ can we find a $(u, v)$ so that $541u + 34v = c$? \\
For every such $c$, what are $u$ and $v$?

\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
	$2 = 14 - (20 - 14) \times 2$, \\
	which gives us $2 = \gcd(20, 14) = (3)14 + (-2)20$. \\

	\textcolor{gray}{\textit{Note to instructors:} You can present the $(20, 14)$ case as an example.}

	\linehack{}

	$(-2)20 + (3)14 = \gcd(20, 14) = 2$ \\
	$(11)541 + (-175)34 = \gcd(541, 34) = 1$

	\linehack{}

	We can find a solution $(u, v)$ when $c$ is any integer multiple of $\gcd(541, 34)$. \\
	If $c = k \times \gcd(541, 34)$, \\
	$u = k \times u_0 = 11k$ and $v = k \times v_0 = -175k$. \\
	(See Part A)

\end{solution}


\vfill
\pagebreak