Finished De Bruijn sections

Advanced/De Bruijn/parts/0 intro.tex

@@ -41,14 +41,13 @@ Consider the same lock, now set with a three-digit binary code.
 How about a four-digit code? How many digits do we need? \par
 
 \begin{instructornote}
-	Don't spend too long on this problem.
+	Don't spend too much time here.
 	Provide a solution at the board once everyone has had a few
-	minutes to think about this.
+	minutes to think about this problem.
 \end{instructornote}
 
 \begin{solution}
-	Interestingly enough, we can only save one digit. \par
-	Any optimal sequence has 15 digits, for example \texttt{0000111101100101000}
+	One example is \texttt{0000 1111 0110 0101 000}
 \end{solution}
 
 \vfill

Advanced/De Bruijn/parts/1 words.tex

@@ -175,12 +175,12 @@ For example, $C_1 = \texttt{0}$, $C_2 = \texttt{011011}$, and $C_3 = \texttt{011
 		\item If $w$ starts with a \texttt{1}, $w$ must appear in $C_k$ by construction.
 
 		\item If $w$ does starts with a \texttt{0} and contains a \texttt{1}, $w$ has the form
-		$\texttt{0}^x\texttt{1}[..\texttt{y}..]$ \par
+		$\texttt{0}^x\texttt{1}\overline{\texttt{y}}$ \par
 		\note{
 			That is, $x$ copies of \texttt{0} followed by a \texttt{1}, followed by \par
-			an arbitrary sequence $\texttt{y}$ with length $(k-x-1)$.
+			an arbitrary sequence $\overline{\texttt{y}}$ with length $(k-x-1)$.
 		} \par
-		Now consider the word $\texttt{1}[..\texttt{y}..]\texttt{0}^x\texttt{1}[..\texttt{y}..]\texttt{0}^{(x-1)}\texttt{1}$. \par
+		Now consider the word $\texttt{1}\overline{\texttt{y}}\texttt{0}^x\texttt{1}\overline{\texttt{y}}\texttt{0}^{(x-1)}\texttt{1}$. \par
 		This is the concatenation of two consecutive binary numbers with $k$ digits, and thus appears in $C_k$.
 		$w$ is a subword of this word, and therefore also appears in $C_k$.
 	\end{itemize}

Advanced/De Bruijn/parts/2 bruijn.tex

@@ -66,6 +66,24 @@ Find the single unique Eulerian cycle in the graph below.
 \end{solution}
 
 \vfill
+
+\theorem{}<eulerexists>
+A directed graph contains an Eulerian cycle iff...
+\begin{itemize}
+	\item There is a path between every pair of nodes, and
+	\item every node has as many \say{in} edges as it has \say{out} edges.
+\end{itemize}
+
+If the a graph contains an Eulerian cycle, it must contain an Eulerian path. \note{(why?)} \par
+Some graphs contain an Eulerian path, but not a cycle. In this case, both conditions above must
+still hold, but the following exceptions are allowed:
+\begin{itemize}
+	\item There may be at most one node where $(\text{number in} - \text{number out}) = 1$
+	\item There may be at most one node where $(\text{number in} - \text{number out}) = -1$
+\end{itemize}
+We won't provide a proof of this theorem today. However, you should convince yourself that it is true:
+if any of these conditions are violated, why do we know that an Eulerian cycle (or path) cannot exist?
+
 \pagebreak
 
 
@@ -77,7 +95,7 @@ We'll call the optimal solution to this problem a \textit{De Bruijn\footnotemark
 \footnotetext{Dutch. Rhymes with \say{De Grown.}}
 
 
-\problem{}
+\problem{}<dbbounds>
 Let $\mathcal{B}_n$ be the length of an order-$n$ De Bruijn word. \par
 Show that the following bounds always hold:
 \begin{itemize}
@@ -96,7 +114,11 @@ Show that the following bounds always hold:
 
 \remark{}
 Now, we'd like to show that $\mathcal{B}_n = 2^n + n - 1$... \par
-That is, that the optimal solution to the subword problem always has $2^n + n - 1$ letters.
+That is, that the optimal solution to the subword problem always has $2^n + n - 1$ letters. \par
+We'll do this by construction: for a given $n$, we want to build a word with length $2^n + n - 1$
+that solves the binary $n$-subword problem.
+
+
 
 \definition{}
 Consider a $n$-length word $w$. \par
@@ -241,16 +263,20 @@ Draw $G_4$.
 	\end{itemize}
 \end{solution}
 
+\problem{}<dbpath>
+Show that $G_4$ always contains an Eulerian path. \par
+\hint{\ref{eulerexists}}
+
 \vfill
 \pagebreak
 
 
-\theorem{}
+\theorem{}<dbeuler>
 We can now easily construct De Bruijn words for a given $n$: \par
 \begin{itemize}
 	\item Construct $G_n$,
-	\item then an Eulerian cycle in $G_n$.
-	\item Finally, construct a De Bruijn by writing the label of our starting vertex,
+	\item find an Eulerian cycle in $G_n$,
+	\item then, construct a De Bruijn word by writing the label of our starting vertex,
 	then appending the label of every edge we travel.
 \end{itemize}
 
@@ -284,4 +310,48 @@ Find De Bruijn words of orders $2$, $3$, and $4$.
 \end{solution}
 
 \vfill
+
+Let's quickly show that the process described in \ref{dbeuler}
+indeed produces a valid De Bruijn word.
+
+\problem{}<dblength>
+How long will a word generated by the above process be?
+
+\begin{solution}
+	A De Bruijn graph has $2^n$ edges, each of which is traversed exactly once.
+	The starting node consists of $n - 1$ letters.
+
+	\vspace{2mm}
+
+	Thus, the resulting word contains $2^n + n - 1$ symbols.
+\end{solution}
+
+\vfill
+
+\problem{}<dbsubset>
+Show that a word generated by the process in \ref{dbeuler}
+contains every possible length-$n$ subword. \par
+In other words, show that $\mathcal{S}_n(w) = 2^n$ for a generated word $w$.
+
+\begin{solution}
+	TODO
+\end{solution}
+
+\vfill
+
+\remark{}
+\begin{itemize}
+	\item We found that \ref{dbeuler} generates a word with length $2^n + n - 1$ in \ref{dblength}, \par
+	\item and we showed that this word always solves the $n$-subword problem in \ref{dbsubset}.
+
+	\item From \ref{dbbounds}, we know that any solution to the binary $n$-subword problem \par
+	must have at least $2^n + n - 1$ letters.
+
+	\item Finally, \ref{dbpath} guarantees that it is possible to generate such a word in any $G_n$.
+\end{itemize}
+
+Thus, we have shown that the process in \ref{dbeuler} generates ideal solutions
+to the $n$-subword problem, and that such solutions always exist.
+We can now conclude that for any $n$, the binary $n$-subword problem may be solved with a word of length $2^n + n - 1$.
+
 \pagebreak

Advanced/De Bruijn/parts/3 line.tex

@@ -97,5 +97,8 @@ Construct $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$. What do you notice?
 
 \begin{solution}
 	After fixing edge labels, we find that
-	$\mathcal{L}(G_2) \cong G_3$, and $\mathcal{L}(G_3) \cong G_4$
-\end{solution}
+	$\mathcal{L}(G_2) \cong G_3$ and $\mathcal{L}(G_3) \cong G_4$
+\end{solution}
+
+\vfill
+\pagebreak