Finished De Bruijn sections

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