Edits

Advanced/De Bruijn/parts/2 bruijn.tex

@@ -96,11 +96,11 @@ We'll call the optimal solution to this problem a \textit{De Bruijn\footnotemark
 
 
 \problem{}<dbbounds>
-Let $\mathcal{B}_n$ be the length of an order-$n$ De Bruijn word. \par
+Let $w$ be the an order-$n$ De Bruijn word, and denote its length with $|w|$. \par
 Show that the following bounds always hold:
 \begin{itemize}
-	\item $\mathcal{B}_n \leq n2^n$
-	\item $\mathcal{B}_n \geq 2^n + n - 1$
+	\item $|w| \leq n2^n$
+	\item $|w| \geq 2^n + n - 1$
 \end{itemize}
 
 \begin{solution}
@@ -113,7 +113,7 @@ Show that the following bounds always hold:
 
 
 \remark{}
-Now, we'd like to show that $\mathcal{B}_n = 2^n + n - 1$... \par
+Now, we'd like to show that the length of a De Bruijn word is always $2^n + n - 1$... \par
 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.
@@ -244,7 +244,7 @@ Draw $G_4$.
 \end{solution}
 
 \vfill
-
+\pagebreak
 
 \problem{}
 \begin{itemize}
@@ -268,8 +268,6 @@ Show that $G_4$ always contains an Eulerian path. \par
 \hint{\ref{eulerexists}}
 
 \vfill
-\pagebreak
-
 
 \theorem{}<dbeuler>
 We can now easily construct De Bruijn words for a given $n$: \par
@@ -310,6 +308,7 @@ Find De Bruijn words of orders $2$, $3$, and $4$.
 \end{solution}
 
 \vfill
+\pagebreak
 
 Let's quickly show that the process described in \ref{dbeuler}
 indeed produces a valid De Bruijn word.
@@ -334,7 +333,17 @@ 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
+	Any length-$n$ subword of $w$ is the concatenation of a vertex label and an edge label.
+	By construction, the next length-$n$ subword is the concatenation of the next vertex and edge
+	in the Eulerian cycle.
+
+	\vspace{2mm}
+
+	This cycle traverses each edge exactly once, so each length-$n$ subword is distinct. \par
+	Since $w$ has length $2^n + n - 1$, there are $2^n$ total subwords. \par
+	These are all different, so $\mathcal{S}_n \geq 2^n$. \par
+	However, $\mathcal{S}_n \leq 2^n$ by \ref{sbounds}, so $\mathcal{S}_n = 2^n$.
+
 \end{solution}
 
 \vfill