Cleanup

Advanced/De Bruijn/parts/0 intro.tex

@@ -3,7 +3,7 @@
 \example{}<lockproblem>
 A certain electronic lock has two buttons: \texttt{0} and \texttt{1}.
 It opens as soon as the correct two-digit code is entered, completely ignoring
-previous inputs.\hspace{-0.5ex}\footnotemark{} For example, if the correct code is \text{10}, the lock will open
+previous inputs. For example, if the correct code is \text{10}, the lock will open
 once the sequence \texttt{010} is entered.
 
 \vspace{2mm}
@@ -19,36 +19,38 @@ Unlock this lock with only 5 keypresses.
 \begin{solution}
 	The sequence \texttt{00110} is guaranteed to unlock this lock.
 \end{solution}
+\vfill
 
+Now, consider the same lock, now set with a three-digit binary code.
+\problem{}
+How many codes are possible?
+\vfill
 
 \problem{}
-Consider the same lock, now set with a three-digit binary code.
+Show that there is no solution with fewer than three keypresses
-\begin{itemize}
+\vfill
-	\item How many codes are possible?
+
-	\item What is the shortest sequence that is guaranteed to unlock the lock? \par
+\problem{}
-	\hint{You'll need 10 digits.}
+What is the shortest sequence that is guaranteed to unlock the lock? \par
-\end{itemize}
+\hint{You'll need 10 digits.}
 
 \begin{solution}
-	\begin{itemize}
+	\texttt{0001110100} will do.
-		\item $2^3 = 8$
-		\item \texttt{0001110100} will do.
-	\end{itemize}
 \end{solution}
 
 
-\problem{}
+%\problem{}
-How about a four-digit code? How many digits do we need? \par
+%How about a four-digit code? How many digits do we need? \par
-
+%
-\begin{instructornote}
+%\begin{instructornote}
-	Don't spend too much time here.
+%	Don't spend too much time here.
-	Provide a solution at the board once everyone has had a few
+%	Provide a solution at the board once everyone has had a few
-	minutes to think about this problem.
+%	minutes to think about this problem.
-\end{instructornote}
+%\end{instructornote}
-
+%
-\begin{solution}
+%\begin{solution}
-	One example is \texttt{0000 1111 0110 0101 000}
+%	One example is \texttt{0000 1111 0110 0101 000}
-\end{solution}
+%\end{solution}
 
 \vfill
 \pagebreak

Advanced/De Bruijn/parts/1 words.tex

@@ -2,24 +2,28 @@
 
 \definition{}
 An \textit{alphabet} is a set of symbols. \par
-For example, $\{\texttt{0}, \texttt{1}\}$ is an alphabet of two symbols, \par
+For example, $\{\texttt{0}, \texttt{1}\}$ is an alphabet of two symbols,
 and $\{\texttt{a}, \texttt{b}, \texttt{c}\}$ is an alphabet of three.
 
 \definition{}
 A \textit{word} over an alphabet $A$ is a sequence of symbols in that alphabet. \par
 For example, $\texttt{00110}$ is a word over the alphabet $\{\texttt{0}, \texttt{1}\}$. \par
-We'll let $\varnothing$ denote the empty word, which exists over every alphabet.
+We'll let $\varnothing$ denote the empty word, which is a valid word over any alphabet.
 
 \definition{}
 Let $v$ and $w$ be words over the same alphabet. \par
 We say $v$ is a \textit{subword} of $w$ if $v$ is contained in $w$. \par
+\note{
+	In other words, $v$ is a subword of $w$ if we can construct $v$ \par
+	by removing a few characters from the start and end of $w$.
+}
 For example, \texttt{11} is a subword of \texttt{011}, but \texttt{00} is not.
 
 \definition{}
-Recall \ref{lockproblem}. From now on, we'll call this the \textit{$n$-subword problem}: \par
+Recall \ref{lockproblem}. Let's generalize this to the \textit{$n$-subword problem}: \par
 Given an alphabet $A$ and a positive integer $n$,
-we want a word over $A$ that contains all possible length-$n$ subwords. \par
+we want a word over $A$ that contains all possible length-$n$ subwords.
-That shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}.
+The shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}.
 
 
 
@@ -55,6 +59,12 @@ Find the following:
 \vfill
 \pagebreak
 
+
+
+
+
+
+
 \problem{}<sbounds>
 Let $w$ be a word over an alphabet of size $k$. \par
 Prove the following:
@@ -87,6 +97,10 @@ Prove the following:
 
 
 
+
+
+
+
 \definition{}
 Let $v$ and $w$ be words over the same alphabet. \par
 The word $vw$ is the word formed by writing $v$ after $w$. \par
@@ -145,6 +159,10 @@ We'll call this the \textit{Fibonacci word} of order $k$.
 
 
 
+
+
+
+
 % C_k is called the "Champernowne word" of order k.
 \problem{}<cword>
 Let $C_k$ denote the word over the alphabet $\{\texttt{0}, \texttt{1}\}$ obtained by \par

Advanced/De Bruijn/parts/2 bruijn.tex

@@ -27,8 +27,8 @@ and five edges (labeled $0, ... , 4$).
 \end{center}
 
 \definition{}
-A \textit{path} in a graph is a sequence of adjacent edges. \par
+A \textit{path} in a graph is a sequence of adjacent edges, \par
-In a directed graph, adjacent edges are those that start and end at the same node. \par
+In a directed graph, edges $a$ and $b$ are adjacent if $a$ ends at the node which $b$ starts at. \par
 \vspace{2mm}
 For example, consider the graph above. \par
 The edges $0$ and $1$ are not adjacent, because $0$ and $1$ both \textit{end} at $b$. \par
@@ -88,6 +88,9 @@ if any of these conditions are violated, why do we know that an Eulerian cycle (
 
 
 
+
+
+
 \definition{}
 Now, consider the $n$-subword problem over $\{\texttt{0}, \texttt{1}\}$. \par
 We'll call the optimal solution to this problem a \textit{De Bruijn\footnotemark{} word} of order $n$. \par
@@ -113,7 +116,7 @@ Show that the following bounds always hold:
 
 
 \remark{}
-Now, we'd like to show that the length of a De Bruijn word is always $2^n + n - 1$... \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.
@@ -189,6 +192,16 @@ $G_2$ and $G_3$ are shown below.
 \vfill
 \pagebreak
 
+
+
+
+
+
+
+
+
+
+
 \problem{}
 Draw $G_4$.
 
@@ -310,6 +323,16 @@ Find De Bruijn words of orders $2$, $3$, and $4$.
 \vfill
 \pagebreak
 
+
+
+
+
+
+
+
+
+
+
 Let's quickly show that the process described in \ref{dbeuler}
 indeed produces a valid De Bruijn word.
 

Advanced/De Bruijn/parts/3 line.tex

@@ -1,7 +1,8 @@
 \section{Line Graphs}
 
 \problem{}
-Given a graph $G$, we can construct its \textit{line graph} (denoted $\mathcal{L}(G)$) by doing the following: \par
+Given a graph $G$, we can construct a graph called the \par
+\textit{line graph} of $G$ (\hspace{0.3ex}denoted $\mathcal{L}(G)$\hspace{0.3ex}) by doing the following: \par
 \begin{itemize}
 	\item Creating a node in $\mathcal{L}(G)$ for each edge in $G$
 	\item Drawing a directed edge between every pair of nodes $a, b$ in $\mathcal{L}(G)$ \par
@@ -93,7 +94,11 @@ Now, relabel the edge from $a$ to $b$ as $\texttt{x}\overline{\texttt{p}}\texttt
 Use these new labels to name nodes in $\mathcal{L}(G_n)$.
 
 \problem{}
-Construct $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$. What do you notice?
+Construct $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$. What do you notice? \par
+\hint{
+	What are $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$? We've seen them before! \par
+	You may need to re-label a few edges.
+}
 
 \begin{solution}
 	After fixing edge labels, we find that

Advanced/De Bruijn/parts/4 sturmian.tex

@@ -275,7 +275,7 @@ Attempt the above construction a few times. Is $w$ a minimal Sturmian word?
 
 
 
-\theorem{}
+\theorem{}<sturmanthm>
 We can construct a miminal Sturmian word of order $n \geq 3$ as follows:
 \begin{itemize}
 	\item Start with $G_2$, create $R_2$ by removing one edge.
@@ -287,6 +287,7 @@ We can construct a miminal Sturmian word of order $n \geq 3$ as follows:
 	\item Construct a word $w$ using the Eulerian path, as before. \par
 	This is a minimal Sturmian word.
 \end{itemize}
+For now, assume this theorem holds. We'll prove it in the next few problems.
 
 \problem{}<sturmianfour>
 Construct a minimal Sturmain word of order 4.
@@ -374,7 +375,7 @@ Construct a minimal Sturmain word of order 5.
 
 
 \problem{}
-Argue that the words we get are mimimal Sturmain words: \par
+Argue that the words we get by \ref{sturmanthm} are mimimal Sturmain words. \par
 That is, the word $w$ has length $2n$ and $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$.
 
 \begin{solution}