Cleanup

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
-In a directed graph, adjacent edges are those that start and end at the same node. \par
+A \textit{path} in a graph is a sequence of adjacent edges, \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 (
 
 
 
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 \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.
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 \vfill
 \pagebreak
 
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 \problem{}
 Draw $G_4$.
 
@@ -310,6 +323,16 @@ Find De Bruijn words of orders $2$, $3$, and $4$.
 \vfill
 \pagebreak
 
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 Let's quickly show that the process described in \ref{dbeuler}
 indeed produces a valid De Bruijn word.