Post-class fixes

Advanced/De Bruijn/parts/2 bruijn.tex

@@ -31,9 +31,9 @@ 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
-$0$ and $2$, however, are: $0$ ends at $b$, and $2$ starts at $b$.
-$[0, 3, 2]$ is a path in the graph above, drawn below. \par
+The edges $1$ and $0$ are adjacent, since you can take edge $0$ after taking edge $1$. \par
+$0$ starts where $1$ ends. \par
+$0$ and $1$, however, are not: $1$ does not start at the edge at which $0$ ends.
 
 
 \definition{}
@@ -81,6 +81,7 @@ still hold, but the following exceptions are allowed:
 	\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}
+\note[Note]{Either both exceptions occur, or neither occurs. Bonus problem: why?}
 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?
 
@@ -276,6 +277,8 @@ Draw $G_4$.
 	\end{itemize}
 \end{solution}
 
+\vfill
+
 \problem{}<dbpath>
 Show that $G_4$ always contains an Eulerian path. \par
 \hint{\ref{eulerexists}}