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Post-class fixes

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Mark 2024-04-08 07:37:59 -07:00
parent f2ba5bf1b3
commit 759e7e05f6
Signed by: Mark
GPG Key ID: C6D63995FE72FD80
4 changed files with 24 additions and 19 deletions
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7
Advanced/De Bruijn/parts/1 words.tex
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@ -51,7 +51,7 @@ Find the following:
\begin{solution} \begin{solution}
In order from $\mathcal{S}_0$ to $\mathcal{S}_6$: In order from $\mathcal{S}_0$ to $\mathcal{S}_6$:
\begin{itemize} \begin{itemize}
\item 1, 2, 3, 3, 3, 2, 1 \item 1, 2, 3, 4, 3, 2, 1
\item 1, 3, 5, 4, 3, 2, 1 \item 1, 3, 5, 4, 3, 2, 1
\end{itemize} \end{itemize}
\end{solution} \end{solution}
@ -167,9 +167,10 @@ We'll call this the \textit{Fibonacci word} of order $k$.
\problem{}<cword> \problem{}<cword>
Let $C_k$ denote the word over the alphabet $\{\texttt{0}, \texttt{1}\}$ obtained by \par Let $C_k$ denote the word over the alphabet $\{\texttt{0}, \texttt{1}\}$ obtained by \par
concatenating the binary representations of the integers $0,~...,~2^k -1$. \par concatenating the binary representations of the integers $0,~...,~2^k -1$. \par
For example, $C_1 = \texttt{0}$, $C_2 = \texttt{011011}$, and $C_3 = \texttt{011011100101110111}$. For example, $C_1 = \texttt{01}$, $C_2 = \texttt{011011}$, and $C_3 = \texttt{011011100101110111}$.
\begin{itemize} \begin{itemize}
\item How many symbols does the word $C_k$ contain? % Good bonus problem, hard to find a closed-form solution
% \item How many symbols does the word $C_k$ contain?
\item Compute $\mathcal{S}_0$, $\mathcal{S}_1$, $\mathcal{S}_2$, and $\mathcal{S}_3$ for $C_3$. \item Compute $\mathcal{S}_0$, $\mathcal{S}_1$, $\mathcal{S}_2$, and $\mathcal{S}_3$ for $C_3$.
\item Show that $\mathcal{S}_k(C_k) = 2^k - 1$. \item Show that $\mathcal{S}_k(C_k) = 2^k - 1$.
\item Show that $\mathcal{S}_n(C_k) = 2^n$ for $n < k$. \item Show that $\mathcal{S}_n(C_k) = 2^n$ for $n < k$.

9
Advanced/De Bruijn/parts/2 bruijn.tex
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@ -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 In a directed graph, edges $a$ and $b$ are adjacent if $a$ ends at the node which $b$ starts at. \par
\vspace{2mm} \vspace{2mm}
For example, consider the graph above. \par 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 The edges $1$ and $0$ are adjacent, since you can take edge $0$ after taking edge $1$. \par
$0$ and $2$, however, are: $0$ ends at $b$, and $2$ starts at $b$. $0$ starts where $1$ ends. \par
$[0, 3, 2]$ is a path in the graph above, drawn below. \par $0$ and $1$, however, are not: $1$ does not start at the edge at which $0$ ends.
\definition{} \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 $(\text{number in} - \text{number out}) = 1$
\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} \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: 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? 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{itemize}
\end{solution} \end{solution}
\vfill
\problem{}<dbpath> \problem{}<dbpath>
Show that $G_4$ always contains an Eulerian path. \par Show that $G_4$ always contains an Eulerian path. \par
\hint{\ref{eulerexists}} \hint{\ref{eulerexists}}

25
Advanced/De Bruijn/parts/3 line.tex
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@ -36,22 +36,23 @@ Have an instructor check your solution.
\begin{center} \begin{center}
\begin{tikzpicture} \begin{tikzpicture}
\begin{scope}[layer = nodes] \begin{scope}[layer = nodes]
\node[main] (0) at (0, 0) {$0$}; \node[main] (1) at (0, 0) {$1$};
\node[main] (1) at (2, -4) {$1$}; \node[main] (4) at (3.5, 1) {$4$};
\node[main] (2) at (0, -2) {$2$}; \node[main] (3) at (2, 0) {$3$};
\node[main] (3) at (2, -2) {$3$}; \node[main] (2) at (2, 2) {$2$};
\node[main] (4) at (2, 0) {$4$}; \node[main] (0) at (0, 2) {$0$};
\end{scope} \end{scope}
\draw[->] \draw[->]
(0) edge[bend left] (2) (0) edge[bend left] (1)
(2) edge[bend left] (0) (1) edge[bend left] (0)
(0) edge (4) (0) edge (2)
(2) edge (3)
(2) edge[bend left] (4)
(4) edge[bend left] (2) (4) edge[bend left] (2)
(2) edge (1) (3) edge (1)
(1) edge[bend left] (3) (4) edge (3)
(3) edge[bend left] (1) (4) edge[loop right] (4)
(3) edge (0)
; ;
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}

2
Advanced/De Bruijn/parts/4 sturmian.tex
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@ -176,7 +176,7 @@ Show that each of the following is possible:
Construct $R_2$ by removing one edge from $G_2$, then construct $\mathcal{L}(R_2)$. \par Construct $R_2$ by removing one edge from $G_2$, then construct $\mathcal{L}(R_2)$. \par
\begin{itemize} \begin{itemize}
\item If this line graph has four edges, set $R_3 = \mathcal{L}(R_2)$. \par \item If this line graph has four edges, set $R_3 = \mathcal{L}(R_2)$. \par
\item If not, remove one edge from $R_2$ so that an Eulerian path still exists \item If not, remove one edge from $\mathcal{L}(R_2)$ so that an Eulerian path still exists
and set $R_3$ to the resulting graph. and set $R_3$ to the resulting graph.
\end{itemize} \end{itemize}
Label each edge in $R_3$ with the last letter of its target node. \par Label each edge in $R_3$ with the last letter of its target node. \par