From 759e7e05f69172ff7d7a7d38370e0867a2334e4b Mon Sep 17 00:00:00 2001 From: Mark Date: Mon, 8 Apr 2024 07:37:59 -0700 Subject: [PATCH] Post-class fixes --- Advanced/De Bruijn/parts/1 words.tex | 7 ++++--- Advanced/De Bruijn/parts/2 bruijn.tex | 9 ++++++--- Advanced/De Bruijn/parts/3 line.tex | 25 +++++++++++++------------ Advanced/De Bruijn/parts/4 sturmian.tex | 2 +- 4 files changed, 24 insertions(+), 19 deletions(-) diff --git a/Advanced/De Bruijn/parts/1 words.tex b/Advanced/De Bruijn/parts/1 words.tex index 19b7409..527cd10 100644 --- a/Advanced/De Bruijn/parts/1 words.tex +++ b/Advanced/De Bruijn/parts/1 words.tex @@ -51,7 +51,7 @@ Find the following: \begin{solution} In order from $\mathcal{S}_0$ to $\mathcal{S}_6$: \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 \end{itemize} \end{solution} @@ -167,9 +167,10 @@ We'll call this the \textit{Fibonacci word} of order $k$. \problem{} 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 -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} - \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 Show that $\mathcal{S}_k(C_k) = 2^k - 1$. \item Show that $\mathcal{S}_n(C_k) = 2^n$ for $n < k$. diff --git a/Advanced/De Bruijn/parts/2 bruijn.tex b/Advanced/De Bruijn/parts/2 bruijn.tex index ff96437..c585343 100644 --- a/Advanced/De Bruijn/parts/2 bruijn.tex +++ b/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{} Show that $G_4$ always contains an Eulerian path. \par \hint{\ref{eulerexists}} diff --git a/Advanced/De Bruijn/parts/3 line.tex b/Advanced/De Bruijn/parts/3 line.tex index 91477cc..0454474 100644 --- a/Advanced/De Bruijn/parts/3 line.tex +++ b/Advanced/De Bruijn/parts/3 line.tex @@ -36,22 +36,23 @@ Have an instructor check your solution. \begin{center} \begin{tikzpicture} \begin{scope}[layer = nodes] - \node[main] (0) at (0, 0) {$0$}; - \node[main] (1) at (2, -4) {$1$}; - \node[main] (2) at (0, -2) {$2$}; - \node[main] (3) at (2, -2) {$3$}; - \node[main] (4) at (2, 0) {$4$}; + \node[main] (1) at (0, 0) {$1$}; + \node[main] (4) at (3.5, 1) {$4$}; + \node[main] (3) at (2, 0) {$3$}; + \node[main] (2) at (2, 2) {$2$}; + \node[main] (0) at (0, 2) {$0$}; \end{scope} \draw[->] - (0) edge[bend left] (2) - (2) edge[bend left] (0) - (0) edge (4) + (0) edge[bend left] (1) + (1) edge[bend left] (0) + (0) edge (2) + (2) edge (3) + (2) edge[bend left] (4) (4) edge[bend left] (2) - (2) edge (1) - (1) edge[bend left] (3) - (3) edge[bend left] (1) - (3) edge (0) + (3) edge (1) + (4) edge (3) + (4) edge[loop right] (4) ; \end{tikzpicture} \end{center} diff --git a/Advanced/De Bruijn/parts/4 sturmian.tex b/Advanced/De Bruijn/parts/4 sturmian.tex index 2d99ee6..9458e74 100644 --- a/Advanced/De Bruijn/parts/4 sturmian.tex +++ b/Advanced/De Bruijn/parts/4 sturmian.tex @@ -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 \begin{itemize} \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. \end{itemize} Label each edge in $R_3$ with the last letter of its target node. \par