Added sturmian section
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\input{parts/1 words}
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\input{parts/2 bruijn}
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\input{parts/3 line}
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\input{parts/4 sturmian}
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\end{document}
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431
Advanced/De Bruijn/parts/4 sturmian.tex
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431
Advanced/De Bruijn/parts/4 sturmian.tex
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\section{Sturmian Words}
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A De Bruijn word is the shortest word that contains all subwords
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of a given length. \par
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Let's now solve a similar problem: given an alphabet, we want to
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construct a word that contains exactly $m$ distinct subwords of
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length $n$.
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\vspace{2mm}
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% TODO: better, intuitive description
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In general, this is a difficult problem. We'll restrict ourselves
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to a special case: \par
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We'd like to find a word that contains exactly $m + 1$ distinct subwords
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of length $m$ for all $m < n$.
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\definition{}
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We say a word $w$ is a \textit{Sturmian word} of order $n$
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if $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$. \par
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We say $w$ is a \textit{minimal} Sturmian word if there is no shorter
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Sturmian word of that order.
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\problem{}
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Show that the length of a Sturmian word of order $n$ is at least $2n$.
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\begin{solution}
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In order to have $n + 1$ subwords of length $n$, a word must have at
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least $(n+1) + (n-1) = 2n$ letters.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Construct $R_3$ by removing four edges from $G_3$. \par
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Show that each of the following is possible:
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\begin{itemize}[itemsep=2mm ]
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\item $R_3$ does not contain an Eulerian path.
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\item $R_3$ contains an Eulerian path, and this path \par
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constructs a word $w$ with $\mathcal{S}_3(w) = 4$
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and $\mathcal{S}_2(w) = 4$.
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\item $R_3$ contains an Eulerian path, and this path \par
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constructs a word $w$ that is a minimal Sturmian word
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of order 3.
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\end{itemize}
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\begin{solution}
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Remove the edges $\texttt{00} \rightarrow \texttt{01}$,
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$\texttt{01} \rightarrow \texttt{10}$,
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$\texttt{10} \rightarrow \texttt{00}$, and
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$\texttt{11} \rightarrow \texttt{11}$:
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(00) edge[loop left] node[label] {$0$} (00)
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(10) edge[bend left] node[label] {$1$} (01)
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(01) edge[bend left] node[label] {$1$} (11)
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(11) edge[bend left] node[label] {$0$} (10)
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;
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\end{tikzpicture}
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\end{center}
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\linehack{}
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Remove the edges $\texttt{00} \rightarrow \texttt{00}$,
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$\texttt{01} \rightarrow \texttt{10}$,
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$\texttt{10} \rightarrow \texttt{01}$, and
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$\texttt{11} \rightarrow \texttt{11}$. \par
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The Eulerian path starting at \texttt{00} produces \texttt{001100},
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where $\mathcal{S}_2 = \mathcal{S}_3 = 4$.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(00) edge[bend left] node[label] {$1$} (01)
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(10) edge[bend left] node[label] {$0$} (00)
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(01) edge[bend left] node[label] {$1$} (11)
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(11) edge[bend left] node[label] {$0$} (10)
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;
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\end{tikzpicture}
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\end{center}
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\linehack{}
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Remove the edges $\texttt{01} \rightarrow \texttt{11}$,
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$\texttt{10} \rightarrow \texttt{00}$,
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$\texttt{11} \rightarrow \texttt{10}$, and
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$\texttt{11} \rightarrow \texttt{11}$. \par
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The Eulerian path starting at \texttt{00} produces \texttt{000101},
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where $\mathcal{S}_0 = 1$, $\mathcal{S}_1 = 2$, $\mathcal{S}_2 = 3$,
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and $\mathcal{S}_3 = 4$. \par
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\texttt{000101} has length $2 \times 3 = 6$, and is thus minimal.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(00) edge[loop left] node[label] {$0$} (00)
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(00) edge[bend left] node[label] {$1$} (01)
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(01) edge[bend left] node[label] {$0$} (10)
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(10) edge[bend left] node[label] {$1$} (01)
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;
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\end{tikzpicture}
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\end{center}
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Note that this graph contains an Eulerian path even though
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\texttt{11} is disconnected. \par
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An Eulerian path needs to visit all \textit{edges}, not all \textit{nodes}!
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<trysturmian>
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Construct $R_2$ by removing one edge from $G_2$, then construct $\mathcal{L}(R_2)$. \par
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\begin{itemize}
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\item If this line graph has four edges, set $R_3 = \mathcal{L}(R_2)$. \par
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\item If not, remove one edge from $R_2$ so that an Eulerian path still exists
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and set $R_3$ to the resulting graph.
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\end{itemize}
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Label each edge in $R_3$ with the last letter of its target node. \par
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Let $w$ be the word generated by an Eulerian path in this graph, as before.
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\vspace{2mm}
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Attempt the above construction a few times. Is $w$ a minimal Sturmian word?
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\begin{solution}
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If $R_2$ is constructed by removing the edge $\texttt{0} \rightarrow \texttt{1}$,
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$\mathcal{L}(R_2)$ is the graph shown below.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(00) edge[loop left] node[label] {$0$} (00)
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(10) edge[bend left] node[label] {$0$} (00)
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(11) edge[bend left] node[label] {$0$} (10)
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(11) edge[loop right] node[label] {$1$} (11)
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;
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\end{tikzpicture}
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\end{center}
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We obtain the Sturmian word \texttt{111000} via the Eulerian path through the nodes
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$\texttt{11} \rightarrow \texttt{11} \rightarrow \texttt{10}
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\rightarrow \texttt{00} \rightarrow \texttt{00}$.
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\linehack{}
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If $R_2$ is constructed by removing the edge $\texttt{0} \rightarrow \texttt{0}$,
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$\mathcal{L}(R_2)$ is the graph pictured below.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(01) edge[bend left] node[label] {$0$} (10)
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(10) edge[bend left] node[label] {$1$} (01)
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(11) edge[bend left] node[label] {$0$} (10)
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(01) edge[bend left] node[label] {$1$} (11)
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(11) edge[loop right] node[label] {$1$} (11)
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;
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\end{tikzpicture}
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\end{center}
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This graph contains five edges, we need to remove one. \par
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To keep an Eulerian path, we can remove any of the following:
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\begin{itemize}
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\item $\texttt{10} \rightarrow \texttt{01}$ to produce \texttt{011101}
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\item $\texttt{01} \rightarrow \texttt{11}$ to produce \texttt{111010}
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\item $\texttt{11} \rightarrow \texttt{10}$ to produce \texttt{010111}
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\item $\texttt{11} \rightarrow \texttt{11}$ to produce \texttt{011010}
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\end{itemize}
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Each of these is a minimal Sturmian word.
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\linehack{}
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The case in which we remove $\texttt{1} \rightarrow \texttt{0}$ in $G_2$ should
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produce a minimal Sturmian word where \texttt{0} and \texttt{1} are interchanged
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in the word produced by removing $\texttt{0} \rightarrow \texttt{1}$.
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\vspace{2mm}
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If we remove $\texttt{1} \rightarrow \texttt{1}$ will produce minimal
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Sturmian words where \texttt{0} and \texttt{1} are interchanged from the words
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produced by removing $\texttt{0} \rightarrow \texttt{0}$.
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\end{solution}
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\vfill
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\pagebreak
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\theorem{}
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We can construct a miminal Sturmian word of order $n \geq 3$ as follows:
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\begin{itemize}
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\item Start with $G_2$, create $R_2$ by removing one edge.
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\item Construct $\mathcal{L}(G_2)$, remove an edge if necessary. \par
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The resulting graph must have an 4 edges and an Eulerian path. Call this $R_3$.
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\item Repeat the previous step to construct a sequence of graphs $R_n$. \par
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$R_{n-1}$ is used to create $R_n$, which has $n + 1$ edges and an Eulerian path. \par
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Label edges with the last letter of their target vertex.
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\item Construct a word $w$ using the Eulerian path, as before. \par
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This is a minimal Sturmian word.
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\end{itemize}
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\problem{}<sturmianfour>
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Construct a minimal Sturmain word of order 4.
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\begin{solution}
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Let $R_3$ be the graph below (see \ref{trysturmian}).
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (00) at (0, 0) {\texttt{00}};
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\node[main] (01) at (2, 1) {\texttt{01}};
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\node[main] (10) at (2, -1) {\texttt{10}};
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\node[main] (11) at (4, 0) {\texttt{11}};
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\end{scope}
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\draw[->]
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(00) edge[loop left] node[label] {$0$} (00)
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(10) edge[bend left] node[label] {$0$} (00)
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(11) edge[bend left] node[label] {$0$} (10)
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(11) edge[loop right] node[label] {$1$} (11)
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;
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\end{tikzpicture}
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\end{center}
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$R_4 = \mathcal{L}(R_3)$ is then as shown below, producing the
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order $4$ minimal Sturman word \texttt{11110000}. Disconnected
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nodes are ommited.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (000) at (0, 0) {\texttt{000}};
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\node[main] (100) at (2, 1) {\texttt{100}};
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\node[main] (110) at (2, -1) {\texttt{110}};
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\node[main] (111) at (4, 0) {\texttt{111}};
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\end{scope}
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\draw[->]
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(000) edge[loop left] node[label] {$0$} (000)
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(100) edge[bend right] node[label] {$0$} (000)
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(110) edge[bend left] node[label] {$0$} (100)
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(111) edge[bend left] node[label] {$0$} (110)
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(11) edge[loop right] node[label] {$1$} (11)
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;
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Construct a minimal Sturmain word of order 5.
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\begin{solution}
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Use $R_4$ from \ref{sturmianfour} to construct $R_5$, shown below.
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\begin{center}
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\begin{tikzpicture}
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\begin{scope}[layer = nodes]
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\node[main] (0000) at (0, 0) {\texttt{0000}};
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\node[main] (1000) at (2, 0) {\texttt{1000}};
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\node[main] (1100) at (4, 0) {\texttt{1100}};
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\node[main] (1110) at (6, 0) {\texttt{1110}};
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\node[main] (1111) at (8, 0) {\texttt{1111}};
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\end{scope}
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\draw[->]
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(1111) edge[loop right] node[label] {$1$} (1111)
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(1111) edge[bend right] node[label] {$0$} (1110)
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(1110) edge[bend left] node[label] {$0$} (1100)
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(1100) edge[bend right] node[label] {$0$} (1000)
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(1000) edge[bend left] node[label] {$0$} (0000)
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(0000) edge[loop left] node[label] {$0$} (0000)
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;
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\end{tikzpicture}
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\end{center}
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This graph generates the minimal Sturmian word \texttt{1111100000}
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Argue that the words we get are mimimal Sturmain words: \par
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That is, the word $w$ has length $2n$ and $\mathcal{S}_m(w) = m + 1$ for all $m \leq n$.
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\begin{solution}
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We proceed by induction. \par
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First, show that we can produce a minimal order 3 Sturmian word: \par
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\vspace{2mm}
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$R_3$ is guaranteed to have four edges with length-$2$ node labels,
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the length of $w$ is $2 \times 3 = 6$. \par
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Trivially, we also have $\mathcal{S}_0 = 1$ and $\mathcal{S}_1 = 2$. \par
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\vspace{2mm}
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There are three vertices of $R_3$ given by the three remaining nodes of $R_2$.
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Each length-2 subword of $w$ will be represented by the label of one of these
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three nodes. Thus, $\mathcal{S}_2(w) \leq 3$. The line graph of a connected graph
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is connected, so an Eulerian path on $R_3$ reaches every node. We thus have that
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$\mathcal{S}_2(w) = 3$.
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\vspace{2mm}
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By construction, the length 3 subwords of $w$ are all distinct, so $\mathcal{S}_3(w) = 4$.
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We thus conclude that $w$ is a minimal order 3 Sturmain word.
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\linehack{}
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Now, we prove our inductive step: \par
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Assume that the process above produces an order $n-1$ minimal Sturmain word $w_{n-1}$. \par
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We want to show that $w_n$ is also a minimal Sturmain word. \par
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\vspace{2mm}
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By construction, $R_n$ has node labels of length $n-1$ and $n+1$ edges. \par
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Thus, $w_n$ has length $2n$.
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\vspace{2mm}
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The only possilble length-$m$ subwords of $w_n$ are those of $w_{n-1}$ for $m < n$. \par
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The line graph of a connected graph is connected, so an Eulerian path on $R_3$ reaches each node.
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Thus, all length-$m$ subwords of $w_{n-1}$ appear in $w_n$.
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\vspace{2mm}
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By our inductive hypothesis, $\mathcal{S}_m(w_n) = m + 1$ for $m < n$. \par
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The length-$n$ subwords of $w_n$ are distinct by construction, and there are
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$n+1$ such subwords.
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\vspace{2mm}
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Thus, $\mathcal{S}_n(w_n) = n + 1$.
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\end{solution}
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\vfill
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\pagebreak
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