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% use [nosolutions] flag to hide solutions.
% use [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering,
shortwarning
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\input{tikzset.tex}
\uptitlel{Advanced 2}
\uptitler{Spring 2024}
\title{De Bruijn Sequences}
\subtitle{Prepared by \githref{Mark} on \today{}}
\begin{document}
\maketitle
\input{parts/0 intro}
\input{parts/1 words}
\input{parts/2 bruijn}
\input{parts/3 line}
\end{document}

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Advanced/De Bruijn/parts/0 intro.tex Normal file
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\section{Introduction}
\example{}<lockproblem>
A certain electronic lock has two buttons: \texttt{0} and \texttt{1}.
It opens as soon as the correct two-digit code is entered, completely ignoring
previous inputs.\hspace{-0.5ex}\footnotemark{} For example, if the correct code is \text{10}, the lock will open
once the sequence \texttt{010} is entered.
\vspace{2mm}
Naturally, there are $2^2 = 4$ possible combinations that open this lock. \par
If don't know the lock's combination, we could try to guess it by trying all four combinations. \par
This would require eight key presses: \texttt{0001101100}.
\problem{}
There is, of course, a better way. \par
Unlock this lock with only 5 keypresses.
\begin{solution}
The sequence \texttt{00110} is guaranteed to unlock this lock.
\end{solution}
\problem{}
Consider the same lock, now set with a three-digit binary code.
\begin{itemize}
\item How many codes are possible?
\item What is the shortest sequence that is guaranteed to unlock the lock? \par
\hint{You'll need 10 digits.}
\end{itemize}
\begin{solution}
\begin{itemize}
\item $2^3 = 8$
\item \texttt{0001110100} will do.
\end{itemize}
\end{solution}
\problem{}
How about a four-digit code? How many digits do we need? \par
\begin{instructornote}
Don't spend too long on this problem.
Provide a solution at the board once everyone has had a few
minutes to think about this.
\end{instructornote}
\begin{solution}
Interestingly enough, we can only save one digit. \par
Any optimal sequence has 15 digits, for example \texttt{0000111101100101000}
\end{solution}
\vfill
\pagebreak

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\section{Words}
\definition{}
An \textit{alphabet} is a set of symbols. \par
For example, $\{\texttt{0}, \texttt{1}\}$ is an alphabet of two symbols, \par
and $\{\texttt{a}, \texttt{b}, \texttt{c}\}$ is an alphabet of three.
\definition{}
A \textit{word} over an alphabet $A$ is a sequence of symbols in that alphabet. \par
For example, $\texttt{00110}$ is a word over the alphabet $\{\texttt{0}, \texttt{1}\}$. \par
We'll let $\varnothing$ denote the empty word, which exists over every alphabet.
\definition{}
Let $v$ and $w$ be words over the same alphabet. \par
We say $v$ is a \textit{subword} of $w$ if $v$ is contained in $w$. \par
For example, \texttt{11} is a subword of \texttt{011}, but \texttt{00} is not.
\definition{}
Recall \ref{lockproblem}. From now on, we'll call this the \textit{$n$-subword problem}: \par
Given an alphabet $A$ and a positive integer $n$, we want a \par
word over $A$ that contains all possible length-$n$ subwords. \par
The shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}.
\problem{}
List all subwords of \texttt{110}. \par
\hint{There are six.}
\begin{solution}
They are $\varnothing$, \texttt{0}, \texttt{1}, \texttt{10}, \texttt{11}, and \texttt{110}.
\end{solution}
\vfill
\definition{}
Let $\mathcal{S}_n(w)$ be the number of subwords of length $n$ in a word $w$.
\problem{}
Find the following:
\begin{itemize}
\item $\mathcal{S}_n(\texttt{101001})$ for $n \in \{0, 1, ..., 6\}$
\item $\mathcal{S}_n(\texttt{abccac})$ for $n \in \{0, 1, ..., 6\}$
\end{itemize}
\begin{solution}
In order from $\mathcal{S}_0$ to $\mathcal{S}_6$:
\begin{itemize}
\item 1, 2, 3, 3, 3, 2, 1
\item 1, 3, 5, 4, 3, 2, 1
\end{itemize}
\end{solution}
\vfill
\pagebreak
\problem{}
Let $w$ be a word over an alphabet of size $k$. \par
Prove the following:
\begin{itemize}
\item $\mathcal{S}_n(w) \leq k^n$
\item $\mathcal{S}_n(w) \geq \mathcal{S}_{n-1}(w) - 1$
\item $\mathcal{S}_n(w) \leq k \times \mathcal{S}_{n-1}(w)$
\end{itemize}
\begin{solution}
\begin{itemize}
\item There are $k$ choices for each of $n$ letters in the subword.
So, there are $k^n$ possible words of length $n$, and $\mathcal{S}_n(w) \leq k^n$.
\item For almost every distinct subword counted by $\mathcal{S}_{n-1}$,
concatenating the next letter creates a distinct length $n$ subword.
The only exception is the last subword with length $n-1$, so
$\mathcal{S}_n(w) \geq \mathcal{S}_{n-1}(w) - 1$
\item For each subword counted by $\mathcal{S}_{n-1}$, there are $k$ possibilities
for the letter that follows in $w$. Each element in the count $\mathcal{S}_n$ comes from
one of $k$ different length $n$ words starting with an element counted by $\mathcal{S}_{n-1}$.
Thus, $\mathcal{S}_n(w) \leq k \times \mathcal{S}_{n-1}(w)$
\end{itemize}
\end{solution}
\vfill
\pagebreak
\definition{}
Let $v$ and $w$ be words over the same alphabet. \par
The word $vw$ is the word formed by writing $v$ after $w$. \par
For example, if $v = \texttt{1001}$ and $w = \texttt{10}$, $vw$ is $\texttt{100110}$.
\problem{}
Let $F_k$ denote the word over the alphabet $\{\texttt{0}, \texttt{1}\}$ obtained from the following relation:
\begin{equation*}
F_0 = \texttt{0}; ~~ F_1 = \texttt{1}; ~~ F_k = F_{k-1}F_{k-2}
\end{equation*}
We'll call this the \textit{Fibonacci word} of order $k$.
\begin{itemize}
\item What are $F_3$, $F_4$, and $F_5$?
\item Compute $\mathcal{S}_0$ through $\mathcal{S}_5$ for $F_5$.
\item Show that the length of $F_k$ is the $(k + 2)^\text{th}$ Fibonacci number. \par
\hint{Induction.}
\end{itemize}
\begin{solution}
\begin{itemize}
\item $F_3 = \texttt{101}$
\item $F_4 = \texttt{10110}$
\item $F_5 = \texttt{10110101}$
\end{itemize}
\linehack{}
\begin{itemize}
\item $\mathcal{S}_0 = 1$
\item $\mathcal{S}_1 = 2$
\item $\mathcal{S}_2 = 3$
\item $\mathcal{S}_3 = 4$
\item $\mathcal{S}_4 = 5$
\item $\mathcal{S}_5 = 4$
\end{itemize}
\linehack
As stated, use induction. The base case is trivial. \par
Let $N_k$ represent the Fibonacci numbers, with $N_0 = 0$, $N_1 = 1$, and $N_{k} = N_{k-1} + N_{k-2}$
\vspace{2mm}
Assume that $F_k$ has length $N_{k+2}$ for all $k \leq n$.
We want to show that $F_{k+1}$ has length $N_{k+3}$. \par
Since $F_{k} = F_{k-1}F_{k-2}$, it has the length $|F_{k-1}| + |F_{k-2}|$. \par
By our assumption, $|F_{k-1}| = N_{k+1}$ and $|F_{k-2}| = N_{k}$. \par
So, $|F_{k}| = |F_{k-1}| + |F_{k-2}| = N_{k+1} + N_{k} = N_{k + 2}$.
\end{solution}
\vfill
\pagebreak
% C_k is called the "Champernowne word" of order k.
\problem{}<cword>
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 $C_3 = \texttt{011011100101110111}$.
\begin{itemize}
\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 $n < k$.
\end{itemize}
\hint{
If $v$ is a subword of $w$ and $w$ is a subword of $u$, $v$ must be a subword of $u$. \par
In other words, the \say{subword} relation is transitive.
}
\begin{solution}
$\mathcal{S}_0 = 1$, $\mathcal{S}_1 = 2$, $\mathcal{S}_2 = 4$, and $\mathcal{S}_3 = 7$.
\linehack{}
First, we show that $\mathcal{S}_k(C_k) = 2^k - 1$. \par
Consider an arbitrary word $w$ of length $k$. We'll consider three cases:
\begin{itemize}
\item If $w$ consists only of zeros, $w$ does not appear in $C_k$.
\item If $w$ starts with a \texttt{1}, $w$ must appear in $C_k$ by construction.
\item If $w$ does starts with a \texttt{0} and contains a \texttt{1}, $w$ has the form
$\texttt{0}^x\texttt{1}[..\texttt{y}..]$ \par
\note{
That is, $x$ copies of \texttt{0} followed by a \texttt{1}, followed by \par
an arbitrary sequence $\texttt{y}$ with length $(k-x-1)$.
} \par
Now consider the word $\texttt{1}[..\texttt{y}..]\texttt{0}^x\texttt{1}[..\texttt{y}..]\texttt{0}^{(x-1)}\texttt{1}$. \par
This is the concatenation of two consecutive binary numbers with $k$ digits, and thus appears in $C_k$.
$w$ is a subword of this word, and therefore also appears in $C_k$.
\end{itemize}
\linehack{}
We can use the above result to conclude that $\mathcal{S}_n(C_k) = 2^n$ for $n < k$: \par
If we take any word of length $n < k$ and repeatedly append \texttt{1} to create a word of length $k$, \par
we end up with a subword of $C_k$ by the reasoning above. \par
Thus, any word of length $n$ is a subword of $w$, of which there are $2^n$.
\end{solution}
\vfill
\problem{}
Convince yourself that $C_{n+1}$ provides a solution to the $n$-subword problem over $\{\texttt{0}, \texttt{1}\}$. \par
\note[Note]{$C_{n+1}$ may or may not be an \textit{optimal} solution---but it is a \textit{valid} solution} \par
Which part of \ref{cword} shows that this is true?
\pagebreak

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\section{De Bruijn Words}
Before we continue, we'll need to review some basic
graph theory.
\definition{}
A \textit{directed graph} consists of nodes and directed edges. \par
An example is shown below. It consists of three vertices (labeled $a, b, c$), \par
and five edges (labeled $0, ... , 4$).
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (2, 0) {$b$};
\node[main] (c) at (4, 0) {$c$};
\end{scope}
\draw[->]
(a) edge node[label] {$0$} (b)
(a) edge[loop above] node[label] {$1$} (a)
(b) edge[bend left] node[label] {$2$} (c)
(b) edge[loop above] node[label] {$3$} (b)
(c) edge[bend left] node[label] {$4$} (b)
;
\end{tikzpicture}
\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
\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
\definition{}
An \textit{Eulerian path} is a path that visits each edge of a graph exactly once. \par
An \textit{Eulerian cycle} is an Eulerian path that starts and ends on the same node.
\problem{}
Find the single unique Eulerian cycle in the graph below.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (2, 0) {$b$};
\node[main] (c) at (4, 0) {$c$};
\end{scope}
\draw[->]
(a) edge[bend left] node[label] {$0$} (b)
(b) edge[bend left] node[label] {$1$} (a)
(b) edge[bend left] node[label] {$2$} (c)
(c) edge[bend left] node[label] {$3$} (b)
(c) edge[loop right] node[label] {$4$} (c)
;
\end{tikzpicture}
\end{center}
\begin{solution}
$24310$ is one way to write this cycle. \par
There are other options, but they're all the same.
\end{solution}
\vfill
\pagebreak
\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
\footnotetext{Dutch. Rhymes with \say{De Grown.}}
\problem{}
Let $\mathcal{B}_n$ be the length of an order-$n$ De Bruijn word. \par
Show that the following bounds always hold:
\begin{itemize}
\item $\mathcal{B}_n \leq n2^n$
\item $\mathcal{B}_n \geq 2^n + n - 1$
\end{itemize}
\begin{solution}
\begin{itemize}
\item There are $2^n$ binary words with length $n$. \par
Concatenate these to get a word with length $n2^n$.
\item A word must have at least $2^n + n - 1$ letters to have $2^n$ subwords with length $n$.
\end{itemize}
\end{solution}
\remark{}
Now, we'd like to show that $\mathcal{B}_n = 2^n + n - 1$... \par
That is, that the optimal solution to the subword problem always has $2^n + n - 1$ letters.
\definition{}
Consider a $n$-length word $w$. \par
The \textit{prefix} of $w$ is the word formed by the first $n-1$ letters of $w$. \par
The \textit{suffix} of $w$ is the word formed by the last $n-1$ letters of $w$. \par
For example, the prefix of the word \texttt{1101} is \texttt{110}, and its suffix is \texttt{101}.
The prefix and suffix of any one-letter word are both $\varnothing$.
\definition{}
A \textit{De Bruijn graph} of order $n$, denoted $G_n$, is constructed as follows:
\begin{itemize}
\item Nodes are created for each word of length $n - 1$.
\item A directed edge is drawn from $a$ to $b$ if the suffix of
$a$ matches the prefix of $b$. \par
Note that a node may have an edge to itself.
\item We label each edge with the last letter of $b$.
\end{itemize}
$G_2$ and $G_3$ are shown below.
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
$G_2$
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (0) at (0, 0) {\texttt{0}};
\node[main] (1) at (2, 0) {\texttt{1}};
\end{scope}
\draw[->]
(0) edge[loop left] node[label] {$0$} (0)
(1) edge[loop right] node[label] {$1$} (1)
(1) edge[bend left] node[label] {$0$} (0)
(0) edge[bend left] node[label] {$1$} (1)
;
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
$G_3$
\begin{tikzpicture}[scale = 0.9]
\begin{scope}[layer = nodes]
\node[main] (00) at (0, 0) {\texttt{00}};
\node[main] (01) at (2, 1) {\texttt{01}};
\node[main] (10) at (2, -1) {\texttt{10}};
\node[main] (11) at (4, 0) {\texttt{11}};
\end{scope}
\draw[->]
(00) edge[loop left] node[label] {$0$} (00)
(11) edge[loop right] node[label] {$1$} (11)
(00) edge[bend left] node[label] {$1$} (01)
(01) edge[bend left] node[label] {$0$} (10)
(10) edge[bend left] node[label] {$1$} (01)
(10) edge[bend left] node[label] {$0$} (00)
(01) edge[bend left] node[label] {$1$} (11)
(11) edge[bend left] node[label] {$0$} (10)
;
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\vfill
\pagebreak
\problem{}
Draw $G_4$.
\begin{solution}
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (7) at (0, 0) {\texttt{111}};
\node[main] (3) at (0, -2) {\texttt{011}};
\node[main] (6) at (2, -2) {\texttt{110}};
\node[main] (4) at (4, -2) {\texttt{100}};
\node[main] (1) at (-4, -4) {\texttt{001}};
\node[main] (5) at (0, -4) {\texttt{101}};
\node[main] (2) at (-2, -4) {\texttt{010}};
\node[main] (0) at (-2, -6) {\texttt{000}};
\end{scope}
\draw[->]
(0) edge[loop left, looseness = 7] node[label] {\texttt{0}} (0)
(7) edge[loop above, looseness = 7] node[label] {\texttt{1}} (7)
(0) edge[out=90,in=-90] node[label] {\texttt{1}} (1)
(1) edge node[label] {\texttt{0}} (2)
(1) edge[out=45,in=-135] node[label] {\texttt{1}} (3)
(2) edge[bend left] node[label] {\texttt{1}} (5)
(3) edge node[label] {\texttt{0}} (6)
(3) edge node[label] {\texttt{1}} (7)
(5) edge[bend left] node[label] {\texttt{0}} (2)
(5) edge node[label] {\texttt{1}} (3)
(6) edge[bend left] node[label] {\texttt{0}} (4)
(6) edge[out=-90,in=0] node[label] {\texttt{1}} (5)
(7) edge[out=0,in=90] node[label] {\texttt{0}} (6)
;
\draw[->, rounded corners = 10mm]
(4) to (4, 2) to node[label] {\texttt{1}} (-4, 2) to (1)
;
\draw[->, rounded corners = 10mm]
(4) to (4, -6) to node[label] {\texttt{0}} (0)
;
\draw[->, rounded corners = 5mm]
(2) to (-2, -5) to node[label] {\texttt{0}} (3, -5) to (3, -2) to (4)
;
\end{tikzpicture}
\end{center}
\begin{instructornote}
This graph also appears as a solution to a different
problem in the DFA handout.
\end{instructornote}
\end{solution}
\vfill
\problem{}
\begin{itemize}
\item Show that $G_n$ has $2^{n-1}$ nodes and $2^n$ edges;
\item that each node has two outgoing edges;
\item and that there are as many edges labeled $0$ as are labeled $1$.
\end{itemize}
\begin{solution}
\begin{itemize}
\item There $2^{n-1}$ binary words of length $n-1$.
\item The suffix of a given word is the prefix of two other words, \par
so there are two edges leaving each node.
\item One of those words will end with one, and the other will end with zero.
\item Our $2^{n-1}$ nodes each have $2$ outgoing edges---we thus have $2^n$ edges in total.
\end{itemize}
\end{solution}
\vfill
\pagebreak
\theorem{}
We can now easily construct De Bruijn words for a given $n$: \par
\begin{itemize}
\item Construct $G_n$,
\item then an Eulerian cycle in $G_n$.
\item Finally, construct a De Bruijn by writing the label of our starting vertex,
then appending the label of every edge we travel.
\end{itemize}
\problem{}
Find De Bruijn words of orders $2$, $3$, and $4$.
\begin{solution}
\begin{itemize}
\item
One Eulerian cycle in $G_2$ starts at node \texttt{0}, and takes the edges labeled $[1, 1, 0, 0]$. \par
We thus have the word \texttt{01100}.
\item
In $G_3$, we have an Eulerian cycle that visits nodes in the following order: \par
$
\texttt{00}
\rightarrow \texttt{01}
\rightarrow \texttt{11}
\rightarrow \texttt{11}
\rightarrow \texttt{10}
\rightarrow \texttt{01}
\rightarrow \texttt{10}
\rightarrow \texttt{00}
\rightarrow \texttt{00}
$\par
This gives us the word \texttt{0011101000}
\item Similarly, we $G_4$ gives us the word \texttt{0001 0011 0101 1110 000}. \par
\note{Spaces have been added for convenience.}
\end{itemize}
\end{solution}
\vfill
\pagebreak

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\section{Line Graphs}
\problem{}
Given a graph $G$, we can construct its \textit{line graph} (denoted $\mathcal{L}(G)$) by doing the following: \par
\begin{itemize}
\item Creating a node in $\mathcal{L}(G)$ for each edge in $G$
\item Drawing a directed edge between every pair of nodes $a, b$ in $\mathcal{L}(G)$ \par
if the corresponding edges in $G$ are adjacent. \par
\note{That is, if edge $b$ in $G$ starts at the node at which $a$ ends.}
\end{itemize}
\problem{}
Draw the line graph for the graph below. \par
Have an instructor check your solution.
\begin{center}
\begin{tikzpicture}
\begin{scope}[layer = nodes]
\node[main] (a) at (0, 0) {$a$};
\node[main] (b) at (2, 0) {$b$};
\node[main] (c) at (4, 0) {$c$};
\end{scope}
\draw[->]
(a) edge[bend left] node[label] {$0$} (b)
(b) edge[bend left] node[label] {$1$} (a)
(b) edge[bend left] node[label] {$2$} (c)
(c) edge[bend left] node[label] {$3$} (b)
(c) edge[loop right] node[label] {$4$} (c)
;
\end{tikzpicture}
\end{center}
\begin{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) {$4$};
\end{scope}
\draw[->]
(0) edge[bend left] (2)
(2) edge[bend left] (0)
(0) edge (4)
(4) edge[bend left] (2)
(2) edge (1)
(1) edge[bend left] (3)
(3) edge[bend left] (1)
(3) edge (0)
;
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\definition{}
We say a graph $G$ is \textit{connected} if there is a path
between any two vertices of $G$.
\problem{}
Show that if $G$ is connected, $\mathcal{L}(G)$ is connected.
\begin{solution}
Let $a, b$ and $x, y$ be nodes in a connected graph $G$ so that an edges $a \rightarrow b$ and
and $x \rightarrow y$ exist. Since $G$ is connected, we can find a path from $b$ to $x$.
The path $a$ to $y$ corresponds to a path in $\mathcal{L}(G)$ between $a \rightarrow b$ and $x \rightarrow y$.
\end{solution}
\vfill
\pagebreak
\definition{}
Consider $\mathcal{L}(G_n)$, where $G_n$ is the $n^\text{th}$ order De Bruijn graph. \par
\vspace{2mm}
We'll need to label the vertices of $\mathcal{L}(G_n)$. To do this, do the following:
\begin{itemize}
\item Let $a$ and $b$ be nodes in $G_n$
\item Let \texttt{x} be the first letter of $a$
\item Let \texttt{y}, the last letter of $b$
\item Let $\overline{\texttt{p}}$ be the prefix/suffix that $a$ and $b$ share. \par
Note that $a = \texttt{x}\overline{\texttt{p}}$ and $b = \overline{\texttt{p}}\texttt{y}$,
\end{itemize}
Now, relabel the edge from $a$ to $b$ as $\texttt{x}\overline{\texttt{p}}\texttt{y}$. \par
Use these new labels to name nodes in $\mathcal{L}(G_n)$.
\problem{}
Construct $\mathcal{L}(G_2)$ and $\mathcal{L}(G_3)$. What do you notice?
\begin{solution}
After fixing edge labels, we find that
$\mathcal{L}(G_2) \cong G_3$, and $\mathcal{L}(G_3) \cong G_4$
\end{solution}

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\usetikzlibrary{arrows.meta}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{patterns}
% We put nodes in a separate layer, so we can
% slightly overlap with paths for a perfect fit
\pgfdeclarelayer{nodes}
\pgfdeclarelayer{path}
\pgfsetlayers{main,nodes}
% Layer settings
\tikzset{
% Layer hack, lets us write
% later = * in scopes.
layer/.style = {
execute at begin scope={\pgfonlayer{#1}},
execute at end scope={\endpgfonlayer}
},
%
% Arrowhead tweak
>={Latex[ width=2mm, length=2mm ]},
%
% Labels inside edges
label/.style = {
rectangle,
% For automatic red background in solutions
fill = \ORMCbgcolor,
draw = none,
rounded corners = 0mm
},
%
% Nodes
main/.style = {
draw,
circle,
fill = white,
line width = 0.35mm
},
%
% Loop tweaks
loop above/.style = {
min distance = 2mm,
looseness = 8,
out = 45,
in = 135
},
loop below/.style = {
min distance = 5mm,
looseness = 10,
out = 315,
in = 225
},
loop right/.style = {
min distance = 5mm,
looseness = 10,
out = 45,
in = 315
},
loop left/.style = {
min distance = 5mm,
looseness = 10,
out = 135,
in = 215
}
}