\section{DFAs}

This week, we will study computational devices called \textit{deterministic finite automata}. \par
A DFA has a simple job: it will either \say{accept} or \say{reject} a string of letters.

\vspace{2mm}

Consider the automaton $A$ shown below:

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[main] (a) at (0, 0) {$a$};
		\node[accept] (b) at (2, 0) {$b$};
		\node[main] (c) at (5, 0) {$c$};
		\node[start] (s) at (-2, 0) {\texttt{start}};
	\end{scope}

	\draw[->]
		(s) edge (a)
		(a) edge node[label] {$1$} (b)
		(a) edge[loop above] node[label] {$0$} (a)
		(b) edge[bend left] node[label] {$0$} (c)
		(b) edge[loop above] node[label] {$1$} (b)
		(c) edge[bend left] node[label] {$0,1$} (b)
	;
\end{tikzpicture}
\end{center}

$A$ takes strings of letters in the alphabet $\{0, 1\}$ and reads them left to right, one letter at a time. \par
Starting in the state $a$, the automaton $A$ will move between states along the edge marked by each letter. \par

\vspace{2mm}

Note that node $b$ has a \say{double edge} in the diagram above. This means that the state $b$ is \textit{accepting}. Any string that makes $A$ end in state $b$ is \textit{accepted}. Similarly, strings that end in states $a$ or $c$ are \textit{rejected}. \par

\vspace{2mm}

For example, consider the string \texttt{1011}. \par
$A$ will go through the states $a - b - c - b - b$ while processing this string. \par


\problem{}
Which of the following strings are accepted by $A$? \\
\begin{itemize}
	\item \texttt{1}
	\item \texttt{1010}
	\item \texttt{1110010}
	\item \texttt{1000100}
\end{itemize}

\vfill



\problem{}
Describe the general form of a string accepted by $A$. \par
\hint{Work backwards from the accepting state, and decide what all the strings must look like at the end in order to be accepted.}

\begin{solution}
	$A$ will accept strings that contain at least one $1$ and end with an even (possibly 0) number of zeroes.
\end{solution}

\vfill
\pagebreak



Now consider the automaton $B$, which uses the alphabet $\{a, b\}$. \par
It starts in the state $s$ and has two accepting states $a_1$ and $b_1$.

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[main] (s) at (0, 0) {$s$};
		\node[accept] (a1) at (-2, -0.5) {$a_1$};
		\node[main] (a2) at (-2, -2.5) {$a_2$};
		\node[accept] (b1) at (2, -0.5) {$b_1$};
		\node[main] (b2) at (2, -2.5) {$b_2$};
		\node[start] (start) at (0, 1) {\texttt{start}};
	\end{scope}

	\clip (-4, -3.5) rectangle (4, 1);

	\draw[->]
		(start) edge (s)
		(s) edge node[label] {\texttt{a}} (a1)
		(a1) edge[loop left] node[label] {\texttt{a}} (a1)
		(a1) edge[bend left] node[label] {\texttt{b}} (a2)
		(a2) edge[bend left] node[label] {\texttt{a}} (a1)
		(a2) edge[loop left] node[label] {\texttt{b}} (a2)
		(s) edge node[label] {\texttt{b}} (b1)
		(b1) edge[loop right] node[label] {\texttt{b}} (b1)
		(b1) edge[bend left] node[label] {\texttt{a}} (b2)
		(b2) edge[bend left] node[label] {\texttt{b}} (b1)
		(b2) edge[loop right] node[label] {\texttt{a}} (b2)
	;
\end{tikzpicture}
\end{center}

\problem{}
Which of the following strings are accepted by $B$:
\begin{itemize}
	\item \texttt{aa}
	\item \texttt{abba}
	\item \texttt{abbba}
	\item \texttt{baabab}
\end{itemize}

\vfill



\problem{}
Describe the strings accepted by $B$.

\begin{solution}
	$B$ accepts strings that start and end with the same letter.
\end{solution}

\vfill
\pagebreak




\definition{}
An \textit{alphabet} is a finite set of symbols. \par

\definition{}
A \textit{string} over an alphabet $Q$ is a finite sequence of symbols from $Q$. \par
We denote the empty string $\varepsilon$. \par


\vspace{2mm}

$Q^*$ is the set of all possible strings over $Q$. \par
For example, $\{\texttt{0}, \texttt{1}\}^*$ is the set $\{\varepsilon, \texttt{0}, \texttt{1}, \texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}, \texttt{000},... \}$ \par
Note that this set contains the empty string.

\definition{}
A \textit{language} over an alphabet $Q$ is a subset of $Q^*$. \\
For example, the language \say{strings of length 2} over $\{\texttt{0}, \texttt{1}\}$ is $\{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$

\definition{}
We say a language $L$ is \textit{recognized} by a DFA if that DFA accepts a string $w$ if and only if $w \in L$.


%\begin{remark}
%A machine, such as DFA or Turing machine, may accept several strings, but it always recognizes only one language. If the machine %accepts no strings, it still recognizes one language — namely, the empty language $\emptyset$.
%\end{remark}


\vspace{8mm}

\problem{}
How many strings of length $n$ are accepted by the automaton $C$?

\begin{center}
\begin{tikzpicture}
	\begin{scope}[layer = nodes]
		\node[main] (0) at (0, 0) {$0$};
		\node[accept] (1) at (3, 0) {$1$};
		\node[main] (2) at (5, 0) {$2$};
		\node[start] (s) at (-2, 0) {\texttt{start}};
	\end{scope}

	\draw[->]
		(s) edge (0)
		(0) edge[loop above] node[label] {\texttt{b}} (0)
		(0) edge[bend left] node[label] {\texttt{a}} (1)
		(1) edge[bend left] node[label] {\texttt{b}} (0)
		(1) edge node[label] {\texttt{a}} (2)
		(2) edge[loop above] node[label] {\texttt{a,b}} (2)
	;
\end{tikzpicture}
\end{center}

\begin{solution}
	If $A_n$ is the number of accepted strings of length $n$, then $A_n = A_{n-1}+A_{n-2}$. \par
	Computing initial conditions, we see that $A_n$ is an $n+2$-th Fibonacci number.
\end{solution}

%\begin{remark}
%Note that all the states in our DFAs $A$, $B$ and $C$ from figures 1, 2, 3 have outgoing symbols for each letter of the alphabet. %Do the same for your DFAs.
%\end{remark}

\vfill
\pagebreak

\problem{}
Draw DFAs that recognize the following languages. In all parts, the alphabet is $\{0, 1\}$:
\begin{itemize}
	\item $\{w~ | ~w~ \text{begins with a \texttt{1} and ends with a \texttt{0}}\}$
	\item $\{w~ | ~w~ \text{contains at least three \texttt{1}s}\}$
	\item $\{w~ | ~w~ \text{contains the substring \texttt{0101} (i.e, $w = x\texttt{0101}y$ for some $x$ and $y$)}\}$
	\item $\{w~ | ~w~ \text{has length at least three and its third symbol is a \texttt{0}}\}$
	\item $\{w~ | ~w~ \text{starts with \texttt{0} and has odd length, or starts with \texttt{1} and has even length}\}$
	\item $\{w~ | ~w~ \text{doesn't contain the substring \texttt{110}}\}$
\end{itemize}

\begin{solution}
	$\{w~ | ~w~ \text{begins with a \texttt{1} and ends with a \texttt{0}}\}$


	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[accept] (0) at (0, 2) {$\phantom{0}$};
			\node[main] (1) at (3, 2) {$\phantom{0}$};
			\node[main] (2) at (0, 0) {$\phantom{0}$};
			\node[main] (3) at (3, 0) {$\phantom{0}$};
			\node[start] (s) at (-2, 0) {\texttt{start}};
		\end{scope}

		\clip (-2, -1) rectangle (4.5, 3);

		\draw[->]
			(s) edge (2)
			(0) edge[loop left] node[label] {\texttt{1}} (0)
			(0) edge[bend left] node[label] {\texttt{1}} (1)
			(1) edge[loop right] node[label] {\texttt{1}} (1)
			(1) edge[bend left] node[label] {\texttt{0}} (0)
			(2) edge[out=90, in=270] node[label] {\texttt{1}} (1)
			(2) edge node[label] {\texttt{0}} (3)
			(3) edge[loop right] node[label] {\texttt{1,0}} (3)
		;
	\end{tikzpicture}
	\end{center}

	\linehack{}
	$\{w~ | ~w~ \text{contains at least three \texttt{1}s}\}$

	\begin{center}


	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (s) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[main] (1) at (2, 0) {$\phantom{0}$};
			\node[main] (2) at (4, 0) {$\phantom{0}$};
			\node[accept] (3) at (6, 0) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(s) edge (0)
			(0) edge[loop above] node[label] {\texttt{0}} (0)
			(1) edge[loop above] node[label] {\texttt{0}} (1)
			(2) edge[loop above] node[label] {\texttt{0}} (2)
			(3) edge[loop above] node[label] {\texttt{0,1}} (3)
			(0) edge node[label] {\texttt{1}} (1)
			(1) edge node[label] {\texttt{1}} (2)
			(2) edge node[label] {\texttt{1}} (3)
		;
	\end{tikzpicture}
	\end{center}

	\linehack{}
	$\{w~ | ~w~ \text{contains the substring \texttt{0101}}\}$

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (s) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[main] (1) at (2, 1) {$\phantom{0}$};
			\node[main] (2) at (4, 1) {$\phantom{0}$};
			\node[main] (3) at (0, 3) {$\phantom{0}$};
			\node[accept] (4) at (2, 3) {$\phantom{0}$};
		\end{scope}

		% Tikz includes invisible handles in picture size.
		% This crops the image to fix sizing.
		\clip (-2, -1.75) rectangle (5, 5.25);

		\draw[->]
			(s) edge (0)
			(0) edge[loop above] node[label] {\texttt{1}} (0)
			(0) edge[bend right] node[label] {\texttt{0}} (1)
			(1) edge[loop above] node[label] {\texttt{0}} (1)
			(1) edge node[label] {\texttt{1}} (2)
			(3) edge[bend right] node[label] {\texttt{0}} (1)
			(3) edge node[label] {\texttt{1}} (4)
			(4) edge[loop above] node[label] {\texttt{0,1}} (4)
		;

		\draw[->, rounded corners = 10mm]
			(2) to (4, 5) to node[label] {\texttt{0}} (0, 5) to (3)
		;

		\draw[->, rounded corners = 10mm]
			(2) to (4, -1.5) to node[label] {\texttt{1}} (0, -1.5) to (0)
		;

	\end{tikzpicture}
	\end{center}

	Notice that after getting two 0's in a row we don't reset to the initial state.

	\pagebreak
	$\{w~ | ~w~ \text{has length at least three and its third symbol is a \texttt{0}}\}$

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (s) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[main] (1) at (2, 0) {$\phantom{0}$};
			\node[main] (2) at (4, 0) {$\phantom{0}$};
			\node[accept] (3) at (6, 1) {$\phantom{0}$};
			\node[accept] (4) at (6, -1) {$\phantom{0}$};
		\end{scope}

		\clip (-2, -2.5) rectangle (7, 2.5);

		\draw[->]
			(s) edge (0)
			(0) edge node[label] {\texttt{0,1}} (1)
			(1) edge node[label] {\texttt{0,1}} (2)
			(2) edge node[label] {\texttt{0}} (3)
			(2) edge node[label] {\texttt{1}} (4)
			(3) edge[loop above] node[label] {\texttt{0,1}} (3)
			(4) edge[loop below] node[label] {\texttt{0,1}} (4)
		;
	\end{tikzpicture}
	\end{center}

	\linehack{}
	$\{w~ | ~w~ \text{starts with \texttt{0} and has odd length, or starts with \texttt{1} and has even length}\}$

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (s) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[accept] (1) at (2, 1) {$\phantom{0}$};
			\node[main] (2) at (4, 1) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(s) edge (0)
			(0) edge node[label] {\texttt{0}} (1)
			(1) edge[bend left] node[label] {\texttt{0,1}} (2)
			(2) edge[bend left] node[label] {\texttt{0,1}} (1)
		;

		\draw[->, rounded corners = 5mm]
			(0) to node[label] {\texttt{1}} (4, 0) to (2)
		;
	\end{tikzpicture}
	\end{center}

	\linehack{}
	$\{w~ | ~w~ \text{doesn't contain the substring \texttt{110}}\}$

	\begin{center}


	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (s) at (-2, 0){\texttt{start}};
			\node[accept] (0) at (0, 0) {$\phantom{0}$};
			\node[accept] (1) at (2, 0) {$\phantom{0}$};
			\node[accept] (2) at (4, 0) {$\phantom{0}$};
			\node[main] (3) at (6, 0) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(s) edge (0)
			(0) edge[loop above] node[label] {\texttt{0}} (0)
			(2) edge[loop above] node[label] {\texttt{1}} (2)
			(3) edge[loop above] node[label] {\texttt{0,1}} (3)
			(0) edge[bend left] node[label] {\texttt{1}} (1)
			(1) edge[bend left] node[label] {\texttt{0}} (0)
			(1) edge node[label] {\texttt{1}} (2)
			(2) edge node[label] {\texttt{0}} (3)
		;
	\end{tikzpicture}
	\end{center}

	Notice that after getting three 1's in a row we don't reset to the initial state.
\end{solution}

\vfill
\pagebreak

\problem{}
Draw a DFA over an alphabet $\{\texttt{a}, \texttt{b}, \texttt{@}, \texttt{.}\}$ recognizing the language of strings of the form \texttt{user@website.domain}, where \texttt{user}, \texttt{website} and \texttt{domain} are nonempty strings over $\{\texttt{a}, \texttt{b}\}$ and \texttt{domain} has length 2 or 3.

\begin{solution}

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (start) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[main] (1) at (0, 2) {$\phantom{0}$};
			\node[main] (2) at (0, 4) {$\phantom{0}$};
			\node[main] (3) at (0, 6) {$\phantom{0}$};
			\node[main] (4) at (0, 8) {$\phantom{0}$};
			\node[main] (5) at (0, 10) {$\phantom{0}$};
			\node[accept] (6) at (0, 12) {$\phantom{0}$};
			\node[accept] (7) at (0, 14) {$\phantom{0}$};

			\node[main] (8) at (5, 7) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(start) edge (0)
			(0) edge node[label] {\texttt{a,b}} (1)
			(1) edge node[label] {\texttt{@}} (2)
			(2) edge node[label] {\texttt{a,b}} (3)
			(3) edge node[label] {\texttt{.}} (4)
			(4) edge node[label] {\texttt{a,b}} (5)
			(5) edge node[label] {\texttt{a,b}} (6)
			(6) edge node[label] {\texttt{a,b}} (7)

			(1) edge[loop left] node[label] {\texttt{a,b}} (1)
			(3) edge[loop left] node[label] {\texttt{a,b}} (3)

			(0) edge[out=0, in=270] node[label] {\texttt{@,.}} (8)
			(1) edge[out=0, in=245] node[label] {\texttt{.}} (8)
			(2) edge[out=0, in=220] node[label] {\texttt{@,.}} (8)
			(3) edge[out=0, in=195] node[label] {\texttt{@}} (8)
			(4) edge[out=0, in=170] node[label] {\texttt{@,.}} (8)
			(5) edge[out=0, in=145] node[label] {\texttt{@,.}} (8)
			(6) edge[out=0, in=120] node[label] {\texttt{@,.}} (8)
			(7) edge[out=0, in=95] node[label] {\texttt{a,b,@,.}} (8)
		;

		\draw[->, rounded corners = 5mm]
			(8) to +(1.5, 1) to node[label] {\texttt{a,b,@,.}} +(1.5, -1) to (8)
		;
	\end{tikzpicture}
	\end{center}

\end{solution}

\vfill
\pagebreak

\problem{}
Draw a state diagram for a DFA over an alphabet of your choice that accepts exactly $f(n)$ strings of length $n$ if \\
\begin{itemize}
	\item $f(n) = n$
	\item $f(n) = n+1$
	\item $f(n) = 3^n$
	\item $f(n) = n^2$
	\item $f(n)$ is a Tribonacci number. \par
	Tribonacci numbers are defined by the sequence $f(0) = 0$, $f(1) = 1$, $f(2) = 1$,
	and $f(n) = f(n-1)+f(n-2)+f(n-3)$ for $n \ge 3$ \par
	\hint{Fibonacci numbers are given by the automaton prohibiting two \texttt{'a'}s in a row.}
\end{itemize}


\begin{solution}

	\textbf{Part 4:} $f(n) = n^2$ \par
	Consider the language of words over $\{0, 1, 2\}$ that have the sum of their digits equal to $2$. \par
	Such words must contain two ones or one two:

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (start) at (-2, 0) {\texttt{start}};
			\node[main] (0) at (0, 0) {$\phantom{0}$};
			\node[accept] (1) at (2, 0) {$\phantom{0}$};
			\node[main] (2) at (0, -2) {$\phantom{0}$};
			\node[main] (3) at (2, -2) {$\phantom{0}$};
		\end{scope}

		\clip (-2, 1.5) rectangle (4, -2.75);

		\draw[->]
			(start) edge (0)

			(0) edge[loop above] node[label] {\texttt{0}} (0)
			(1) edge[loop above] node[label] {\texttt{0}} (1)
			(2) edge[loop left] node[label] {\texttt{0}} (2)
			(3) edge[loop right] node[label] {\texttt{0,1,2}} (3)

			(0) edge node[label] {\texttt{2}} (1)
			(0) edge node[label] {\texttt{1}} (2)
			(1) edge node[label] {\texttt{1,2}} (3)
			(2) edge node[label] {\texttt{1}} (1)
			(2) edge node[label] {\texttt{2}} (3)
		;
	\end{tikzpicture}
	\end{center}

	\linehack{}


	\textbf{Part 5:} Tribonacci numbers \par
	Using the hint, we get the following automaton. \par
	It rejects all strings with three \texttt{'a'}s in a row.

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (start) at (-2, 0) {\texttt{start}};
			\node[accept] (0) at (0, 0) {$\phantom{0}$};
			\node[accept] (1) at (0, 2) {$\phantom{0}$};
			\node[accept] (2) at (2, 0) {$\phantom{0}$};
			\node[main] (3) at (4, 0) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(start) edge (0)
			(0) edge[loop below] node[label] {\texttt{b}} (0)
			(3) edge[loop above] node[label] {\texttt{a,b}} (3)
			(0) edge[bend left] node[label] {\texttt{a}} (1)
			(1) edge[bend left] node[label] {\texttt{b}} (0)
			(1) edge[bend left] node[label] {\texttt{a}} (2)
			(2) edge node[label] {\texttt{a}} (3)
			(2) edge node[label] {\texttt{b}} (0)
		;
	\end{tikzpicture}
	\end{center}

	This automaton rejects all strings with three \texttt{'a'}s in a row. If we count accepted strings, we get the Tribonacci numbers with an offest: $f(0) = 1$, $f(1) = 2$, $f(2)=4$, ... \par

	\pagebreak

	We can fix this by adding a node and changing the start state:

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (start) at (1, -2) {\texttt{start}};
			\node[accept] (0) at (0, 0) {$\phantom{0}$};
			\node[accept] (1) at (0, 2) {$\phantom{0}$};
			\node[accept] (2) at (2, 0) {$\phantom{0}$};
			\node[main] (3) at (4, 0) {$\phantom{0}$};
			\node[main] (4) at (3, -2) {$\phantom{0}$};
		\end{scope}

		\draw[->]
			(start) edge (4)
			(2) edge node[label] {\texttt{b}} (0)
			(4) edge node[label] {\texttt{b}} (2)
			(4) edge node[label] {\texttt{a}} (3)

			(0) edge[loop below] node[label] {\texttt{b}} (0)
			(3) edge[loop above] node[label] {\texttt{a,b}} (3)
			(0) edge[bend left] node[label] {\texttt{a}} (1)
			(1) edge[bend left] node[label] {\texttt{b}} (0)
			(1) edge[bend left] node[label] {\texttt{a}} (2)
			(2) edge node[label] {\texttt{a}} (3)
			(2) edge node[label] {\texttt{b}} (0)
		;
	\end{tikzpicture}
	\end{center}
\end{solution}

\vfill
\pagebreak

% \problem{}
% Draw a DFA over an alphabet $\{a, b, c\}$, accepting all the suffixes of the string $abbc$ (including $\varepsilon$) and only them.


\problem{}
	Draw a DFA recognizing the language of strings over $\{\texttt{0}, \texttt{1}\}$ in which \texttt{0} is the third digit from the end. \par
	Prove that any such DFA must have at least 8 states.

\begin{solution}

	\textbf{Part 1:} \par
	Index the states by three-letter suffixes \texttt{000}, \texttt{001}, ..., \texttt{111}. All strings that end with letters $d_1d_2d_3$ will end up in the state $d_1d_2d_3$. We accept all states that start with a \texttt{0}. \par
	Note that we can start at any node if we ignore strings with fewer than three letters.

	\begin{center}
	\begin{tikzpicture}
		\begin{scope}[layer = nodes]
			\node[start] (start) at (-2, 0) {\texttt{start}};
			\node[main] (7) at (0, 0) {\texttt{111}};
			\node[accept] (3) at (0, -2) {\texttt{011}};
			\node[main] (6) at (2, -2) {\texttt{110}};
			\node[main] (4) at (4, -2) {\texttt{100}};
			\node[accept] (1) at (-4, -4) {\texttt{001}};
			\node[main] (5) at (0, -4) {\texttt{101}};
			\node[accept] (2) at (-2, -4) {\texttt{010}};
			\node[accept] (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)

			(start) edge (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}

	\linehack{}

	\textbf{Part 2:} \par
	Strings \texttt{000}, \texttt{001}, ..., \texttt{111} must lead to pairwise different states. \par

	\vspace{2mm}

	Assume \texttt{101} and \texttt{010} lead to the same state. Append a \texttt{1} to the end of the string. \par
	\texttt{101} will become \texttt{011}, and \texttt{010} will become \texttt{101}. These must be different states, since we accept \texttt{011} and reject \texttt{101}. We now have a contradiction: one edge cannot lead to two states!

	\vspace{2mm}

	\texttt{101} and \texttt{010} must thus correspond to distinct states. \par
	We can repeat this argument for any other pair of strings. \par

\end{solution}

\vfill
\pagebreak