<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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[-&gt;]
		(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 $A$? \par
\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[-&gt;]
		(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{}&lt;SameStartAndEnd&gt;
Describe the strings accepted by $B$.

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

\vfill
\pagebreak

Before we continue, let's properly define all the words we've been using in the problems above.

\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'll denote the empty string $\varepsilon$. \par


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

\definition{}
The language \textit{recognized} by a DFA is the set of strings that the DFA accepts.

\vspace{5mm}

\problem{}&lt;fibonacci&gt;
How many strings of length $n$ are accepted by the automaton below? \par
\hint{Induction.}

\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[-&gt;]
		(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}

\vfill

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

		\draw[-&gt;, 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[-&gt;]
			(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[-&gt;]
			(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[-&gt;, 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[-&gt;]
			(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[-&gt;]
			(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[-&gt;, 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

%\problem{}
%Construct a DFA that accepts a binary integer iff it is divisible by two.
%\vfill


\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 \par
\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 letters \say{\texttt{a}} 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[-&gt;]
			(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[-&gt;]
			(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 offset: $f(0) = 1$, $f(1) = 2$, $f(2)=4$, ... \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[-&gt;]
			(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[-&gt;]
			(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[-&gt;, rounded corners = 10mm]
			(4) to (4, 2) to node[label] {\texttt{1}} (-4, 2) to (1)
		;

		\draw[-&gt;, rounded corners = 10mm]
			(4) to (4, -6) to node[label] {\texttt{0}} (0)
		;

		\draw[-&gt;, 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

\problem{}&lt;divsix&gt;
Construct a DFA that accepts an binary integer if and only if it is divisible by six. \par
Strings are read from most to least significant digit.
(that is, 18 will be read as 1,0,0,1,0)

\vfill

\problem{}
Construct a DFA that satisfies \ref{divsix}, but reads digits in the opposite order.

\vfill
\pagebreak</pre></body></html>