diff --git a/Advanced/DFAs/main.tex b/Advanced/DFAs/main.tex
index 4ef0529..036574c 100755
--- a/Advanced/DFAs/main.tex
+++ b/Advanced/DFAs/main.tex
@@ -1,13 +1,10 @@
 % use [nosolutions] flag to hide solutions.
 % use [solutions] flag to show solutions.
 \documentclass[
-	solutions,
-	%shortwarning
+	solutions
 ]{../../resources/ormc_handout}
 
-\include{tikxset.tex}
-
-\tikzset{loop above/.style={min distance=5mm,looseness=10}}
+\input{tikxset.tex}
 
 \begin{document}
 
@@ -19,7 +16,6 @@
 
 
 		\input{parts/0 DFA.tex}
-
-
+		%\input{parts/1 regular.tex}
 
 \end{document}
\ No newline at end of file
diff --git a/Advanced/DFAs/parts/0 DFA.tex b/Advanced/DFAs/parts/0 DFA.tex
index d1b2512..2799677 100644
--- a/Advanced/DFAs/parts/0 DFA.tex	
+++ b/Advanced/DFAs/parts/0 DFA.tex	
@@ -13,9 +13,11 @@ Consider the automaton $A$ shown below:
 		\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)
@@ -25,12 +27,17 @@ Consider the automaton $A$ shown below:
 \end{tikzpicture}
 \end{center}
 
-$A$ always starts in the state $q_1$. This is called the \textit{start state}. \par
-It takes strings using letters in the alphabet $\{0, 1\}$ and reads them left to right, moving between states along the edges marked by each letter.
+$A$ 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
 
-For example, consider the string \texttt{1011}. Processing this string, $A$ will go through the states $q_1 - q_2 - q_3 - q_2 - q_2$. \par
-Note that $q_2$ has a circle in the diagram above. This means that the state $q_2$ is \textit{accepting}, and that all the strings which end up in it are \textit{accepted}. Similarly, states $q_1$ and $q_3$ are \textit{rejecting} and the strings which end up there are \textit{rejected}.
+\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{}
@@ -47,7 +54,7 @@ Which of the following strings are accepted by $A$? \\
 
 
 \problem{}
-Describe the general form of a string accepted by $A$.
+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}
@@ -70,9 +77,13 @@ It starts in the state $s$ and has two accepting states $a_1$ and $b_1$.
 		\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)
@@ -87,9 +98,6 @@ It starts in the state $s$ and has two accepting states $a_1$ and $b_1$.
 \end{tikzpicture}
 \end{center}
 
-
-
-
 \problem{}
 Which of the following strings are accepted by $B$:
 \begin{itemize}
@@ -107,7 +115,7 @@ Which of the following strings are accepted by $B$:
 Describe the strings accepted by $B$.
 
 \begin{solution}
-	They are strings that start and end with the same letter.
+	$B$ accepts strings that start and end with the same letter.
 \end{solution}
 
 \vfill
@@ -135,7 +143,7 @@ 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 $A$ if that DFA accepts a string $w$ iff $w \in L$.
+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}
@@ -154,20 +162,23 @@ How many strings of length $n$ are accepted by the automaton $C$?
 		\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[->]
-		(a) edge[loop above] node[label] {\texttt{b}} (a)
-		(a) edge[bend left] node[label] {\texttt{a}} (b)
-		(b) edge[bend left] node[label] {\texttt{b}} (a)
-		(b) edge node[label] {\texttt{a}} (c)
-		(c) edge[loop above] node[label] {\texttt{a, b}} (c)
+		(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 $A_n = A_{n-1}+A_{n-2}$. Together with initial conditions, we see that $A_n$ is an $n+2$-th Fibonacci number.
+	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}
@@ -178,7 +189,7 @@ How many strings of length $n$ are accepted by the automaton $C$?
 \pagebreak
 
 \problem{}
-Draw DFAs that recognize the following languages. In all parts, the alphabet is $\{0,1\}$:
+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 $\{w~ | ~w~ \text{contains at least three \texttt{1}s}\}$
@@ -188,82 +199,440 @@ Draw DFAs that recognize the following languages. In all parts, the alphabet is
 	\item $\{w~ | ~w~ \text{doesn't contain the substring \texttt{110}}\}$
 \end{itemize}
 
-
 \begin{solution}
-	%\part{a}  \includegraphics[width=0.3\linewidth]{6a.png}
-	%\part{b}  \includegraphics[width=0.4\linewidth]{6b.png}
-	%\part{c}  \includegraphics[width=0.3\linewidth]{6c.png}
+	$\{w~ | ~w~ \text{begins with a \texttt{1} and ends with a \texttt{0}}\}$
+
+
+	\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}
 
-	\medskip
 	Notice that after getting two 0's in a row we don't reset to the initial state.
-	%\part{d}  \includegraphics[width=0.4\linewidth]{6d.png}
-	%\part{e}  \includegraphics[width=0.3\linewidth]{6e.png}
-	%\part{f}  \includegraphics[width=0.4\linewidth]{6f.png}
 
-	\medskip
+	\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
-
-\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}
-%\includegraphics[width=0.9\linewidth]{Email.png}
-\end{solution}
-
 \vfill
 \pagebreak
 
 \problem{}
-Draw a state diagram for a DFA over an alphabet of your choice that recognizes exactly $f(n)$ strings of length $n$ if \\
+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
-	\textit{Tribonacci numbers} are defined by the sequence $f(0) = 0$, $f(1) = 1$, $f(2) = 1$,
+	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.}
+	\hint{Fibonacci numbers are given by the automaton prohibiting two \texttt{'a'}s in a row.}
 \end{itemize}
 
 
 \begin{solution}
-	\begin{itemize}
-		\item You would need to have an alphabet with three letters.
-		\item Consider the language of words over $\{0, 1, 2\}$ having the sum of digits equal to $2$, so they contain two 1's or one 2. 	%\includegraphics[width=0.5\linewidth]{NSqrd.png}
-		\item Following the hint gives the automaton %\includegraphics[width=0.5\linewidth]{Trib1.png}
-		\item For this automaton $f(n)$ gives Tribonacci numbers with a shift: $f(0)=1$, $f(1)=2$, $f(2)=4$, $f(3)=7$. To account for the shift one can move the starting state in, e.g., this fashion:
-		%\includegraphics[width=0.5\linewidth]{Trib2.png}
-	\end{itemize}
+
+	\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.
-%
-% \com{TD}{Something suffix automaton}
 
 
 \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}
+\begin{solution}
 
 	\textbf{Part 1:} \par
-	Index the states by triples of digits \texttt{000}, \texttt{001}, ..., \texttt{111}. All strings which end by 3 digits $d_1d_2d_3$ will end up in the state $d_1d_2d_3$. The starting state will be \texttt{111}. The transitions from $d_1d_2d_3$ by \texttt{0} and \texttt{1} will lead to $d_2d_3\texttt{0}$ and $d_2d_3\texttt{1}$, respectively. Accepting states are states with indices starting with \texttt{0}.
+	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.
 
-	%\includegraphics[width=0.7\linewidth]{9.png}
+	\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} should lead to pairwise different states since they differ in $i$-th position and after completing them with $i-1$ digit, they will need to be in different states.
+	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
diff --git a/Advanced/DFAs/tikxset.tex b/Advanced/DFAs/tikxset.tex
index aed6432..911338c 100644
--- a/Advanced/DFAs/tikxset.tex
+++ b/Advanced/DFAs/tikxset.tex
@@ -17,29 +17,63 @@
 		execute at end scope={\endpgfonlayer}
 	},
 	%
-	% Arrowhead tweaks
+	% Arrowhead tweak
 	>={Latex[ width=2mm, length=2mm ]},
+	%
+	% Labels inside edges
 	label/.style = {
-		circle,
+		rectangle,
 		% For automatic red background in solutions
 		fill = \ORMCbgcolor,
-		draw = none
+		draw = none,
+		rounded corners = 0mm
 	},
 	%
 	% Nodes
 	main/.style = {
 		draw,
 		circle,
-		fill = white
+		fill = white,
+		line width = 0.35mm
 	},
 	accept/.style = {
 		draw,
 		circle,
 		fill = white,
 		double,
+		double distance = 0.5mm,
+		line width = 0.35mm
 	},
-	hatch/.style = {
-		pattern=north west lines,
-		pattern color=gray
+	start/.style = {
+		draw,
+		rectangle,
+		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
 	}
 }
\ No newline at end of file