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\section*{Handout 3 (Automata)}

Every formal language course I know of bombards you first with

automata and then to a much, much smaller extend with regular

expressions. As you can see, this course is turned upside

down: regular expressions come first. The reason is that

regular expressions are easier to reason about and the notion

of derivatives, although already quite old, only became more

widely known rather recently. Still let us in this lecture

have a closer look at automata and their relation to regular

expressions. This will help us with understanding why the

regular expression matchers in Python and Ruby are so slow

with certain regular expressions. The central definition

is:\medskip 

\noindent 

A \emph{deterministic finite automaton} (DFA), say $A$, is

defined by a four-tuple written $A(Q, q_0, F, \delta)$ where

\begin{itemize}

\item $Q$ is a finite set of states,

\item $q_0 \in Q$ is the start state,

\item $F \subseteq Q$ are the accepting states, and

\item $\delta$ is the transition function.

\end{itemize}

\noindent The transition function determines how to

``transition'' from one state to the next state with respect

to a character. We have the assumption that these transition

functions do not need to be defined everywhere: so it can be

the case that given a character there is no next state, in

which case we need to raise a kind of ``failure exception''. A

typical example of a DFA is

\begin{center}

\begin{tikzpicture}[>=stealth',very thick,auto,

                    every state/.style={minimum size=0pt,

                    inner sep=2pt,draw=blue!50,very thick,

                    fill=blue!20},scale=2]

\node[state,initial]  (q_0)  {$q_0$};

\node[state] (q_1) [right=of q_0] {$q_1$};

\node[state] (q_2) [below right=of q_0] {$q_2$};

\node[state] (q_3) [right=of q_2] {$q_3$};

\node[state, accepting] (q_4) [right=of q_1] {$q_4$};

\path[->] (q_0) edge node [above]  {$a$} (q_1);

\path[->] (q_1) edge node [above]  {$a$} (q_4);

\path[->] (q_4) edge [loop right] node  {$a, b$} ();

\path[->] (q_3) edge node [right]  {$a$} (q_4);

\path[->] (q_2) edge node [above]  {$a$} (q_3);

\path[->] (q_1) edge node [right]  {$b$} (q_2);

\path[->] (q_0) edge node [above]  {$b$} (q_2);

\path[->] (q_2) edge [loop left] node  {$b$} ();

\path[->] (q_3) edge [bend left=95, looseness=1.3] node [below]  {$b$} (q_0);

\end{tikzpicture}

\end{center}

\noindent In this graphical notation, the accepting state

$q_4$ is indicated with double circles. Note that there can be

more than one accepting state. It is also possible that a DFA

has no accepting states at all, or that the starting state is

also an accepting state. In the case above the transition

function is defined everywhere and can be given as a table as

follows:

\[

\begin{array}{lcl}

(q_0, a) &\rightarrow& q_1\\

(q_0, b) &\rightarrow& q_2\\

(q_1, a) &\rightarrow& q_4\\

(q_1, b) &\rightarrow& q_2\\

(q_2, a) &\rightarrow& q_3\\

(q_2, b) &\rightarrow& q_2\\

(q_3, a) &\rightarrow& q_4\\

(q_3, b) &\rightarrow& q_0\\

(q_4, a) &\rightarrow& q_4\\

(q_4, b) &\rightarrow& q_4\\

\end{array}

\]

We need to define the notion of what language is accepted by

an automaton. For this we lift the transition function

$\delta$ from characters to strings as follows:

\[

\begin{array}{lcl}

\hat{\delta}(q, [])        & \dn & q\\

\hat{\delta}(q, c\!::\!s) & \dn & \hat{\delta}(\delta(q, c), s)\\

\end{array}

\]

\noindent This lifted transition function is often called

``delta-hat''. Given a string, we start in the starting state

and take the first character of the string, follow to the next

sate, then take the second character and so on. Once the

string is exhausted and we end up in an accepting state, then

this string is accepted by the automaton. Otherwise it is not

accepted. So $s$ is in the \emph{language accepted by the

automaton} $A(Q, q_0, F, \delta)$ iff

\[

\hat{\delta}(q_0, s) \in F 

\]

\noindent I let you think about a definition that describes

the set of strings accepted by an automaton.

While with DFAs it will always clear that given a character

what the next state is (potentially none), it will be useful

to relax this restriction. That means we have several

potential successor states. We even allow ``silent

transitions'', also called epsilon-transitions. They allow us

to go from one state to the next without having a character

consumed. We label such silent transition with the letter

$\epsilon$. The resulting construction is called a

\emph{non-deterministic finite automaton} (NFA) given also as

a four-tuple $A(Q, q_0, F, \rho)$ where

\begin{itemize}

\item $Q$ is a finite set of states

\item $q_0$ is a start state

\item $F$ are some accepting states with $F \subseteq Q$, and

\item $\rho$ is a transition relation.

\end{itemize}

\noindent

Two typical examples of NFAs are

\begin{center}

\begin{tabular}[t]{c@{\hspace{9mm}}c}

\begin{tikzpicture}[scale=0.7,>=stealth',very thick,

                             every state/.style={minimum size=0pt,draw=blue!50,very thick,fill=blue!20},]

\node[state,initial]  (q_0)  {$q_0$};

\node[state] (q_1) [above=of q_0] {$q_1$};

\node[state, accepting] (q_2) [below=of q_0] {$q_2$};

\path[->] (q_0) edge node [left]  {$\epsilon$} (q_1);

\path[->] (q_0) edge node [left]  {$\epsilon$} (q_2);

\path[->] (q_0) edge [loop right] node  {$a$} ();

\path[->] (q_1) edge [loop above] node  {$a$} ();

\path[->] (q_2) edge [loop below] node  {$b$} ();

\end{tikzpicture} &

\raisebox{20mm}{

\begin{tikzpicture}[scale=0.7,>=stealth',very thick,

                             every state/.style={minimum size=0pt,draw=blue!50,very thick,fill=blue!20},]

\node[state,initial]  (r_1)  {$r_1$};

\node[state] (r_2) [above=of r_1] {$r_2$};

\node[state, accepting] (r_3) [right=of r_1] {$r_3$};

\path[->] (r_1) edge node [below]  {$b$} (r_3);

\path[->] (r_2) edge [bend left] node [above]  {$a$} (r_3);

\path[->] (r_1) edge [bend left] node  [left] {$\epsilon$} (r_2);

\path[->] (r_2) edge [bend left] node  [right] {$a$} (r_1);

\end{tikzpicture}}

\end{tabular}

\end{center}

\noindent There are, however, a number of points you should

note. Every DFA is a NFA, but not vice versa. The $\rho$ in

NFAs is a transition \emph{relation} (DFAs have a transition

function). The difference between a function and a relation is

that a function has always a single output, while a relation

gives, roughly speaking, several outputs. Look at the NFA on

the right-hand side above: if you are currently in the state

$r_2$ and you read a character $a$, then you can transition to

either $r_1$ \emph{or} $r_3$. Which route you take is not

determined. This means if we need to decide whether a string

is accepted by a NFA, we might have to explore all

possibilities. Also there is the special silent transition in

NFAs. As mentioned already this transition means you do not

have to ``consume'' any part of the input string, but

``silently'' change to a different state. In the left picture,

for example, if you are in the starting state, you can 

silently move either to $q_1$ or $q_2$.

The reason for introducing NFAs is that there is a relatively

simple (recursive) translation of regular expressions into

NFAs. Consider the simple regular expressions $\varnothing$,

$\epsilon$ and $c$. They can be translated as follows:

\begin{center}

\begin{tabular}[t]{l@{\hspace{10mm}}l}

\raisebox{1mm}{$\varnothing$} & 

\begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node[state, initial]  (q_0)  {$\mbox{}$};

\end{tikzpicture}\\\\

\raisebox{1mm}{$\epsilon$} & 

\begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node[state, initial, accepting]  (q_0)  {$\mbox{}$};

\end{tikzpicture}\\\\

\raisebox{2mm}{$c$} & 

\begin{tikzpicture}[scale=0.7,>=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node[state, initial]  (q_0)  {$\mbox{}$};

\node[state, accepting]  (q_1)  [right=of q_0] {$\mbox{}$};

\path[->] (q_0) edge node [below]  {$c$} (q_1);

\end{tikzpicture}\\\\

\end{tabular}

\end{center}

\noindent The case for the sequence regular expression $r_1

\cdot r_2$ is as follows: We are given by recursion two

automata representing $r_1$ and $r_2$ respectively. 

\begin{center}

\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node[state, initial]  (q_0)  {$\mbox{}$};

\node (r_1)  [right=of q_0] {$\ldots$};

\node[state, accepting]  (t_1)  [right=of r_1] {$\mbox{}$};

\node[state, accepting]  (t_2)  [above=of t_1] {$\mbox{}$};

\node[state, accepting]  (t_3)  [below=of t_1] {$\mbox{}$};

\node[state, initial]  (a_0)  [right=2.5cm of t_1] {$\mbox{}$};

\node (b_1)  [right=of a_0] {$\ldots$};

\node[state, accepting]  (c_1)  [right=of b_1] {$\mbox{}$};

\node[state, accepting]  (c_2)  [above=of c_1] {$\mbox{}$};

\node[state, accepting]  (c_3)  [below=of c_1] {$\mbox{}$};

\begin{pgfonlayer}{background}

\node (1) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (q_0) (r_1) (t_1) (t_2) (t_3)] {};

\node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (a_0) (b_1) (c_1) (c_2) (c_3)] {};

\node [yshift=2mm] at (1.north) {$r_1$};

\node [yshift=2mm] at (2.north) {$r_2$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent The first automaton has some accepting states. We

obtain an automaton for $r_1\cdot r_2$ by connecting these

accepting states with $\epsilon$-transitions to the starting

state of the second automaton. By doing so we make them

non-accepting like so:

\begin{center}

\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node[state, initial]  (q_0)  {$\mbox{}$};

\node (r_1)  [right=of q_0] {$\ldots$};

\node[state]  (t_1)  [right=of r_1] {$\mbox{}$};

\node[state]  (t_2)  [above=of t_1] {$\mbox{}$};

\node[state]  (t_3)  [below=of t_1] {$\mbox{}$};

\node[state]  (a_0)  [right=2.5cm of t_1] {$\mbox{}$};

\node (b_1)  [right=of a_0] {$\ldots$};

\node[state, accepting]  (c_1)  [right=of b_1] {$\mbox{}$};

\node[state, accepting]  (c_2)  [above=of c_1] {$\mbox{}$};

\node[state, accepting]  (c_3)  [below=of c_1] {$\mbox{}$};

\path[->] (t_1) edge node [above, pos=0.3]  {$\epsilon$} (a_0);

\path[->] (t_2) edge node [above]  {$\epsilon$} (a_0);

\path[->] (t_3) edge node [below]  {$\epsilon$} (a_0);

\begin{pgfonlayer}{background}

\node (3) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (q_0) (c_1) (c_2) (c_3)] {};

\node [yshift=2mm] at (3.north) {$r_1\cdot r_2$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent The case for the choice regular expression $r_1 +

r_2$ is slightly different: We are given by recursion two

automata representing $r_1$ and $r_2$ respectively. 

\begin{center}

\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node at (0,0)  (1)  {$\mbox{}$};

\node[state, initial]  (2)  [above right=16mm of 1] {$\mbox{}$};

\node[state, initial]  (3)  [below right=16mm of 1] {$\mbox{}$};

\node (a)  [right=of 2] {$\ldots$};

\node[state, accepting]  (a1)  [right=of a] {$\mbox{}$};

\node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};

\node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};

\node (b)  [right=of 3] {$\ldots$};

\node[state, accepting]  (b1)  [right=of b] {$\mbox{}$};

\node[state, accepting]  (b2)  [above=of b1] {$\mbox{}$};

\node[state, accepting]  (b3)  [below=of b1] {$\mbox{}$};

\begin{pgfonlayer}{background}

\node (1) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (2) (a1) (a2) (a3)] {};

\node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (3) (b1) (b2) (b3)] {};

\node [yshift=3mm] at (1.north) {$r_1$};

\node [yshift=3mm] at (2.north) {$r_2$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent Each automaton has a single start state and

potentially several accepting states. We obtain a NFA for the

regular expression $r_1 + r_2$ by introducing a new starting

state and connecting it with an $\epsilon$-transition to the

two starting states above, like so

\begin{center}

\hspace{2cm}\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node at (0,0) [state, initial]  (1)  {$\mbox{}$};

\node[state]  (2)  [above right=16mm of 1] {$\mbox{}$};

\node[state]  (3)  [below right=16mm of 1] {$\mbox{}$};

\node (a)  [right=of 2] {$\ldots$};

\node[state, accepting]  (a1)  [right=of a] {$\mbox{}$};

\node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};

\node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};

\node (b)  [right=of 3] {$\ldots$};

\node[state, accepting]  (b1)  [right=of b] {$\mbox{}$};

\node[state, accepting]  (b2)  [above=of b1] {$\mbox{}$};

\node[state, accepting]  (b3)  [below=of b1] {$\mbox{}$};

\path[->] (1) edge node [above]  {$\epsilon$} (2);

\path[->] (1) edge node [below]  {$\epsilon$} (3);

\begin{pgfonlayer}{background}

\node (3) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (1) (a2) (a3) (b2) (b3)] {};

\node [yshift=3mm] at (3.north) {$r_1+ r_2$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent 

Finally for the $*$-case we have an automaton for $r$

\begin{center}

\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node at (0,0)  (1)  {$\mbox{}$};

\node[state, initial]  (2)  [right=16mm of 1] {$\mbox{}$};

\node (a)  [right=of 2] {$\ldots$};

\node[state, accepting]  (a1)  [right=of a] {$\mbox{}$};

\node[state, accepting]  (a2)  [above=of a1] {$\mbox{}$};

\node[state, accepting]  (a3)  [below=of a1] {$\mbox{}$};

\begin{pgfonlayer}{background}

\node (1) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (2) (a1) (a2) (a3)] {};

\node [yshift=3mm] at (1.north) {$r$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent and connect its accepting states to a new starting

state via $\epsilon$-transitions. This new starting state is

also an accepting state, because $r^*$ can recognise the

empty string. This gives the following automaton for $r^*$:

\begin{center}

\begin{tikzpicture}[node distance=3mm,

                             >=stealth',very thick, every state/.style={minimum size=3pt,draw=blue!50,very thick,fill=blue!20},]

\node at (0,0) [state, initial,accepting]  (1)  {$\mbox{}$};

\node[state]  (2)  [right=16mm of 1] {$\mbox{}$};

\node (a)  [right=of 2] {$\ldots$};

\node[state]  (a1)  [right=of a] {$\mbox{}$};

\node[state]  (a2)  [above=of a1] {$\mbox{}$};

\node[state]  (a3)  [below=of a1] {$\mbox{}$};

\path[->] (1) edge node [above]  {$\epsilon$} (2);

\path[->] (a1) edge [bend left=45] node [above]  {$\epsilon$} (1);

\path[->] (a2) edge [bend right] node [below]  {$\epsilon$} (1);

\path[->] (a3) edge [bend left=45] node [below]  {$\epsilon$} (1);

\begin{pgfonlayer}{background}

\node (2) [rounded corners, inner sep=1mm, thick, draw=black!60, fill=black!20, fit= (1) (a2) (a3)] {};

\node [yshift=3mm] at (2.north) {$r^*$};

\end{pgfonlayer}

\end{tikzpicture}

\end{center}

\noindent This construction of a NFA from a regular expression

was invented by Ken Thompson in 1968.

What is interesting that for every NFA we can find a DFA which

recognises the same language. This can be done by the

\emph{subset construction}. Consider again the NFA on the 

left, consisting of nodes labeled $0$, $1$ and $2$. 

\begin{center}

\begin{tabular}{c@{\hspace{10mm}}c}

\begin{tikzpicture}[scale=0.7,>=stealth',very thick,

                    every state/.style={minimum size=0pt,

                    draw=blue!50,very thick,fill=blue!20},

                    baseline=0mm]

\node[state,initial]  (q_0)  {$0$};

\node[state] (q_1) [above=of q_0] {$1$};

\node[state, accepting] (q_2) [below=of q_0] {$2$};

\path[->] (q_0) edge node [left]  {$\epsilon$} (q_1);

\path[->] (q_0) edge node [left]  {$\epsilon$} (q_2);

\path[->] (q_0) edge [loop right] node  {$a$} ();

\path[->] (q_1) edge [loop above] node  {$a$} ();

\path[->] (q_2) edge [loop below] node  {$b$} ();

\end{tikzpicture}

&

\begin{tabular}{r|cl}

nodes & $a$ & $b$\\

\hline

$\varnothing\phantom{\star}$ & $\varnothing$ & $\varnothing$\\

$\{0\}\phantom{\star}$       & $\{0,1,2\}$   & $\{2\}$\\

$\{1\}\phantom{\star}$       & $\{1\}$       & $\varnothing$\\

$\{2\}\star$  & $\varnothing$ & $\{2\}$\\

$\{0,1\}\phantom{\star}$     & $\{0,1,2\}$   & $\{2\}$\\

$\{0,2\}\star$ & $\{0,1,2\}$   & $\{2\}$\\

$\{1,2\}\star$ & $\{1\}$       & $\{2\}$\\

s: $\{0,1,2\}\star$ & $\{0,1,2\}$ & $\{2\}$\\

\end{tabular}

\end{tabular}

\end{center}

\noindent The nodes of the DFA are given by calculating all

subsets of the set of nodes of the NFA (seen in the nodes

column on the right). The table shows the transition function

for the DFA. The first row states that $\varnothing$ is the

sink node which has transitions for $a$ and $b$ to itself.

The next three lines are calculated as follows: 

\begin{itemize}

\item suppose you calculate the entry for the transition for

      $a$ and the node $\{0\}$

\item start from the node $0$ in the NFA

\item do as many $\epsilon$-transition as you can obtaining a

      set of nodes, in this case $\{0,1,2\}$

\item filter out all notes that do not allow an $a$-transition

      from this set, this excludes $2$ which does not permit a

      $a$-transition

\item from the remaining set, do as many $\epsilon$-transition

      as you can, this yields $\{0,1,2\}$     

\item the resulting set specifies the transition from $\{0\}$

      when given an $a$ 

\end{itemize}

\noindent Similarly for the other entries in the rows for

$\{0\}$, $\{1\}$ and $\{2\}$. The other rows are calculated by

just taking the union of the single node entries. For example

for $a$ and $\{0,1\}$ we need to take the union of $\{0,1,2\}$

(for $0$) and $\{1\}$ (for $1$). The starting state of the DFA

can be calculated from the starting state of the NFA, that is

$0$, and then do as many $\epsilon$-transitions as possible.

This gives $\{0,1,2\}$ which is the starting state of the DFA.

One terminal states in the DFA are given by all sets that

contain a $2$, which is the terminal state of the NFA. This

completes the subset construction.

There are two points to note: One is that the resulting DFA

contains a number of ``dead'' nodes that are never reachable

from the starting state (that is that the calculated DFA in

this example is not a minimal DFA). Such dead nodes can be

safely removed without changing the language that is

recognised by the DFA. Another point is that in some cases the

subset construction produces a DFA that does \emph{not}

contain any dead nodes\ldots{}that means it calculates a

minimal DFA. Which in turn means that in some cases the number

of nodes by going from NFAs to DFAs exponentially increases,

namely by $2^n$ (which is the number of subsets you can form

for $n$ nodes).