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\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}<1>[t]
\frametitle{%
\begin{tabular}{@ {}c@ {}}
\\[-3mm]
\LARGE Automata and \\[-2mm]
\LARGE Formal Languages (2)\\[3mm]
\end{tabular}}
%\begin{center}
%\includegraphics[scale=0.3]{pics/ante1.jpg}\hspace{5mm}
%\includegraphics[scale=0.31]{pics/ante2.jpg}\\
%\footnotesize\textcolor{gray}{Antikythera automaton, 100 BC (Archimedes?)}
%\end{center}
\normalsize
\begin{center}
\begin{tabular}{ll}
Email: & christian.urban at kcl.ac.uk\\
Office: & S1.27 (1st floor Strand Building)\\
Slides: & KEATS
\end{tabular}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Languages\end{tabular}}
A \alert{language} is a set of strings.\bigskip
A \alert{regular expression} specifies a set of strings, or language.
\end{frame}}
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Strings\end{tabular}}
Different ways of writing strings:
\begin{center}
\begin{tabular}{ccc}
\bl{\consolas"$hello$"}\quad{} & \bl{$[h, e, l, l, o]$} & \quad\bl{$h\!::\!e\!::\!l\!::\!l\!::\!o\!::\!N\!il$}\bigskip\\
\bl{\consolas ""} & \bl{$[]$} & \bl{$N$\!$il$}
\end{tabular}
\end{center}\pause
The concatenation operation on strings and sets of strings:
\begin{center}
\begin{tabular}{l}
\bl{{\consolas"$f\!oo$"}$\;@\;${\consolas"$bar$"}$\;=\;${\consolas"$f\!oobar$"}}\medskip\\
\bl{$A \;@\; B \dn \{ s_1 @ s_2 \mid s_1 \in A \wedge s_2 \in B\}$}
\end{tabular}
\end{center}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Regular Expressions\end{tabular}}
Their inductive definition:
\begin{textblock}{6}(2,6.5)
\begin{tabular}{@ {}rrl@ {\hspace{13mm}}l}
\bl{r} & \bl{$::=$} & \bl{$\varnothing$} & null\\
& \bl{$\mid$} & \bl{$\epsilon$} & empty string / {\consolas""} / $[]$\\
& \bl{$\mid$} & \bl{c} & character\\
& \bl{$\mid$} & \bl{r$_1$ $\cdot$ r$_2$} & sequence\\
& \bl{$\mid$} & \bl{r$_1$ + r$_2$} & alternative / choice\\
& \bl{$\mid$} & \bl{r$^*$} & star (zero or more)\\
\end{tabular}
\end{textblock}
\only<2->{
\begin{textblock}{9}(4,0.5)
\begin{tikzpicture}
\draw (0,0) node[inner sep=2mm,fill=cream, ultra thick, draw=red, rounded corners=2mm]
{\normalsize\color{darkgray}
\begin{minipage}{9cm}
\hspace{5mm}\mbox{{\lstset{language=Scala}\fontsize{8}{10}\selectfont
\texttt{\lstinputlisting{../progs/app51.scala}}}}
\end{minipage}};
\end{tikzpicture}
\end{textblock}}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Meaning of a\\[-2mm] Regular Expression\end{tabular}}
\begin{textblock}{15}(1,4)
\begin{tabular}{@ {}rcl}
\bl{$L(\varnothing)$} & \bl{$\dn$} & \bl{$\varnothing$}\\
\bl{$L(\epsilon)$} & \bl{$\dn$} & \bl{$\{$""$\}$}\\
\bl{$L(c)$} & \bl{$\dn$} & \bl{$\{"c"\}$}\\
\bl{$L(r_1 + r_2)$} & \bl{$\dn$} & \bl{$L(r_1) \cup L(r_2)$}\\
\bl{$L(r_1 \cdot r_2)$} & \bl{$\dn$} & \bl{$L(r_1) \,@\, L(r_2)$}\\
\bl{$L(r^*)$} & \bl{$\dn$} & \bl{$\bigcup_{n \ge 0} L(r)^n$}\\
\end{tabular}\bigskip
\only<2->{
\hspace{5mm}\textcolor{blue}{$L(r)^0 \;\dn\; \{""\}$}\\
\textcolor{blue}{$L(r)^{n+1} \;\dn\; L(r) \,@\, L(r)^n$}}
\end{textblock}
\only<1->{
\begin{textblock}{6}(9,12)\small
\bl{$L$} is a function from regular expressions to sets of strings\\
\bl{$L$ : Rexp $\Rightarrow$ Set$[$String$]$}
\end{textblock}}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\large
\begin{center}
What is \bl{$L(a^*)$}?
\end{center}
\end{frame}}
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\newcommand{\YES}{\textcolor{gray}{yes}}
\newcommand{\NO}{\textcolor{gray}{no}}
\newcommand{\FORALLR}{\textcolor{gray}{$\forall$ r.}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Reg Exp Equivalences\end{tabular}}
\begin{center}
\begin{tabular}{l@ {\hspace{7mm}}rcl@ {\hspace{7mm}}l}
&\bl{$(a + b) + c$} & \bl{$\equiv^?$} & \bl{$a + (b + c)$} & \onslide<2->{\YES}\\
&\bl{$a + a$} & \bl{$\equiv^?$} & \bl{$a$} & \onslide<3->{\YES}\\
&\bl{$(a \cdot b) \cdot c$} & \bl{$\equiv^?$} & \bl{$a \cdot (b \cdot c)$} & \onslide<4->{\YES}\\
&\bl{$a \cdot a$} & \bl{$\equiv^?$} & \bl{$a$} & \onslide<5->{\NO}\\
&\bl{$\epsilon^*$} & \bl{$\equiv^?$} & \bl{$\epsilon$} & \onslide<6->{\YES}\\
&\bl{$\varnothing^*$} & \bl{$\equiv^?$} & \bl{$\varnothing$} & \onslide<7->{\NO}\\
\FORALLR &\bl{$r \cdot \epsilon$} & \bl{$\equiv^?$} & \bl{$r$} & \onslide<8->{\YES}\\
\FORALLR &\bl{$r + \epsilon$} & \bl{$\equiv^?$} & \bl{$r$} & \onslide<9->{\NO}\\
\FORALLR &\bl{$r + \varnothing$} & \bl{$\equiv^?$} & \bl{$r$} & \onslide<10->{\YES}\\
\FORALLR &\bl{$r \cdot \varnothing$} & \bl{$\equiv^?$} & \bl{$r$} & \onslide<11->{\NO}\\
&\bl{$c \cdot (a + b)$} & \bl{$\equiv^?$} & \bl{$(c \cdot a) + (c \cdot b)$} & \onslide<12->{\YES}\\
&\bl{$a^*$} & \bl{$\equiv^?$} & \bl{$\epsilon + (a \cdot a^*)$} & \onslide<13->{\YES}
\end{tabular}
\end{center}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Specification\\[-1mm] of Matching\end{tabular}}
\large
a regular expression \bl{r} matches a string \bl{s} is defined as
\begin{center}
\bl{s $\in$ $L$(r)}\\
\end{center}\bigskip\bigskip\pause
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}\bl{$(a?\{n\}) \cdot a\{n\}$}\end{tabular}}
\mbox{}\\[-13mm]
\begin{tikzpicture}[y=.2cm, x=.3cm]
%axis
\draw (0,0) -- coordinate (x axis mid) (30,0);
\draw (0,0) -- coordinate (y axis mid) (0,30);
%ticks
\foreach \x in {0,5,...,30}
\draw (\x,1pt) -- (\x,-3pt)
node[anchor=north] {\x};
\foreach \y in {0,5,...,30}
\draw (1pt,\y) -- (-3pt,\y)
node[anchor=east] {\y};
%labels
\node[below=0.6cm] at (x axis mid) {\bl{a}s};
\node[rotate=90, left=1.2cm] at (y axis mid) {secs};
%plots
\draw[color=blue] plot[mark=*, mark options={fill=white}]
file {re-python.data};
\draw[color=red] plot[mark=triangle*, mark options={fill=white} ]
file {re1.data};
\draw[color=green] plot[mark=square*, mark options={fill=white} ]
file {re2.data};
\draw[color=brown] plot[mark=pentagon*, mark options={fill=white} ]
file {re-ruby.data};
%legend
\begin{scope}[shift={(4,20)}]
\draw[color=blue] (0,0) --
plot[mark=*, mark options={fill=white}] (0.25,0) -- (0.5,0)
node[right]{\small Python};
\draw[yshift=-\baselineskip, color=brown] (0,0) --
plot[mark=pentagon*, mark options={fill=white}] (0.25,0) -- (0.5,0)
node[right]{\small Ruby};
\end{scope}
\end{tikzpicture}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}A Matching Algorithm\end{tabular}}
\small
\ldots{}whether a regular expression can match the empty string:
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}
\bl{$nullable(\varnothing)$} & \bl{$\dn$} & \bl{$f\!\/alse$}\\
\bl{$nullable(\epsilon)$} & \bl{$\dn$} & \bl{$true$}\\
\bl{$nullable (c)$} & \bl{$\dn$} & \bl{$f\!alse$}\\
\bl{$nullable (r_1 + r_2)$} & \bl{$\dn$} & \bl{$nullable(r_1) \vee nullable(r_2)$} \\
\bl{$nullable (r_1 \cdot r_2)$} & \bl{$\dn$} & \bl{$nullable(r_1) \wedge nullable(r_2)$} \\
\bl{$nullable (r^*)$} & \bl{$\dn$} & \bl{$true$} \\
\end{tabular}
\end{center}
\only<2->{
\begin{textblock}{9}(3.4,10)
\begin{tikzpicture}
\draw (0,0) node[inner sep=2mm,fill=cream, ultra thick, draw=red, rounded corners=2mm]
{\normalsize\color{darkgray}
\begin{minipage}{9cm}
\hspace{5mm}\mbox{{\lstset{language=Scala}\fontsize{8}{10}\selectfont
\texttt{\lstinputlisting{../progs/app5.scala}}}}
\end{minipage}};
\end{tikzpicture}
\end{textblock}}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Derivative of a Rexp\end{tabular}}
\large
If \bl{r} matches the string \bl{c::s}, what is a regular expression that matches \bl{s}?\bigskip\bigskip\bigskip\bigskip
\small
\bl{$der\,c\,r$} gives the answer
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Derivative of a Rexp (2)\end{tabular}}
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {\hspace{-10mm}}l@ {}}
\bl{$der\, c\, (\varnothing)$} & \bl{$\dn$} & \bl{$\varnothing$} & \\
\bl{$der\, c\, (\epsilon)$} & \bl{$\dn$} & \bl{$\varnothing$} & \\
\bl{$der\, c\, (d)$} & \bl{$\dn$} & \bl{if $c = d$ then $\epsilon$ else $\varnothing$} & \\
\bl{$der\, c\, (r_1 + r_2)$} & \bl{$\dn$} & \bl{$der\, c\, r_1 + der\, c\, r_2$} & \\
\bl{$der\, c\, (r_1 \cdot r_2)$} & \bl{$\dn$} & \bl{if $nullable (r_1)$}\\
& & \bl{then $(der\,c\,r_1) \cdot r_2 + der\, c\, r_2$}\\
& & \bl{else $(der\, c\, r_1) \cdot r_2$}\\
\bl{$der\, c\, (r^*)$} & \bl{$\dn$} & \bl{$(der\,c\,r) \cdot (r^*)$} &\smallskip\\\pause
\bl{$der\!s\, [] r$} & \bl{$\dn$} & \bl{$r$} & \\
\bl{$der\!s\, (c\!::\!s) r$} & \bl{$\dn$} & \bl{$der\!s\,s\,(der\,c\,r)$} & \\
\end{tabular}
\end{center}
\only<3->{
\begin{textblock}{10.5}(2,5)
\begin{tikzpicture}
\draw (0,0) node[inner sep=2mm,fill=cream, ultra thick, draw=red, rounded corners=2mm]
{\normalsize\color{darkgray}
\begin{minipage}{10.5cm}
\hspace{5mm}\mbox{{\lstset{language=Scala}\fontsize{8}{10}\selectfont
\texttt{\lstinputlisting{../progs/app6.scala}}}}
\end{minipage}};
\end{tikzpicture}
\end{textblock}}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}The Rexp Matcher\end{tabular}}
{\lstset{language=Scala}\fontsize{8}{10}\selectfont
\texttt{\lstinputlisting{../progs/app7.scala}}}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[t]
\frametitle{\begin{tabular}{c}Proofs about Rexp\end{tabular}}
Remember their inductive definition:\\[5cm]
\begin{textblock}{6}(5,5)
\begin{tabular}{@ {}rrl}
\bl{r} & \bl{$::=$} & \bl{$\varnothing$}\\
& \bl{$\mid$} & \bl{$\epsilon$} \\
& \bl{$\mid$} & \bl{c} \\
& \bl{$\mid$} & \bl{r$_1$ $\cdot$ r$_2$}\\
& \bl{$\mid$} & \bl{r$_1$ + r$_2$} \\
& \bl{$\mid$} & \bl{r$^*$} \\
\end{tabular}
\end{textblock}
If we want to prove something, say a property \bl{$P$(r)}, for all regular expressions \bl{r} then \ldots
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Rexp (2)\end{tabular}}
\begin{itemize}
\item \bl{$P$} holds for \bl{$\varnothing$}, \bl{$\epsilon$} and \bl{c}\bigskip
\item \bl{$P$} holds for \bl{r$_1$ + r$_2$} under the assumption that \bl{$P$} already
holds for \bl{r$_1$} and \bl{r$_2$}.\bigskip
\item \bl{$P$} holds for \bl{r$_1$ $\cdot$ r$_2$} under the assumption that \bl{$P$} already
holds for \bl{r$_1$} and \bl{r$_2$}.
\item \bl{$P$} holds for \bl{r$^*$} under the assumption that \bl{$P$} already
holds for \bl{r}.
\end{itemize}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Rexp (3)\end{tabular}}
Assume \bl{$P(r)$} is the property:
\begin{center}
\bl{nullable(r)} if and only if \bl{"" $\in$ $L$(r)}
\end{center}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Strings\end{tabular}}
If we want to prove something, say a property \bl{$P$(s)}, for all strings \bl{s} then \ldots\bigskip
\begin{itemize}
\item \bl{$P$} holds for the empty string, and\medskip
\item \bl{$P$} holds for the string \bl{c::s} under the assumption that \bl{$P$}
already holds for \bl{s}
\end{itemize}
\end{frame}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Proofs about Strings (2)\end{tabular}}
Let \bl{Der c A} be the set defined as
\begin{center}
\bl{Der c A $\dn$ $\{$ s $|$ c::s $\in$ A$\}$ }
\end{center}
Assume that \bl{$L$(der c r) = Der c ($L$(r))}. Prove that
\begin{center}
\bl{matcher(r, s) if and only if s $\in$ $L$(r)}
\end{center}
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Regular Languages\end{tabular}}
A language (set of strings) is \alert{regular} iff there exists
a regular expression that recognises all its strings.
\end{frame}}
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\mode<presentation>{
\begin{frame}[c]
\frametitle{\begin{tabular}{c}Automata\end{tabular}}
A deterministic finite automaton consists of:
\begin{itemize}
\item a set of states
\item one of these states is the start state
\item some states are accepting states, and
\item there is transition function\medskip
\small
which takes a state as argument and a character and produces a new state\smallskip\\
this function might not always be defined
\end{itemize}
\end{frame}}
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\end{document}
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