\documentclass[dvipsnames,14pt,t]{beamer}
\usepackage{../slides}
\usepackage{../graphics}
\usepackage{../langs}
\usepackage{../data}

\hfuzz=220pt 

\lstset{language=Scala,
        style=mystyle,
        numbersep=0pt,
        numbers=none,
        xleftmargin=0mm}

\pgfplotsset{compat=1.12}

\newcommand{\bl}[1]{\textcolor{blue}{#1}}     

% beamer stuff 
\renewcommand{\slidecaption}{CFL 02, King's College London}


\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{%
  \begin{tabular}{@ {}c@ {}}
  \\[-3mm]
  \LARGE Compilers and \\[-1mm] 
  \LARGE Formal Languages (2)\\[3mm] 
  \end{tabular}}

  \normalsize
  \begin{center}
  \begin{tabular}{ll}
  Email:  & christian.urban at kcl.ac.uk\\
  Office: & N\liningnums{7.07} (North Wing, Bush House)\\
  Slides: & KEATS (also homework is there)
  \end{tabular}
  \end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
  \frametitle{\Large
    Lets Implement an Efficient\\[-2mm] 
    Regular Expression Matcher}

\footnotesize
\begin{center}
  {\normalsize Graphs: \bl{$a^{?\{n\}} \cdot a^{\{n\}}$} and strings \bl{$\underbrace{\,a\ldots a\,}_{n}$}}\medskip\\
\begin{tabular}{@{}cc@{}}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,0.0)}},
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=33,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=4.5cm, 
    legend entries={Python,Ruby},  
    legend pos=north west,
    legend cell align=left]
\addplot[blue,mark=*, mark options={fill=white}] table {re-python.data};
\addplot[brown,mark=pentagon*, mark options={fill=white}] table {re-ruby.data};  
\end{axis}
\end{tikzpicture}
&
\begin{tikzpicture}
  \begin{axis}[
    xlabel={$n$},
    x label style={at={(1.1,0.05)}},
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5000,...,10000},
    xmax=11500,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=4.5cm,
    legend entries={Scala V2,Scala V3},
    legend pos=north east,
    legend cell align=left]
\addplot[green,mark=square*,mark options={fill=white}] table {re2.data};
\addplot[black,mark=square*,mark options={fill=white}] table {re3.data};
\end{axis}
\end{tikzpicture}
\end{tabular}
\end{center}

\small
In the handouts is a similar graph for \bl{$(a^*)^* \cdot b$} and Java 8.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Evil Regular Expressions}

\begin{itemize}
\item \alert{R}egular \alert{e}xpression \alert{D}enial \alert{o}f \alert{S}ervice (ReDoS)\bigskip
\item Evil regular expressions\medskip
\begin{itemize}
\item \bl{$a^{?\{n\}} \cdot a^{\{n\}}$}
\item \bl{$(a^*)^*$}
\item \bl{$([a$\,-\,$z]^+)^*$}
\item \bl{$(a + a \cdot a)^*$}
\item \bl{$(a + a^?)^*$}
\end{itemize}

\item sometimes also called \alert{catastrophic backtracking}\bigskip
\item \small\ldots I hope you have watched the video by the
  StackExchange engineer  
\end{itemize}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Languages}

\begin{itemize}

\item A \alert{\bf Language} is a set of strings, for example\medskip
\begin{center}
\bl{$\{[], hello, \textit{foobar}, a, abc\}$}
\end{center}\bigskip

\item \alert{\bf Concatenation} of strings and languages

\begin{center}
\begin{tabular}{rcl}
\bl{$\textit{foo}\;@\;bar$} & \bl{$=$} & \bl{$\textit{foobar}$}\medskip\\
\bl{$A\;@\;B$} & \bl{$\dn$} & \bl{$\{ s_1\,@\,s_2 \;\mid\; s_1 \in A \wedge s_2 \in B\}$}
\end{tabular}
\end{center}
\bigskip

\small
\item [] For example \bl{$A = \{foo, bar\}$}, \bl{$B = \{a, b\}$}

\[
\bl{A \,@\, B = \{fooa, foob, bara, barb\}}
\]

\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{The Power Operation}

\begin{itemize}
\item The \alert{\textbf{\boldmath$n$th Power}} of a language:

\begin{center}
\begin{tabular}{lcl}
\bl{$A^0$}    & \bl{$\dn$} & \bl{$\{[]\}$}\\
\bl{$A^{n+1}$} & \bl{$\dn$} & \bl{$A \,@\, A^n$}
\end{tabular}
\end{center}\bigskip

\item[] For example

\begin{center}
\begin{tabular}{lcl@{\hspace{10mm}}l}
\bl{$A^4$} & \bl{$=$} & \bl{$A \,@\, A \,@\, A \,@\, A$} & \bl{$(@\,\{[]\})$}\\
\bl{$A^1$} & \bl{$=$} & \bl{$A$} & \bl{$(@\,\{[]\})$}\\
\bl{$A^0$} & \bl{$=$} & \bl{$\{[]\}$}\\
\end{tabular}
\end{center}

\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

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\begin{frame}[c]
\frametitle{Homework Question}

\begin{itemize}
\item Say \bl{$A = \{[a],[b],[c],[d]\}$}.\bigskip

\begin{center}
How many strings are in \bl{$A^4$}?
\end{center}\bigskip\medskip\pause


\begin{center}
\begin{tabular}{l}
What if \bl{$A = \{[a],[b],[c],[]\}$};\\ 
how many strings are then in \bl{$A^4$}?
\end{tabular}
\end{center}
\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

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\begin{frame}[c]
\frametitle{The Star Operation}

\begin{itemize}
\item The \alert{\bf Kleene Star} of a \underline{language}:
\bigskip

\begin{center}
\begin{tabular}{c}
\bl{$A\star \dn \bigcup_{0\le n} A^n$}
\end{tabular}
\end{center}\bigskip

\item[] This expands to 

\[
\bl{A^0 \cup A^1 \cup A^2 \cup A^3 \cup A^4 \cup \ldots}
\]

or

\small
\[
\bl{\{[]\} \;\cup\; A \;\cup\; A\,@\,A \;\cup\; 
  A\,@\,A\,@\,A \;\cup\; A\,@\,A\,@\,A\,@\,A \cup \ldots}
\]

\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

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\begin{frame}[c]
\frametitle{
    The Meaning of a\\[-2mm] 
    Regular Expression}

\begin{textblock}{15}(1,4)
 \begin{tabular}{rcl}
 \bl{$L(\ZERO)$}  & \bl{$\dn$} & \bl{$\{\}$}\\
 \bl{$L(\ONE)$}     & \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{$\{ s_1 \,@\, s_2 \;|\; s_1 \in L(r_1) \wedge s_2 \in L(r_2) \}$}\\
 \bl{$L(r^*)$}           & \bl{$\dn$} & \bl{$(L(r))\star \quad\dn \bigcup_{0 \le n} L(r)^n$}\\
  \end{tabular}
\end{textblock}

\begin{textblock}{9}(6,12)\small
\bl{$L$} is a function from regular expressions to 
sets of strings (languages):\smallskip\\
\bl{$L$ : Rexp $\Rightarrow$ Set$[$String$]$}
\end{textblock}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{\begin{tabular}{c}\\[3cm]\huge\alert{Questions?}\end{tabular}}

\bigskip
homework (written exam 80\%)\\
coursework (20\%; first one today)\\
submission Fridays @ 18:00 -- accepted until Mondays
\mbox{}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Semantic Derivative\\[5mm]}

  

\begin{itemize}
\item The \alert{\bf Semantic Derivative} of a 
\underline{language}\\ w.r.t.~to a character \bl{$c$}:
\bigskip

\begin{center}
\bl{$\Der\,c\,A \dn \{ s \;|\;  c\!::\!s \in A\}$ } 
\end{center}\bigskip

For \bl{$A = \{\textit{foo}, \textit{bar}, \textit{frak}\}$} then

\begin{center}
\bl{\begin{tabular}{l@{\hspace{2mm}}c@{\hspace{2mm}}l}
$\Der\,f\,A$ & $=$ & $\{\textit{oo}, \textit{rak}\}$\\
$\Der\,b\,A$ & $=$ &  $\{\textit{ar}\}$\\  
$\Der\,a\,A$ & $=$ & $\{\}$\pause
\end{tabular}}
\end{center}

\small
We can extend this definition to strings
\[
\bl{\Ders\,s\,A = \{s'\;|\;s\,@\,s' \in A\}}
\]

\end{itemize}
 
\end{frame}
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\begin{frame}[c]
\frametitle{The Specification\\ for Matching}

\begin{bubble}[10cm]
\large
A regular expression \bl{$r$} matches a string~\bl{$s$} 
provided

\begin{center}
\bl{$s \in L(r)$}\\ 
\end{center}
\end{bubble}\bigskip\bigskip

\ldots and the point of the this lecture is 
to decide this problem as fast as possible 
(unlike Python, Ruby, Java etc)

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Regular Expressions}

Their inductive definition:


\begin{textblock}{6}(2,7.5)
  \begin{tabular}{@ {}rrl@ {\hspace{13mm}}l}
  \bl{$r$} & \bl{$::=$}  & \bl{$\ZERO$}  & nothing\\
         & \bl{$\mid$} & \bl{$\ONE$}       & empty string / \pcode{""} / $[]$\\
         & \bl{$\mid$} & \bl{$c$}              & single 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->{\footnotesize
\begin{textblock}{9}(2,0.5)
\begin{bubble}[9.8cm]
\lstinputlisting{../progs/app01.scala}
\end{bubble}
\end{textblock}}  
  
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{
 When Are Two Regular\\[-1mm]
 Expressions Equivalent?}

\large
\begin{center}
\bl{$r_1 \equiv r_2 \;\;\dn\;\; L(r_1) = L(r_2)$}
\end{center}  
  
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Concrete Equivalences}

\begin{center}
\bl{\begin{tabular}{rcl}
$(a + b)  + c$ & $\equiv$ & $a + (b + c)$\\
$a + a$        & $\equiv$ & $a$\\
$a + b$        & $\equiv$ & $b + a$\\
$(a \cdot b) \cdot c$ & $\equiv$ & $a \cdot (b \cdot c)$\\
$c \cdot (a + b)$ & $\equiv$ & $(c \cdot a) + (c \cdot b)$\bigskip\bigskip\\\pause
$a \cdot a$       & $\not\equiv$ & $a$\\
$a + (b \cdot c)$ & $\not\equiv$ & $(a + b) \cdot (a + c)$\\
\end{tabular}}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Corner Cases}

\begin{center}
\bl{\begin{tabular}{rcl}
$a \cdot \ZERO$ & $\not\equiv$ & $a$\\
$a + \ONE$ & $\not\equiv$ & $a$\\
$\ONE$ & $\equiv$ & $\ZERO^*$\\
$\ONE^*$ & $\equiv$ & $\ONE$\\
$\ZERO^*$ & $\not\equiv$ & $\ZERO$\\
\end{tabular}}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Simplification Rules}
 
\begin{center}
\bl{\begin{tabular}{rcl}
$r + \ZERO$  & $\equiv$ & $r$\\
$\ZERO + r$  & $\equiv$ & $r$\\
$r \cdot \ONE$ & $\equiv$ & $r$\\
$\ONE \cdot r$     & $\equiv$ & $r$\\
$r \cdot \ZERO$ & $\equiv$ & $\ZERO$\\
$\ZERO \cdot r$ & $\equiv$ & $\ZERO$\\
$r + r$ & $\equiv$ & $r$
\end{tabular}}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Another Homework\\[-2mm] Question}

\begin{itemize}
\item How many basic regular expressions are there to match
  the string \bl{$abcd$}\,?\pause
\item How many if they cannot include
  \bl{$\ONE$} and \bl{$\ZERO$}\/?\pause
\item How many if they are also not
  allowed to contain stars?\pause
\item How many if they are also
      not allowed to contain \bl{$\_ + \_$}\/? 
\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{\mbox{Brzozowski's Algorithm (1)}}


\ldots{}whether a regular expression can match the empty string:
\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}}
\bl{$nullable(\ZERO)$}    & \bl{$\dn$} & \bl{\textit{false}}\\
\bl{$nullable(\ONE)$}       & \bl{$\dn$} & \bl{\textit{true}}\\
\bl{$nullable (c)$}             & \bl{$\dn$} & \bl{\textit{false}}\\
\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{\textit{true}}\\
\end{tabular}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{The Derivative of a Rexp}

\large
If \bl{$r$} matches the string \bl{$c\!::\!s$}, what is a regular 
expression that matches just \bl{$s$}?\bigskip\bigskip\bigskip\bigskip

\small
\bl{$\der\,c\,r$} gives the answer, Brzozowski 1964
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{The Derivative of a Rexp}

\begin{center}
\begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {\hspace{-10mm}}l@ {}}
  \bl{$\der\, c\, (\ZERO)$}      & \bl{$\dn$} & \bl{$\ZERO$} & \\
  \bl{$\der\, c\, (\ONE)$}           & \bl{$\dn$} & \bl{$\ZERO$} & \\
  \bl{$\der\, c\, (d)$}                     & \bl{$\dn$} & \bl{if $c = d$ then $\ONE$ else $\ZERO$} & \\
  \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{$(\der\,c\,r) \cdot (r^*)$} &\bigskip\\\pause

  \bl{$\ders\, []\, r$}     & \bl{$\dn$} & \bl{$r$} & \\
  \bl{$\ders\, (c\!::\!s)\, r$} & \bl{$\dn$} & \bl{$\ders\,s\,(\der\,c\,r)$} & \\
  \end{tabular}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Examples}

Given \bl{$r \dn ((a \cdot b) + b)^*$} what is

\begin{center}
\begin{tabular}{l}
\bl{$\der\,a\,r =\;?$}\\
\bl{$\der\,b\,r =\;?$}\\
\bl{$\der\,c\,r =\;?$}
\end{tabular}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{\mbox{The Brzozowski Algorithm}}

\begin{center}
\begin{tabular}{l} 
\bl{$\textit{matches}\,r\,s \dn \textit{nullable}(\ders\;s\;r)$}  
\end{tabular}
\end{center}


\end{frame}
%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Brzozowski: An Example}

Does \bl{$r_1$} match \bl{$abc$}?
\begin{center}
\begin{tabular}{@{}rl@{}l@{}}
Step 1: & build derivative of \bl{$a$} and \bl{$r_1$} & \bl{$(r_2 = \der\,a\,r_1)$}\smallskip\\
Step 2: & build derivative of \bl{$b$} and \bl{$r_2$} & \bl{$(r_3 = \der\,b\,r_2)$}\smallskip\\
Step 3: & build derivative of \bl{$c$} and \bl{$r_3$} & \bl{$(r_4 = \der\,c\,r_3)$}\smallskip\\
Step 4: & the string is exhausted: & \bl{($nullable(r_4))$}\\
        & test whether \bl{$r_4$} can recognise\\
        & the empty string\medskip\\
Output: & result of the test\\
        & $\Rightarrow \bl{\textit{true}} \,\text{or}\, 
                       \bl{\textit{false}}$\\        
\end{tabular}
\end{center}


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{The Idea of the Algorithm}

If we want to recognise the string \bl{$abc$} with regular 
expression \bl{$r_1$} then\medskip

\begin{enumerate}
\item \bl{$\Der\,a\,(L(r_1))$}\pause
\item \bl{$\Der\,b\,(\Der\,a\,(L(r_1)))$}\pause
\item \bl{$\Der\,c\,(\Der\,b\,(\Der\,a\,(L(r_1))))$}\bigskip
\item finally we test whether the empty string is in this 
set; same for  \bl{$\Ders\,abc\,(L(r_1))$}.\medskip
\end{enumerate}

The matching algorithm works similarly, just over regular expressions instead of sets.
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Oops\ldots \;\bl{$a^{?\{n\}} \cdot a^{\{n\}}$}}

\begin{center}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,0.0)}},
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=31,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=7cm,
    height=7cm, 
    legend entries={Python,Ruby,Scala V1},  
    legend pos=outer north east,
    legend cell align=left  
]
\addplot[blue,mark=*, mark options={fill=white}] 
  table {re-python.data};
\addplot[brown,mark=pentagon*, mark options={fill=white}] 
  table {re-ruby.data};  
\addplot[red,mark=triangle*,mark options={fill=white}] 
  table {re1.data};  
\end{axis}
\end{tikzpicture}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{A Problem\medskip}

We represented the ``n-times'' \bl{$a^{\{n\}}$} as a 
sequence regular expression:

\begin{center}
\begin{tabular}{rl}
1: & \bl{$a$}\\
2: & \bl{$a\cdot a$}\\
3: & \bl{$a\cdot a\cdot a$}\\
& \ldots\\
13: & \bl{$a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a$}\\
& \ldots\\
20:
\end{tabular}
\end{center}

This problem is aggravated with \bl{$a^?$} being represented 
as \bl{$a + \ONE$}.
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Solving the Problem}

What happens if we extend our regular expressions with explicit 
constructors

\begin{center}
\begin{tabular}{rcl}
\bl{$r$} & \bl{$::=$} & \bl{\ldots}\\
             & \bl{$\mid$} & \bl{$r^{\{n\}}$}\\
             & \bl{$\mid$} & \bl{$r^?$} 
\end{tabular}
\end{center}

What is their meaning?\\
What are the cases for \bl{$nullable$} and \bl{$\der$}?
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Brzozowski: \bl{$a^{?\{n\}} \cdot a^{\{n\}}$}}

\begin{center}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,0.0)}},   
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,200,...,1100},
    xmax=1200,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=8cm,
    height=6.5cm, 
    legend entries={Python,Ruby,Scala V1,Scala V2},  
    legend pos=outer north east,
    legend cell align=left  
]
\addplot[blue,mark=*, mark options={fill=white}] table {re-python.data};
\addplot[brown,mark=pentagon*, mark options={fill=white}] table {re-ruby.data};  
\addplot[red,mark=triangle*,mark options={fill=white}] table {re1.data};  
\addplot[green,mark=square*,mark options={fill=white}] table {re2.data};
\end{axis}
\end{tikzpicture}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Examples}

Recall the example of \bl{$r \dn ((a \cdot b) + b)^*$} with

\begin{center}
\begin{tabular}{l}
\bl{$\der\,a\,r = ((\ONE \cdot b) + \ZERO) \cdot r$}\\
\bl{$\der\,b\,r = ((\ZERO \cdot b) + \ONE)\cdot r$}\\
\bl{$\der\,c\,r = ((\ZERO \cdot b) + \ZERO)\cdot r$}
\end{tabular}
\end{center}

What are these regular expressions equivalent to?

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Simplification Rules}

\begin{center}
\bl{\begin{tabular}{rcl}
$r + \ZERO$       & $\Rightarrow$ & $r$\\
$\ZERO + r$       & $\Rightarrow$ & $r$\\
$r \cdot \ONE$    & $\Rightarrow$ & $r$\\
$\ONE \cdot r$    & $\Rightarrow$ & $r$\\
$r \cdot \ZERO$   & $\Rightarrow$ & $\ZERO$\\
$\ZERO \cdot r$   & $\Rightarrow$ & $\ZERO$\\
$r + r$           & $\Rightarrow$ & $r$
\end{tabular}}
\end{center}

\footnotesize
\lstinputlisting[firstline=22,lastline=26]{../progs/app6.scala}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]

\footnotesize
\lstinputlisting[firstline=1,lastline=21]{../progs/app6.scala}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Brzozowski: \bl{$a^{?\{n\}} \cdot a^{\{n\}}$}}

\begin{center}
\begin{tikzpicture}
\begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,0.0)}},     
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5000,...,10000},
    xmax=11000,
    ymax=35,
    ytick={0,5,...,30},
    scaled ticks=false,
    axis lines=left,
    width=9cm,
    height=7cm,
    legend entries={Scala V2,Scala V3},  
    legend pos=north east
]
\addplot[green,mark=square*,mark options={fill=white}] table {re2.data};
\addplot[black,mark=square*,mark options={fill=white}] table {re3.data};
\end{axis}
\end{tikzpicture}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
  \frametitle{
    Another Example\\[-1mm]
    in Java \liningnums{8} and Python}

\begin{center}
\begin{tikzpicture}
  \begin{axis}[
    xlabel={$n$},
    x label style={at={(1.05,0.0)}},
    ylabel={time in secs},
    enlargelimits=false,
    xtick={0,5,...,30},
    xmax=33,
    ymax=35,
    ytick={0,10,...,30},
    scaled ticks=false,
    axis lines=left,
    width=9cm,
    height=5cm, 
    legend entries={Java 8, Python},  
    legend pos=north west,
    legend cell align=left]
\addplot[blue,mark=*, mark options={fill=white}] table {re-python2.data};
\addplot[cyan,mark=*, mark options={fill=white}] table {re-java.data};
\end{axis}
\end{tikzpicture}
\end{center}

Regex: \bl{$(a^*)^* \cdot b$}

Strings of the form \bl{$\underbrace{\,a\ldots a\,}_{n}$}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Same Example in Java \liningnums{9}+}

\begin{center}
\begin{tikzpicture}
  \begin{axis}[
    xlabel={$n$},
    x label style={at={(1.09,-0.15)}},
    ylabel={time in secs},
    scaled x ticks=false,
    enlargelimits=false,
    xtick distance=10000,
    xmax=44000, 
    ytick={0,10,...,30}, 
    ymax=35, 
    axis lines=left,
    width=9cm,
    height=5cm, 
    legend entries={Java \liningnums{9}+},
    legend pos=north west,
    legend cell align=left]
\addplot[blue,mark=square*,mark options={fill=white}] table {re-java9.data};
\end{axis}
\end{tikzpicture}
\end{center}

Regex: \bl{$(a^*)^* \cdot b$}

Strings of the form \bl{$\underbrace{\,a\ldots a\,}_{n}$}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{and with Brzozowski}

\begin{center}
\begin{tikzpicture}
  \begin{axis}[
    xlabel={$n$},
    x label style={at={(1.09,0.0)}},
    ylabel={time in secs},
    scaled x ticks=false,
    xtick distance=2000000,
    enlargelimits=false,
    xmax=6400000, 
    ytick={0,10,...,30},
    ymax=35,
    axis lines=left,
    width=9cm,
    height=5cm, 
    legend entries={Scala V3},
    legend pos=north west,
    legend cell align=left]
%\addplot[green,mark=square*,mark options={fill=white}] table {re2a.data};
\addplot[black,mark=square*,mark options={fill=white}] table {re3a.data};
\end{axis}
\end{tikzpicture}
\end{center}

Regex: \bl{$(a^*)^* \cdot b$}

Strings of the form \bl{$\underbrace{\,a\ldots a\,}_{n}$}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{\mbox{What is good about this Alg.}}

\begin{itemize}
\item extends to most regular expressions, for example
\bl{$\sim r$} (next slide)

\item is easy to implement in a functional language (slide after)

\item the algorithm is already quite old; there is still
  work to be done to use it as a tokenizer (that is relatively new work)

\item we can prove its correctness\ldots
\end{itemize}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Negation of Regular Expr's}

\begin{itemize}
\item \bl{$\sim{}r$}  \hspace{6mm} (everything that \bl{$r$} cannot recognise)\medskip
\item \bl{$L(\sim{}r) \dn UNIV - L(r)$}\medskip
\item \bl{$nullable (\sim{}r) \dn not\, (nullable(r))$}\medskip
\item \bl{$der\,c\,(\sim{}r) \dn \;\sim{}(der\,c\,r)$}
\end{itemize}\bigskip\pause

Used often for recognising comments:

\[
\bl{/ \cdot * \cdot (\sim{}([a\mbox{-}z]^* \cdot * \cdot / \cdot [a\mbox{-}z]^*)) \cdot * \cdot /}
\]

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Coursework}

\underline{Strand 1:}

\begin{itemize}
\item Submission on Friday 12 October\\accepted until Monday 15 @ 18:00\medskip
\item source code needs to be submitted as well\medskip
\item you can re-use my Scala code from KEATS \\
  or use any programming language you like\medskip
\item \small https://nms.kcl.ac.uk/christian.urban/ProgInScala2ed.pdf\normalsize
\end{itemize}  

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   
  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[t]
\frametitle{Proofs about Rexps}

Remember their inductive definition:\\[5cm]

\begin{textblock}{6}(5,5)
  \begin{tabular}{@ {}rrl}
  \bl{$r$} & \bl{$::=$}  & \bl{$\ZERO$}\\
         & \bl{$\mid$} & \bl{$\ONE$}     \\
         & \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}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Rexp (2)}

\begin{itemize}
\item \bl{$P$} holds for \bl{$\ZERO$}, \bl{$\ONE$} 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{$r_2$}.\bigskip
\item \bl{$P$} holds for \bl{$r^*$} under the assumption that \bl{$P$} already
holds for \bl{$r$}.
\end{itemize}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Rexp (3)}

Assume \bl{$P(r)$} is the property:

\begin{center}
\bl{$nullable(r)$} if and only if \bl{$[] \in L(r)$}
\end{center}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Rexp (4)}

\begin{center}
\bl{\begin{tabular}{r@{\hspace{2mm}}c@{\hspace{2mm}}l}
$rev(\ZERO)$   & $\dn$ & $\ZERO$\\
$rev(\ONE)$      & $\dn$ & $\ONE$\\
$rev(c)$             & $\dn$ & $c$\\
$rev(r_1 + r_2)$     & $\dn$ & $rev(r_1) + rev(r_2)$\\
$rev(r_1 \cdot r_2)$ & $\dn$ & $rev(r_2) \cdot rev(r_1)$\\
$rev(r^*)$           & $\dn$ & $rev(r)^*$\\
\end{tabular}}
\end{center}

We can prove

\begin{center}
\bl{$L(rev(r)) = \{s^{-1} \;|\; s \in L(r)\}$}
\end{center}

by induction on \bl{$r$}.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{\begin{tabular}{c}
  Correctness Proof\\[-1mm] for our Matcher\end{tabular}}

\begin{itemize}
\item We started from

\begin{center}
\begin{tabular}{rp{4cm}}
  & \bl{$s \in L(r)$}\medskip\\
\bl{$\Leftrightarrow$} & \bl{$[] \in \Ders\,s\,(L(r))$}\pause  
\end{tabular}
\end{center}

\item if we can show \bl{$\Ders\, s\,(L(r)) = L(\ders\,s\,r)$} we 
have

\begin{center}
\begin{tabular}{rp{4cm}}
\bl{$\Leftrightarrow$} & \bl{$[] \in L(\ders\,s\,r)$}\medskip\\
\bl{$\Leftrightarrow$} & \bl{$nullable(\ders\,s\,r)$}\medskip\\
\bl{$\dn$} & \bl{$matches\,s\,r$}
\end{tabular}
\end{center}

\end{itemize}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Rexp (5)}

Let \bl{$Der\,c\,A$} be the set defined as

\begin{center}
\bl{$\Der\,c\,A \dn \{ s \;|\;  c\!::\!s \in A\}$ } 
\end{center}

We can prove

\begin{center}
\bl{$L(\der\,c\,r) = \Der\,c\,(L(r))$}
\end{center}

by induction on \bl{$r$}.

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Strings}

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}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Proofs about Strings (2)}

We can then prove

\begin{center}
\bl{$\Ders\,s\,(L(r)) = L(\ders\,s\,r)$}  
\end{center}


We can finally prove

\begin{center}
\bl{$matches\,s\,r$}  if and only if  \bl{$s \in L(r)$} 
\end{center}


\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[c]
\frametitle{Epilogue}

\footnotesize
\begin{center}
\begin{tabular}{cc}
\begin{tikzpicture}
\begin{axis}[
    title={\small{}Graph: \bl{$a^{?\{n\}} \cdot a^{\{n\}}$}},
    xlabel={$n$},
    x label style={at={(1.04,0.0)}},
    ylabel={time in secs},
    enlargelimits=false,
    xmax=7100000,
    ytick={0,5,...,30},
    ymax=33,
    %scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=5cm, 
    legend entries={Scala V3, Scala V4},
    legend style={at={(0.1,-0.2)},anchor=north}]
\addplot[black,mark=square*,mark options={fill=white}] table {re3.data};
\addplot[purple,mark=square*,mark options={fill=white}] table {re4.data};
\end{axis}
\end{tikzpicture}
&
\begin{tikzpicture}
\begin{axis}[
    title={\small{}Graph: \bl{$(a^*)^* \cdot b$}},
    xlabel={$n$},
    x label style={at={(1.09,0.0)}},
    ylabel={time in secs},
    enlargelimits=false,
    xmax=8100000,
    ytick={0,5,...,30},
    ymax=33,
    %scaled ticks=false,
    axis lines=left,
    width=5cm,
    height=5cm, 
    legend entries={Scala V3, Scala V4},
    legend style={at={(0.1,-0.2)},anchor=north}]
\addplot[black,mark=square*,mark options={fill=white}] table {re3a.data};
\addplot[purple,mark=square*,mark options={fill=white}] table {re4a.data};
\end{axis}
\end{tikzpicture}
\end{tabular}
\end{center}

\only<2->{\footnotesize
\begin{textblock}{9}(0.5,8)
\begin{bubble}[11.8cm]
\lstinputlisting[numbers=none]{../progs/app52.scala}
\end{bubble}
\end{textblock}} 
 

\end{frame}

\end{document}

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