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

%\newcommand{\dn}{\stackrel{\mbox{\scriptsize def}}{=}}% for definitions

\mbox{}\\[-18mm]\mbox{}

{\lstset{language=Scala}\fontsize{10}{12}\selectfont

\texttt{\lstinputlisting[xleftmargin=0mm]{../progs/pow.scala}}}

\small

\mbox{}\\[5mm]

%\begin{textblock}{10}(3,5)

\begin{tikzpicture}[scale=1.5,

                    node distance=1cm,

                    every node/.style={minimum size=7mm}]

\node (r1)  {\bl{$r_1$}};

\node (r2) [right=of r1] {\bl{$r_2$}};

\draw[->,line width=1mm]  (r1) -- (r2) node[above,midway] {\bl{$der\,a$}};

\node (r3) [right=of r2] {\bl{$r_3$}};

\draw[->,line width=1mm]  (r2) -- (r3) node[above,midway] {\bl{$der\,b$}};

\node (r4) [right=of r3] {\bl{$r_4$}};

\draw[->,line width=1mm]  (r3) -- (r4) node[above,midway] {\bl{$der\,c$}};

\draw (r4) node[anchor=west] {\;\raisebox{3mm}{\bl{$nullable$}}};

\node (v4) [below=of r4] {\bl{$v_4$}};

\draw[->,line width=1mm]  (r4) -- (v4);

\node (v3) [left=of v4] {\bl{$v_3$}};

\draw[->,line width=1mm]  (v4) -- (v3) node[below,midway] {\bl{$inj\,c$}};

\node (v2) [left=of v3] {\bl{$v_2$}};

\draw[->,line width=1mm]  (v3) -- (v2) node[below,midway] {\bl{$inj\,b$}};

\node (v1) [left=of v2] {\bl{$v_1$}};

\draw[->,line width=1mm]  (v2) -- (v1) node[below,midway] {\bl{$inj\,a$}};

\draw[->,line width=0.5mm]  (r3) -- (v3);

\draw[->,line width=0.5mm]  (r2) -- (v2);

\draw[->,line width=0.5mm]  (r1) -- (v1);

\draw (r4) node[anchor=north west] {\;\raisebox{-8mm}{\bl{$mkeps$}}};

\end{tikzpicture}

%\end{textblock}

\begin{center}

\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

  \\[-10mm]

  \bl{$inj\,(c)\,c\,Empty$} & \bl{$\dn$}  & \bl{$Char\,c$}\\

  \bl{$inj\,(r_1 + r_2)\,c\,Left(v)$} & \bl{$\dn$}  & \bl{$Left(inj\,r_1\,c\,v)$}\\

  \bl{$inj\,(r_1 + r_2)\,c\,Right(v)$} & \bl{$\dn$}  & \bl{$Right(inj\,r_2\,c\,v)$}\\

  \bl{$inj\,(r_1 \cdot r_2)\,c\,Seq(v_1,v_2)$} & \bl{$\dn$}  & \bl{$Seq(inj\,r_1\,c\,v_1,v_2)$}\\

  \bl{$inj\,(r_1 \cdot r_2)\,c\,Left(Seq(v_1,v_2))$} & \bl{$\dn$}  & \bl{$Seq(inj\,r_1\,c\,v_1,v_2)$}\\

  \bl{$inj\,(r_1 \cdot r_2)\,c\,Right(v)$} & \bl{$\dn$}  & \bl{$Seq(mkeps(r_1),inj\,r_2\,c\,v)$}\\

  \bl{$inj\,(r^*)\,c\,Seq(v,vs)$} & \bl{$\dn$}  & \bl{$inj\,r\,c\,v\,::\,vs$}\\

\end{tabular}

\end{center}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

\newcommand{\qq}{\mbox{\texttt{"}}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[c]

\frametitle{CFGs}

A \alert{context-free} grammar (CFG) \bl{$G$} consists of:

\begin{itemize}

\item a finite set of nonterminal symbols (upper case)

\item a finite terminal symbols or tokens (lower case)

\item a start symbol (which must be a nonterminal)

\item a set of rules

\begin{center}

\bl{$A \rightarrow \text{rhs}_1 | \text{rhs}_2 | \ldots$}

\end{center}

where \bl{rhs} are sequences involving terminals and nonterminals (can also be empty).\medskip\pause

\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Hierarchy of Languages\end{tabular}}

Recall that languages are sets of strings.

[rect/.style={draw=black!50, top color=white,bottom color=black!20, rectangle, very thick, rounded corners}]

\end{frame}}

\mode<presentation>{

\frametitle{\begin{tabular}{c}Arithmetic Expressions\end{tabular}}

A grammar for arithmetic expressions and numbers:

\begin{center}

\bl{\begin{tabular}{lcl}

$E$ & $\rightarrow$ &  $E \cdot + \cdot E \;|\;E \cdot * \cdot E \;|\;( \cdot E \cdot ) \;|\;N$ \\

$N$ & $\rightarrow$ & $N \cdot N \;|\; 0 \;|\; 1 \;|\: \ldots \;|\; 9$ 

\end{tabular}}

\end{center}

Unfortunately it is left-recursive (and ambiguous).\medskip\\

A problem for \alert{recursive descent parsers} (e.g.~parser combinators).

\bigskip\pause

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[t]

\frametitle{\begin{tabular}{c}Numbers\end{tabular}}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ &  $N \cdot N \;|\; 0 \;|\; 1 \;|\; \ldots \;|\; 9$\\

\end{tabular}}

A non-left-recursive, non-ambiguous grammar for numbers:

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ &  $0 \cdot N \;|\;1 \cdot N \;|\;\ldots\;|\; 0 \;|\; 1 \;|\; \ldots \;|\; 9$\\

\end{tabular}}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Operator Precedences\end{tabular}}

To disambiguate

\begin{center}

\bl{\begin{tabular}{lcl}

$E$ & $\rightarrow$ &  $E \cdot + \cdot E \;|\;E \cdot * \cdot E \;|\;( \cdot E \cdot ) \;|\;N$ \\

\end{tabular}}

\end{center}

Decide on how many precedence levels, say\medskip\\

\hspace{5mm}highest for \bl{$()$}, medium for \bl{*}, lowest for \bl{+}

\begin{center}

\bl{\begin{tabular}{lcl}

$E_{low}$ & $\rightarrow$ & $E_{med} \cdot + \cdot E_{low} \;|\; E_{med}$ \\

$E_{med}$ & $\rightarrow$ & $E_{hi} \cdot * \cdot E_{med} \;|\; E_{hi}$\\

$E_{hi}$ & $\rightarrow$ &  $( \cdot E_{low} \cdot ) \;|\;N$ \\

\end{tabular}}

\end{center}\pause

\small What happens with \bl{$1 + 3  + 4$}?

\end{frame}}

\mode<presentation>{

\frametitle{\begin{tabular}{c}Removing Left-Recursion\end{tabular}}

The rule for numbers is directly left-recursive:

$N$ & $\rightarrow$ & $N \cdot N \;|\; 0 \;|\; 1\;\;\;\;(\ldots)$ 

\end{tabular}}

\end{center}

Translate

\begin{center}

\begin{tabular}{ccc}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $N \cdot \alpha$\\

 &  $\;|\;$ & $\beta$\\

 \\ 

\end{tabular}} 

& {\Large$\Rightarrow$} &

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $\beta \cdot N'$\\

$N'$ & $\rightarrow$ & $\alpha \cdot N'$\\

 &  $\;|\;$ & $\epsilon$ 

\end{tabular}

Which means

\begin{center}

\bl{\begin{tabular}{lcl}

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1 \cdot N'$\\

$N'$ & $\rightarrow$ & $N \cdot N' \;|\; \epsilon$\\

\end{tabular}}

\end{center}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Chomsky Normal Form\end{tabular}}

All rules must be of the form

\begin{center}

\bl{$A \rightarrow a$}

\end{center}

\bl{$A \rightarrow B\cdot C$}

\end{center}

No rule can contain \bl{$\epsilon$}.

\end{frame}}

\mode<presentation>{

\frametitle{\begin{tabular}{c}$\epsilon$-Removal\end{tabular}}

\item If \bl{$A\rightarrow \alpha \cdot B \cdot \beta$} and \bl{$B \rightarrow \epsilon$} are in the grammar,

then add \bl{$A\rightarrow \alpha \cdot \beta$} (iterate if necessary).

\item Throw out all \bl{$B \rightarrow \epsilon$}.

\small

\begin{tabular}{ccc}

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;\epsilon$\\

\\ 

\\

\\

\\

\\

\end{tabular}} &

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

\\

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;N\;|\;\epsilon$\\

\\

$N$ & $\rightarrow$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $\rightarrow$ & $N \cdot N'\;|\;N$\\

\pause\normalsize

\begin{center}

\bl{\begin{tabular}{l@{\hspace{1mm}}c@{\hspace{1mm}}l}

$N$ & $\rightarrow$ & $0 \cdot N\;|\; 1\cdot N\;|\;0\;|\;1$\\

\end{tabular}}

\end{center}

\end{frame}}

\mode<presentation>{

\frametitle{\begin{tabular}{c}CYK Algorithm\end{tabular}}

If grammar is in Chomsky normalform \ldots

\bl{\begin{tabular}{@ {}lcl@ {}}

$S$ & $\rightarrow$ &  $N\cdot P$ \\

$P$ & $\rightarrow$ &  $V\cdot N$ \\

$N$ & $\rightarrow$ &  $N\cdot N$ \\

$N$ & $\rightarrow$ &  $\texttt{students} \;|\; \texttt{Jeff} \;|\; \texttt{geometry} \;|\; \texttt{trains} $ \\

$V$ & $\rightarrow$ &  $\texttt{trains}$ 

\bl{\texttt{Jeff trains geometry students}}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}CYK Algorithm\end{tabular}}

\begin{itemize}

\item fastest possible algorithm for recognition problem

\item runtime is \bl{$O(n^3)$}\bigskip

\item grammars need to be transferred into CNF

\end{itemize}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

\mode<presentation>{

\bl{\begin{tabular}{@{}lcl@{}}

\textit{Stmt} & $\rightarrow$ &  $\texttt{skip}$\\

              & $|$ & \textit{Id}\;\texttt{:=}\;\textit{AExp}\\

              & $|$ & \texttt{if}\; \textit{BExp} \;\texttt{then}\; \textit{Block} \;\texttt{else}\; \textit{Block}\\

              & $|$ & \texttt{while}\; \textit{BExp} \;\texttt{do}\; \textit{Block}\\

              & $|$ & \texttt{read}\;\textit{Id}\\

              & $|$ & \texttt{write}\;\textit{Id}\\

              & $|$ & \texttt{write}\;\textit{String}\medskip\\

\textit{Stmts} & $\rightarrow$ &  \textit{Stmt} \;\texttt{;}\; \textit{Stmts}\\

              & $|$ & \textit{Stmt}\medskip\\

\textit{Block} & $\rightarrow$ &  \texttt{\{}\,\textit{Stmts}\,\texttt{\}}\\

                & $|$ & \textit{Stmt}\medskip\\

\textit{AExp} & $\rightarrow$ & \ldots\\

\textit{BExp} & $\rightarrow$ & \ldots\\

\end{frame}}

\mbox{\lstinputlisting[language=while]{../progs/fib.while}}

\frametitle{\begin{tabular}{c}An Interpreter\end{tabular}}

\item \bl{\text{eval}(stmt, env)}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Interpreter\end{tabular}}

\begin{center}

\bl{\begin{tabular}{@{}lcl@{}}

$\text{eval}(n, E)$ & $\dn$ & $n$\\

$\text{eval}(x, E)$ & $\dn$ & $E(x)$ \;\;\;\textcolor{black}{lookup \bl{$x$} in \bl{$E$}}\\

$\text{eval}(a_1 + a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) + \text{eval}(a_2, E)$\\

$\text{eval}(a_1 - a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) - \text{eval}(a_2, E)$\\

$\text{eval}(a_1 * a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) * \text{eval}(a_2, E)$\bigskip\\

$\text{eval}(a_1 = a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) = \text{eval}(a_2, E)$\\

$\text{eval}(a_1\,!\!= a_2, E)$ & $\dn$ & $\neg(\text{eval}(a_1, E) = \text{eval}(a_2, E))$\\

$\text{eval}(a_1 < a_2, E)$ & $\dn$ & $\text{eval}(a_1, E) < \text{eval}(a_2, E)$\

\end{tabular}}

\end{center}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Interpreter (2)\end{tabular}}

\begin{center}

\bl{\begin{tabular}{@{}lcl@{}}

$\text{eval}(\text{skip}, E)$ & $\dn$ & $E$\\

$\text{eval}(x:=a, E)$ & $\dn$ & \bl{$E(x \mapsto \text{eval}(a, E))$}\\

\multicolumn{3}{@{}l@{}}{$\text{eval}(\text{if}\;b\;\text{then}\;cs_1\;\text{else}\;cs_2 , E) \dn$}\\

\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{if}\;\text{eval}(b,E)\;\text{then}\;

\text{eval}(cs_1,E)$}\\

\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\phantom{\text{if}\;\text{eval}(b,E)\;}\text{else}\;\text{eval}(cs_2,E)$}\\

\multicolumn{3}{@{}l@{}}{$\text{eval}(\text{while}\;b\;\text{do}\;cs, E) \dn$}\\

\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{if}\;\text{eval}(b,E)$}\\

\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{then}\;

\text{eval}(\text{while}\;b\;\text{do}\;cs, \text{eval}(cs,E))$}\\

\multicolumn{3}{@{}l@{}}{\hspace{2cm}$\text{else}\; E$}\\

$\text{eval}(\text{write}\; x, E)$ & $\dn$ & $\{\;\text{println}(E(x))\; ;\;E\;\}$\\

\end{tabular}}

\end{center}

\end{frame}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[c]

\frametitle{\begin{tabular}{c}Test Program\end{tabular}}

\mbox{}\\[-18mm]\mbox{}

{\lstset{language=While}%%\fontsize{10}{12}\selectfont

\texttt{\lstinputlisting{../progs/loops.while}}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\mode<presentation>{

\begin{frame}[t]

\frametitle{\begin{tabular}{c}Interpreted Code\end{tabular}}

\begin{center}

\begin{tikzpicture}

\begin{axis}[axis x line=bottom, axis y line=left, xlabel=n, ylabel=secs, legend style=small]

\addplot+[smooth] file {interpreted.data};

\end{axis}