% !TEX program = xelatex

\usepackage{../slides}

\usepackage{../graphics}

\usepackage{../langs}

\usepackage{../data}

\usepackage{../grammar}

% beamer stuff 

\renewcommand{\slidecaption}{CFL 06, King's College London}

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

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

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

\begin{document}

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\begin{frame}[t]

\frametitle{%

  \begin{tabular}{@ {}c@ {}}

  \\[-3mm]

  \LARGE Compilers and \\[-2mm] 

  \end{tabular}}

  \normalsize

  \begin{center}

  \begin{tabular}{ll}

    Email:  & christian.urban at kcl.ac.uk\\

    Slides \& Progs: & KEATS (also homework is there)\\  

  \end{tabular}

  \end{center}

\end{frame}

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% \begin{frame}[c]

% \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}

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

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\begin{frame}[t]

\frametitle{Starting Symbol}

\bl{\begin{plstx}[margin=1cm]

  : \meta{S} ::= \meta{A}\cdot\meta{S}\cdot\meta{B} |

                 \meta{B}\cdot\meta{S}\cdot\meta{A} | \epsilon\\

  : \meta{A} ::= a | \epsilon\\

  : \meta{B} ::= b\\  

\end{plstx}}

\end{frame}

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\begin{frame}[c]

\frametitle{Hierarchy of Languages}

Recall that languages are sets of strings.

\begin{center}

\begin{tikzpicture}

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

\draw (0,0) node [rect, text depth=39mm, text width=68mm] {all languages};

\draw (0,-0.4) node [rect, text depth=28.5mm, text width=64mm] {decidable languages};

\draw (0,-0.85) node [rect, text depth=17mm] {context sensitive languages};

\draw (0,-1.14) node [rect, text depth=9mm, text width=50mm] {context-free languages};

\draw (0,-1.4) node [rect] {regular languages};

\end{tikzpicture}

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{Parser Combinators}

Atomic parsers, for example

\begin{center}

\bl{$1::rest \;\Rightarrow\; \{(1, rest)\}$} 

\end{center}\bigskip

\begin{itemize}

\item you consume one or more tokens from the\\ 

  input (stream)

\item also works for characters and strings

\end{itemize}

\end{frame}

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\begin{frame}[c]

Alternative parser (code \bl{$p\;|\;q$})\bigskip

\begin{itemize}

\item apply \bl{$p$} and also \bl{$q$}; then combine 

  the outputs

\end{itemize}

\begin{center}

\large \bl{$p(\text{input}) \cup q(\text{input})$}

\end{center}

\end{frame}

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\begin{frame}[c]

Sequence parser (code \bl{$p\sim q$})\bigskip

\begin{itemize}

\item apply first \bl{$p$} producing a set of pairs

\item then apply \bl{$q$} to the unparsed parts

\item then combine the results:\medskip 

\begin{center}

((output$_1$, output$_2$), unparsed part)

\end{center}

\end{itemize}

\begin{center}

\begin{tabular}{l}

\large \bl{$\{((o_1, o_2), u_2) \;|\;$}\\[2mm] 

\large\mbox{}\hspace{15mm} \bl{$(o_1, u_1) \in p(\text{input}) \wedge$}\\[2mm]

\large\mbox{}\hspace{15mm} \bl{$(o_2, u_2) \in q(u_1)\}$}

\end{tabular}

\end{center}

\end{frame}

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\begin{frame}[c]

Function parser (code \bl{$p \Rightarrow f\;$})\bigskip

\begin{itemize}

\item apply \bl{$p$} producing a set of pairs

\item then apply the function \bl{$f$} to each first component

\end{itemize}

\begin{center}

\begin{tabular}{l}

\large \bl{$\{(f(o_1), u_1) \;|\; (o_1, u_1) \in p(\text{input})\}$}

\end{tabular}

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{Types of Parsers}

\begin{itemize}

\item {\bf Sequencing}: if \bl{$p$} returns results of type \bl{$T$}, and \bl{$q$} results of type \bl{$S$},

then \bl{$p \sim q$} returns results of type

\begin{center}

\bl{$T \times S$}

\end{center}\pause

\item {\bf Alternative}: if \bl{$p$} returns results of type \bl{$T$} then  \bl{$q$} \alert{must} also have results of type \bl{$T$},

and \bl{$p \;|\; q$} returns results of type

\begin{center}

\bl{$T$}

\end{center}\pause

\item {\bf Semantic Action}: if \bl{$p$} returns results of type \bl{$T$} and \bl{$f$} is a function from

\bl{$T$} to \bl{$S$}, then

\bl{$p \Rightarrow f$} returns results of type

\begin{center}

\bl{$S$}

\end{center}

\end{itemize}

\end{frame}

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\begin{frame}[c]

\frametitle{Two Grammars}

Which languages are recognised by the following two grammars?

\bl{\begin{plstx}[margin=3cm]

: \meta{S} ::= \liningnums{1}\cdot\meta{S}\cdot \meta{S} | \epsilon\\

\end{plstx}}

\bl{\begin{plstx}[margin=3cm]

: \meta{U} ::= \liningnums{1}\cdot\meta{U} | \epsilon\\

\end{plstx}}

\end{frame}

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\begin{frame}[t]

\frametitle{Ambiguous Grammars}

\begin{center}

\begin{tikzpicture}

\begin{axis}[xlabel={\pcode{1}s},ylabel={time in secs},

    enlargelimits=false,

    xtick={0,100,...,1000},

    xmax=1050,

    ymax=33,

    ytick={0,5,...,30},

    scaled ticks=false,

    axis lines=left,

    width=11cm,

    height=7cm, 

    legend entries={unambiguous,ambiguous},  

    legend pos=north east,

    legend cell align=left,

    x tick label style={font=\small,/pgf/number format/1000 sep={}}]

\addplot[blue,mark=*, mark options={fill=white}] 

  table {s-grammar1.data};

\only<2>{

  \addplot[red,mark=triangle*, mark options={fill=white}] 

  table {s-grammar2.data};}  

\end{axis}

\end{tikzpicture}

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{Arithmetic Expressions}

A grammar for arithmetic expressions and numbers:

\bl{\begin{plstx}[margin=1cm]

    : \meta{E} ::= \meta{E} \cdot + \cdot \meta{E}

    | \meta{E} \cdot * \cdot \meta{E}

    | ( \cdot \meta{E} \cdot )  | \meta{N}\\

    : \meta{N} ::= \meta{N} \cdot \meta{N} |

    0 | 1 | \ldots | 9\\

\end{plstx}}

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

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

combinators).  \bigskip\pause

\end{frame}

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\begin{frame}[t]

\frametitle{Numbers}

\bl{\begin{plstx}[margin=1cm]

    : \meta{N} ::= \meta{N} \cdot \meta{N} |

    0 | 1 | \ldots | 9\\

\end{plstx}}

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

\bl{\begin{plstx}[margin=1cm]

    : \meta{N} ::= 0 \cdot \meta{N} | 1 \cdot \meta{N} | \ldots |

    0 | 1 | \ldots | 9\\

\end{plstx}}

\end{frame}

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\begin{frame}[c]

\frametitle{Removing Left-Recursion}

The rule for numbers is directly left-recursive:

\begin{center}

\bl{\begin{tabular}{lcl}

$\meta{N}$ & $::=$ & $\meta{N} \cdot \meta{N} \;|\; 0 \;|\; 1\;\;\;\;(\ldots)$ 

\end{tabular}}

\end{center}

Translate

\begin{center}

\begin{tabular}{ccc}

\bl{\begin{tabular}{lcl}

$\meta{N}$ & $::=$ & $\meta{N} \cdot \alpha$\\

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

 \\ 

\end{tabular}} 

& {\Large$\Rightarrow$} &

\bl{\begin{tabular}{lcl}

$\meta{N}$ & $::=$ & $\beta \cdot \meta{N}'$\\

$\meta{N}'$ & $::=$ & $\alpha \cdot \meta{N}'$\\

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

\end{tabular}}

\end{tabular}

\end{center}\pause

Which means in this case:

\begin{center}

\bl{\begin{tabular}{lcl}

$\meta{N}$ & $\rightarrow$ & $0 \cdot \meta{N}' \;|\; 1 \cdot \meta{N}'$\\

$\meta{N}'$ & $\rightarrow$ & $\meta{N} \cdot \meta{N}' \;|\; \epsilon$\\

\end{tabular}}

\end{center}

\end{frame}

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%\begin{frame}[c]

%\frametitle{Operator Precedences}

%

%

%To disambiguate

%

%\begin{center}

%\bl{\begin{tabular}{lcl}

%$\meta{E}$ & $::=$ &  $\meta{E} \cdot + \cdot \meta{E} \;|\;\meta{E} \cdot * \cdot \meta{E} \;|\;( \cdot \meta{E} \cdot ) \;|\;\meta{N}$ \\

%\end{tabular}}

%\end{center}

%

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

%highest for \bl{$()$}, medium for \bl{*}, lowest for \bl{+}

%

%\begin{center}

%\bl{\begin{tabular}{lcl}

%$\meta{E}_{low}$ & $::=$ & $\meta{E}_{med} \cdot + \cdot \meta{E}_{low} \;|\; \meta{E}_{med}$ \\

%$\meta{E}_{med}$ & $::=$ & $\meta{E}_{hi} \cdot * \cdot \meta{E}_{med} \;|\; \meta{E}_{hi}$\\

%$\meta{E}_{hi}$ & $::=$ &  $( \cdot \meta{E}_{low} \cdot ) \;|\;\meta{N}$ \\

%\end{tabular}}

%\end{center}\pause

%

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

%\end{frame}

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\begin{frame}[c]

\frametitle{Chomsky Normal Form}

All rules must be of the form

\begin{center}

\bl{$\meta{A} ::= a$}

\end{center}

or

\begin{center}

\bl{$\meta{A} ::= \meta{B}\cdot \meta{C}$}

\end{center}

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

\end{frame}

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\begin{frame}[c]

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

\begin{enumerate}

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

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

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

\end{enumerate}

\small

\begin{center}

\begin{tabular}{ccc}

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

$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'$\\

$N'$ & $::=$ & $N \cdot N'\;|\;\epsilon$\\

\\ 

\\

\\

\\

\\

\end{tabular}} &

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

\\

$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $::=$ & $N \cdot N'\;|\;N\;|\;\epsilon$\\

\\

$N$ & $::=$ & $0 \cdot N' \;|\; 1\cdot N'\;|\;0\;|\;1$\\

$N'$ & $::=$ & $N \cdot N'\;|\;N$\\

\end{tabular}}

\end{tabular}

\end{center}

\pause\normalsize

\begin{center}

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

$N$ & $::=$ & $0 \cdot N\;|\; 1\cdot N\;|\;0\;|\;1$\\

\end{tabular}}

\end{center}

\end{frame}

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\begin{frame}[c]

\frametitle{CYK Algorithm}

If grammar is in Chomsky normalform \ldots