% !TEX program = xelatex

\documentclass{article}

\usepackage{../style}

\usepackage{../langs}

\usepackage{../grammar}

% epsilon and left-recursion elimination

% http://www.mollypages.org/page/grammar/index.mp

%% parsing scala files

%% https://scalameta.org/

% chomsky normal form transformation

% https://youtu.be/FNPSlnj3Vt0

% Language hierachy is about complexity

%    How hard is it to recognise an element in a language

% Pratt parsing

% https://matklad.github.io/2020/04/13/simple-but-powerful-pratt-parsing.html

% https://www.oilshell.org/blog/2017/03/31.html

\begin{document}

\section*{Handout 5 (Grammars \& Parser)}

\begin{center}

$(((()()))())$  \hspace{10mm} $(((()()))()))$

\end{center}

\noindent Not being able to solve such recognition problems is

a serious limitation. In order to solve such recognition

problems, we need more powerful techniques than regular

expressions. We will in particular look at \emph{context-free

languages}. They include the regular languages as the picture

below about language classes shows:

\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=30mm, text width=46mm] {\small all languages};

\draw (0,-0.4) node [rect, text depth=20mm, text width=44mm] {\small decidable languages};

\draw (0,-0.65) node [rect, text depth=13mm] {\small context sensitive languages};

\draw (0,-0.84) node [rect, text depth=7mm, text width=35mm] {\small context-free languages};

\draw (0,-1.05) node [rect] {\small regular languages};

\end{tikzpicture}

\end{center}

\noindent Each ``bubble'' stands for sets of languages (remember

languages are sets of strings). As indicated the set of regular

languages is fully included inside the context-free languages,

meaning every regular language is also context-free, but not vice

versa. Below I will let you think, for example, what the context-free

grammar is for the language corresponding to the regular expression

$(aaa)^*a$.

Because of their convenience, context-free languages play an important

role in `day-to-day' text processing and in programming

languages. Context-free in this setting means that ``words'' have one

meaning only and this meaning is independent from the context

the ``words'' appear in. For example ambiguity issues like

\begin{center}

\tt Time flies like an arrow. Fruit flies like bananas.

\end{center}  

\noindent

from natural languages were the meaning of \emph{flies} depends on the

surrounding \emph{context} are avoided as much as possible. Here is

an interesting video about C++ being not a context-free language

\begin{center}

\url{https://www.youtube.com/watch?v=OzK8pUu4UfM}

\end{center}  

Context-free languages are usually specified by grammars. For example

a grammar for well-parenthesised expressions can be given as follows:

\begin{plstx}[margin=3cm]

: \meta{P} ::=  ( \cdot  \meta{P} \cdot ) \cdot \meta{P}

             | \epsilon\\ 

\end{plstx}

\noindent 

or a grammar for recognising strings consisting of ones is

\begin{plstx}[margin=3cm]

: \meta{O} ::= 1 \cdot  \meta{O} 

             | 1\\

\end{plstx}

In general grammars consist of finitely many rules built up

from \emph{terminal symbols} (usually lower-case letters) and

\emph{non-terminal symbols} (upper-case letters written in

bold like \meta{A}, \meta{N} and so on). Rules have

the shape

\begin{plstx}[margin=3cm]

: \meta{NT} ::= rhs\\

\end{plstx}

\noindent where on the left-hand side is a single non-terminal

and on the right a string consisting of both terminals and

non-terminals including the $\epsilon$-symbol for indicating

the empty string. We use the convention to separate components

on the right hand-side by using the $\cdot$ symbol, as in the

grammar for well-parenthesised expressions. We also use the

convention to use $|$ as a shorthand notation for several

rules. For example

\begin{plstx}[margin=3cm]

: \meta{NT} ::= rhs_1

   | rhs_2\\

\end{plstx}

\noindent means that the non-terminal \meta{NT} can be replaced by

either $\textit{rhs}_1$ or $\textit{rhs}_2$. If there are more

than one non-terminal on the left-hand side of the rules, then

we need to indicate what is the \emph{starting} symbol of the

grammar. For example the grammar for arithmetic expressions

can be given as follows

\begin{plstx}[margin=3cm,one per line]

\mbox{\rm (1)}: \meta{E} ::= \meta{N}\\

\mbox{\rm (2)}: \meta{E} ::= \meta{E} \cdot + \cdot \meta{E}\\

\mbox{\rm (3)}: \meta{E} ::= \meta{E} \cdot - \cdot \meta{E}\\

\mbox{\rm (4)}: \meta{E} ::= \meta{E} \cdot * \cdot \meta{E}\\

\mbox{\rm (5)}: \meta{E} ::= ( \cdot \meta{E} \cdot )\\

\mbox{\rm (6\ldots)}: \meta{N} ::= \meta{N} \cdot \meta{N} 

  \mid 0 \mid 1 \mid \ldots \mid 9\\

\end{plstx}

\noindent where \meta{E} is the starting symbol. A

\emph{derivation} for a grammar starts with the starting

symbol of the grammar and in each step replaces one

non-terminal by a right-hand side of a rule. A derivation ends

with a string in which only terminal symbols are left. For

example a derivation for the string $(1 + 2) + 3$ is as

follows:

\begin{center}

\begin{tabular}{lll@{\hspace{2cm}}l}

\meta{E} & $\rightarrow$ & $\meta{E}+\meta{E}$          & by (2)\\

       & $\rightarrow$ & $(\meta{E})+\meta{E}$     & by (5)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+\meta{E}$   & by (2)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+\meta{N}$   & by (1)\\

       & $\rightarrow$ & $(\meta{E}+\meta{E})+3$   & by (6\dots)\\

       & $\rightarrow$ & $(\meta{N}+\meta{E})+3$   & by (1)\\

       & $\rightarrow^+$ & $(1+2)+3$ & by (1, 6\ldots)\\

\end{tabular} 

\end{center}

\noindent where on the right it is indicated which 

grammar rule has been applied. In the last step we

merged several steps into one.

The \emph{language} of a context-free grammar $G$

with start symbol $S$ is defined as the set of strings

derivable by a derivation, that is

\begin{center}

$\{c_1\ldots c_n \;|\; S \rightarrow^* c_1\ldots c_n \;\;\text{with all} \; c_i \;\text{being non-terminals}\}$

\end{center}

\noindent

A \emph{parse-tree} encodes how a string is derived with the starting

symbol on top and each non-terminal containing a subtree for how it is

replaced in a derivation. The parse tree for the string $(1 + 23)+4$ is

as follows:

\begin{center}

\begin{tikzpicture}[level distance=8mm, black]

  \node {\meta{E}}

    child {node {\meta{E} } 

       child {node {$($}}

       child {node {\meta{E} }       

         child {node {\meta{E} } child {node {\meta{N} } child {node {$1$}}}}

         child {node {$+$}}

         child {node {\meta{E} } 

            child {node {\meta{N} } child {node {$2$}}}

            child {node {\meta{N} } child {node {$3$}}}

            } 

        }

       child {node {$)$}}

     }

     child {node {$+$}}

     child {node {\meta{E} }

        child {node {\meta{N} } child {node {$4$}}}

     };

\end{tikzpicture}

\end{center}

\noindent We are often interested in these parse-trees since

they encode the structure of how a string is derived by a

grammar. 

Before we come to the problem of constructing such parse-trees, we need

to consider the following two properties of grammars. A grammar is

\emph{left-recursive} if there is a derivation starting from a

non-terminal, say \meta{NT} which leads to a string which again starts

with \meta{NT}. This means a derivation of the form.

\begin{center}

$\meta{NT} \rightarrow \ldots \rightarrow \meta{NT} \cdot \ldots$

\end{center}

\noindent It can be easily seen that the grammar above for arithmetic

expressions is left-recursive: for example the rules $\meta{E}

\rightarrow \meta{E}\cdot + \cdot \meta{E}$ and $\meta{N} \rightarrow

\meta{N}\cdot \meta{N}$ show that this grammar is left-recursive. But

note that left-recursiveness can involve more than one step in the

derivation. The problem with left-recursive grammars is that some

algorithms cannot cope with them: with left-recursive grammars they will

fall into a loop. Fortunately every left-recursive grammar can be

transformed into one that is not left-recursive, although this

transformation might make the grammar less ``human-readable''. For

example if we want to give a non-left-recursive grammar for numbers we

might specify

\begin{center}

$\meta{N} \;\;\rightarrow\;\; 0\;|\;\ldots\;|\;9\;|\;

1\cdot \meta{N}\;|\;2\cdot \meta{N}\;|\;\ldots\;|\;9\cdot \meta{N}$

\end{center}

\noindent Using this grammar we can still derive every number

string, but we will never be able to derive a string of the

form $\meta{N} \to \ldots \to \meta{N} \cdot \ldots$.

The other property we have to watch out for is when a grammar

is \emph{ambiguous}. A grammar is said to be ambiguous if

there are two parse-trees for one string. Again the grammar

for arithmetic expressions shown above is ambiguous. While the

shown parse tree for the string $(1 + 23) + 4$ is unique, this

is not the case in general. For example there are two parse

trees for the string $1 + 2 + 3$, namely

\begin{center}

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

\begin{tikzpicture}[level distance=8mm, black]

  \node {\meta{E} }

    child {node {\meta{E} } child {node {\meta{N} } child {node {$1$}}}}

    child {node {$+$}}

    child {node {\meta{E} }

       child {node {\meta{E} } child {node {\meta{N} } child {node {$2$}}}}

       child {node {$+$}}

       child {node {\meta{E} } child {node {\meta{N} } child {node {$3$}}}}

    }

    ;

\end{tikzpicture} 

&

\begin{tikzpicture}[level distance=8mm, black]

  \node {\meta{E} }

    child {node {\meta{E} }

       child {node {\meta{E} } child {node {\meta{N} } child {node {$1$}}}}

       child {node {$+$}}

       child {node {\meta{E} } child {node {\meta{N} } child {node {$2$}}}} 

    }

    child {node {$+$}}

    child {node {\meta{E} } child {node {\meta{N} } child {node {$3$}}}}

    ;

\end{tikzpicture}

\end{tabular} 

\end{center}

\noindent In particular in programming languages we will try to avoid

ambiguous grammars because two different parse-trees for a string mean a

program can be interpreted in two different ways. In such cases we have

to somehow make sure the two different ways do not matter, or

disambiguate the grammar in some other way (for example making the $+$

left-associative). Unfortunately already the problem of deciding whether

a grammar is ambiguous or not is in general undecidable. But in simple

instance (the ones we deal with in this module) one can usually see when

a grammar is ambiguous.

\subsection*{Removing Left-Recursion}

Let us come back to the problem of left-recursion and consider the 

following grammar for binary numbers:

\begin{plstx}[margin=1cm]

: \meta{B} ::= \meta{B} \cdot \meta{B} | 0 | 1\\

\end{plstx}

\noindent

It is clear that this grammar can create all binary numbers, but

it is also clear that this grammar is left-recursive. Giving this

grammar as is to parser combinators will result in an infinite 

loop. Fortunately, every left-recursive grammar can be translated

into one that is not left-recursive with the help of some

transformation rules. Suppose we identified the ``offensive'' 

rule, then we can separate the grammar into this offensive rule

and the ``rest'':

\begin{plstx}[margin=1cm]

  : \meta{B} ::= \underbrace{\meta{B} \cdot \meta{B}}_{\textit{lft-rec}} 

  | \underbrace{0 \;\;|\;\; 1}_{\textit{rest}}\\

\end{plstx}

\noindent

To make the idea of the transformation clearer, suppose the left-recursive

rule is of the form $\meta{B}\alpha$ (the left-recursive non-terminal 

followed by something called $\alpha$) and the ``rest'' is called $\beta$.

That means our grammar looks schematically as follows

\begin{plstx}[margin=1cm]

  : \meta{B} ::= \meta{B} \cdot \alpha | \beta\\

\end{plstx}

\noindent

To get rid of the left-recursion, we are required to introduce 

a new non-terminal, say $\meta{B'}$ and transform the rule

as follows:

\begin{plstx}[margin=1cm]

  : \meta{B} ::= \beta \cdot \meta{B'}\\

  : \meta{B'} ::= \alpha \cdot \meta{B'} | \epsilon\\

\end{plstx}

\noindent

In our example of binary numbers we would after the transformation 

end up with the rules

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B'} | 1 \cdot \meta{B'}\\

  : \meta{B'} ::= \meta{B} \cdot \meta{B'} | \epsilon\\

\end{plstx}

\noindent

A little thought should convince you that this grammar still derives

all the binary numbers (for example 0 and 1 are derivable because $\meta{B'}$

can be $\epsilon$). Less clear might be why this grammar is non-left recursive.

For $\meta{B'}$ it is relatively clear because we will never be 

able to derive things like

\begin{center}

$\meta{B'} \rightarrow\ldots\rightarrow \meta{B'}\cdot\ldots$

\end{center}  

\noindent

because there will always be a $\meta{B}$ in front of a $\meta{B'}$, and

$\meta{B}$ now has always a $0$ or $1$ in front, so a $\meta{B'}$ can

never be in the first place. The reasoning is similar for $\meta{B}$:

the $0$ and $1$ in the rule for $\meta{B}$ ``protect'' it from becoming

left-recursive. This transformation does not mean the grammar is the

simplest left-recursive grammar for binary numbers. For example the

following grammar would do as well

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B} | 1 \cdot \meta{B} | 0 | 1\\

\end{plstx}

\noindent

The point is that we can in principle transform every left-recursive

grammar into one that is non-left-recursive one. This explains why often

the following grammar is used for arithmetic expressions:

\begin{plstx}[margin=1cm]

  : \meta{E} ::= \meta{T} | \meta{T} \cdot + \cdot \meta{E} |  \meta{T} \cdot - \cdot \meta{E}\\

  : \meta{T} ::= \meta{F} | \meta{F} \cdot * \cdot \meta{T}\\

  : \meta{F} ::= num\_token | ( \cdot \meta{E} \cdot )\\

\end{plstx}

\noindent

In this grammar all $\meta{E}$xpressions, $\meta{T}$erms and $\meta{F}$actors

are in some way protected from being left-recusive. For example if you

start $\meta{E}$ you can derive another one by going through $\meta{T}$, then 

$\meta{F}$, but then $\meta{E}$ is protected by the open-parenthesis.

\subsection*{Removing $\epsilon$-Rules and CYK-Algorithm}

I showed above that the non-left-recursive grammar for binary numbers is

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B'} | 1 \cdot \meta{B'}\\

  : \meta{B'} ::= \meta{B} \cdot \meta{B'} | \epsilon\\

\end{plstx}

\noindent

The transformation made the original grammar non-left-recursive, but at

the expense of introducing an $\epsilon$ in the second rule. Having an

explicit $\epsilon$-rule is annoying to, not in terms of looping, but in

terms of efficiency. The reason is that the $\epsilon$-rule always

applies but since it recognises the empty string, it does not make any

progress with recognising a string. Better are rules like $( \cdot

\meta{E} \cdot )$ where something of the input is consumed. Getting

rid of $\epsilon$-rules is also important for the CYK parsing algorithm,

which can give us an insight into the complexity class of parsing.

It turns out we can also by some generic transformations eliminate

$\epsilon$-rules from grammars. Consider again the grammar above for

binary numbers where have a rule $\meta{B'} ::= \epsilon$. In this case

we look for rules of the (generic) form \mbox{$\meta{A} :=

\alpha\cdot\meta{B'}\cdot\beta$}. That is there are rules that use

$\meta{B'}$ and something ($\alpha$) is in front of $\meta{B'}$ and

something follows ($\beta$). Such rules need to be replaced by

additional rules of the form \mbox{$\meta{A} := \alpha\cdot\beta$}.

In our running example there are the two rules for $\meta{B}$ which

fall into this category

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B'} | 1 \cdot \meta{B'}\\

\end{plstx} 

\noindent To follow the general scheme of the transfromation,

the $\alpha$ is either is either $0$ or $1$, and the $\beta$ happens

to be empty. SO we need to generate new rules for the form 

\mbox{$\meta{A} := \alpha\cdot\beta$}, which in our particular 

example means we obtain

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B'} | 1 \cdot \meta{B'} | 0 | 1\\

\end{plstx} 

\noindent

Unfortunately $\meta{B'}$ is also used in the rule 

\begin{plstx}[margin=1cm]

  : \meta{B'} ::= \meta{B} \cdot \meta{B'}\\

\end{plstx}

\noindent

For this we repeat the transformation, giving 

\begin{plstx}[margin=1cm]

  : \meta{B'} ::= \meta{B} \cdot \meta{B'} | \meta{B}\\

\end{plstx}

\noindent

In this case $\alpha$ was substituted with $\meta{B}$ and $\beta$

was again empty. Once no rule is left over, we can simply throw

away the $\epsilon$ rule.  This gives the grammar

\begin{plstx}[margin=1cm]

  : \meta{B} ::= 0 \cdot \meta{B'} | 1 \cdot \meta{B'} | 0 | 1\\

  : \meta{B'} ::= \meta{B} \cdot \meta{B'} | \meta{B}\\

\end{plstx}

\noindent

I let you think about whether this grammar can still recognise all

binary numbers and whether this grammar is non-left-recursive. The

precise statement for the transformation of removing $\epsilon$-rules is

that if the original grammar was able to recognise only non-empty

strings, then the transformed grammar will be equivalent (matching the

same set of strings); if the original grammar was able to match the

empty string, then the transformed grammar will be able to match the

same strings, \emph{except} the empty string. So the  $\epsilon$-removal

does not preserve equivalence of grammars, but the small defect with the

empty string is not important for practical purposes.

So why are these transformations all useful? Well apart from making the 

parser combinators work (remember they cannot deal with left-recursion and

are inefficient with $\epsilon$-rules), a second reason is that they help

with getting any insight into the complexity of the parsing problem. 

The parser combinators are very easy to implement, but are far from the 

most efficient way of processing input (they can blow up exponentially

with ambiguous grammars). The question remains what is the best possible

complexity for parsing? It turns out that this is $O(n^3)$ for context-free

languages. 

To answer the question about complexity, let me describe next the CYK

algorithm (named after the authors Cocke–Younger–Kasami). This algorithm

works with grammars that are in \emph{Chomsky normalform}. In Chomsky

normalform all rules must be of the form $\meta{A} ::= a$, where $a$ is 

a terminal, or $\meta{A} ::= \meta{B}\cdot \meta{C}$, where $\meta{B}$ and

$\meta{B}$ need to be non-terminals. And no rule can contain $\epsilon$.

The following grammar is in Chomsky normalform:

\begin{plstx}[margin=1cm]

  : \meta{S} ::= \meta{N}\cdot \meta{P}\\

  : \meta{P} ::= \meta{V}\cdot \meta{N}\\

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

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

  : \meta{N} ::= \texttt{student} | \texttt{trainer} | \texttt{team} 

                | \texttt{trains}\\