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\begin{document}
\section*{Handout 6}
While regular expressions are very useful for lexing and for recognising
many patterns in strings (like email addresses), they have their limitations. For
example there is no regular expression that can recognise the language
$a^nb^n$. Another example for which there exists no regular expression is the language of well-parenthesised
expressions. In languages like Lisp, which use parentheses rather
extensively, it might be of interest whether the following two expressions
are well-parenthesised (the left one is, the right one is not):
\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 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] {all languages};
\draw (0,-0.4) node [rect, text depth=20mm, text width=44mm] {decidable languages};
\draw (0,-0.65) node [rect, text depth=13mm] {context sensitive languages};
\draw (0,-0.84) node [rect, text depth=7mm, text width=35mm] {context-free languages};
\draw (0,-1.05) node [rect] {regular languages};
\end{tikzpicture}
\end{center}
\noindent
Context-free languages play an important role in `day-to-day' text processing and in
programming languages. Context-free languages are usually specified by grammars.
For example a grammar for well-parenthesised expressions is
\begin{center}
$P \;\;\rightarrow\;\; ( \cdot P \cdot ) \cdot P \;|\; \epsilon$
\end{center}
\noindent
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). Rules
have the shape
\begin{center}
$NT \;\;\rightarrow\;\; \textit{rhs}$
\end{center}
\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{center}
$NT \;\;\rightarrow\;\; \textit{rhs}_1 \;|\; \textit{rhs}_2$
\end{center}
\noindent
means that the non-terminal $NT$ 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{center}
\begin{tabular}{lcl}
$E$ & $\rightarrow$ & $N$ \\
$E$ & $\rightarrow$ & $E \cdot + \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot - \cdot E$ \\
$E$ & $\rightarrow$ & $E \cdot * \cdot E$ \\
$E$ & $\rightarrow$ & $( \cdot E \cdot )$\\
$N$ & $\rightarrow$ & $N \cdot N \;|\; 0 \;|\; 1 \;|\: \ldots \;|\; 9$
\end{tabular}
\end{center}
\noindent
where $E$ is the starting symbol. A \emph{derivation} for a grammar
starts with the staring 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}
$E$ & $\rightarrow$ & $E+E$\\
& $\rightarrow$ & $(E)+E$\\
& $\rightarrow$ & $(E+E)+E$\\
& $\rightarrow$ & $(E+E)+N$\\
& $\rightarrow$ & $(E+E)+3$\\
& $\rightarrow$ & $(N+E)+3$\\
& $\rightarrow^+$ & $(1+2)+3$\\
\end{tabular}
\end{center}
\noindent
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 {$E$}
child {node {$E$}
child {node {$($}}
child {node {$E$}
child {node {$E$} child {node {$N$} child {node {$1$}}}}
child {node {$+$}}
child {node {$E$}
child {node {$N$} child {node {$2$}}}
child {node {$N$} child {node {$3$}}}
}
}
child {node {$)$}}
}
child {node {$+$}}
child {node {$E$}
child {node {$N$} 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
$NT$ which leads to a string which again starts with $NT$. This means a derivation of the
form.
\begin{center}
$NT \rightarrow \ldots \rightarrow NT \cdot \ldots$
\end{center}
\noindent
It can be easily seen that the grammar above for arithmetic expressions is left-recursive:
for example the rules $E \rightarrow E\cdot + \cdot E$ and $N \rightarrow N\cdot N$
show that this grammar is left-recursive. Some algorithms cannot cope with left-recursive
grammars. 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}
$N \;\;\rightarrow\;\; 0\;|\;\ldots\;|\;9\;|\;1\cdot N\;|\;2\cdot N\;|\;\ldots\;|\;9\cdot N$
\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 $\ldots \rightarrow N \cdot \ldots$.
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 {$E$}
child {node {$E$} child {node {$N$} child {node {$1$}}}}
child {node {$+$}}
child {node {$E$}
child {node {$E$} child {node {$N$} child {node {$2$}}}}
child {node {$+$}}
child {node {$E$} child {node {$N$} child {node {$3$}}}}
}
;
\end{tikzpicture}
&
\begin{tikzpicture}[level distance=8mm, black]
\node {$E$}
child {node {$E$}
child {node {$E$} child {node {$N$} child {node {$1$}}}}
child {node {$+$}}
child {node {$E$} child {node {$N$} child {node {$2$}}}}
}
child {node {$+$}}
child {node {$E$} child {node {$N$} 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.
Let us now turn to the problem of generating a parse-tree for a grammar and string.
In what follows we explain \emph{parser combinators}, because they are easy
to implement and closely resemble grammar rules. Imagine that a grammar
describes the strings of natural numbers, such as the grammar $N$ shown above.
For all such strings we want to generate the parse-trees or later on we actually
want to extract the meaning of these strings, that is the concrete integers ``behind''
these strings. In Scala the parser combinators will be functions of type
\begin{center}
\texttt{I $\Rightarrow$ Set[(T, I)]}
\end{center}
\noindent
that is they take as input something of type \texttt{I}, typically a list of tokens or a string,
and return a set of pairs. The first component of these pairs corresponds to what the
parser combinator was able to process from the input and the second is the unprocessed
part of the input. As we shall see shortly, a parser combinator might return more than one such pair,
with the idea that there are potentially several ways how to interpret the input. As a concrete
example, consider the case where the input is of type string, say the string
\begin{center}
\tt\Grid{iffoo\VS testbar}
\end{center}
\noindent
We might have a parser combinator which tries to interpret this string as a keyword (\texttt{if}) or
an identifier (\texttt{iffoo}). Then the output will be the set
\begin{center}
$\left\{ \left(\texttt{\Grid{if}}\,,\, \texttt{\Grid{foo\VS testbar}}\right),
\left(\texttt{\Grid{iffoo}}\,,\, \texttt{\Grid{\VS testbar}}\right) \right\}$
\end{center}
\noindent
where the first pair means the parser could recognise \texttt{if} from the input and leaves
the rest as `unprocessed' as the second component of the pair; in the other case
it could recognise \texttt{iffoo} and leaves \texttt{\VS testbar} as unprocessed. If the parser
cannot recognise anything from the input then parser combinators just return the empty
set $\varnothing$. This will indicate something ``went wrong''.
The main attraction is that we can easily build parser combinators out of smaller components
following very closely the structure of a grammar. In order to implement this in an object
oriented programming language, like Scala, we need to specify an abstract class for parser
combinators. This abstract class requires the implementation of the function
\texttt{parse} taking an argument of type \texttt{I} and returns a set of type
\mbox{\texttt{Set[(T, I)]}}.
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
abstract class Parser[I, T] {
def parse(ts: I): Set[(T, I)]
def parse_all(ts: I): Set[T] =
for ((head, tail) <- parse(ts); if (tail.isEmpty))
yield head
}
\end{lstlisting}
\end{center}
\noindent
From the function \texttt{parse} we can then ``centrally'' derive the function \texttt{parse\_all},
which just filters out all pairs whose second component is not empty (that is has still some
unprocessed part). The reason is that at the end of parsing we are only interested in the
results where all the input has been consumed and no unprocessed part is left.
One of the simplest parser combinators recognises just a character, say $c$,
from the beginning of strings. Its behaviour is as follows:
\begin{itemize}
\item if the head of the input string starts with a $c$, it returns
the set $\{(c, \textit{tail of}\; s)\}$
\item otherwise it returns the empty set $\varnothing$
\end{itemize}
\noindent
The input type of this simple parser combinator for characters is
\texttt{String} and the output type \mbox{\texttt{Set[(Char, String)]}}.
The code in Scala is as follows:
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
case class CharParser(c: Char) extends Parser[String, Char] {
def parse(sb: String) =
if (sb.head == c) Set((c, sb.tail)) else Set()
}
\end{lstlisting}
\end{center}
\noindent
The \texttt{parse} function tests whether the first character of the
input string \texttt{sb} is equal to \texttt{c}. If yes, then it splits the
string into the recognised part \texttt{c} and the unprocessed part
\texttt{sb.tail}. In case \texttt{sb} does not start with \texttt{c} then
the parser returns the empty set (in Scala \texttt{Set()}).
More interesting are the parser combinators that build larger parsers
out of smaller component parsers. For example the alternative
parser combinator is as follows.
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
class AltParser[I, T]
(p: => Parser[I, T],
q: => Parser[I, T]) extends Parser[I, T] {
def parse(sb: I) = p.parse(sb) ++ q.parse(sb)
}
\end{lstlisting}
\end{center}
\noindent
The types of this parser combinator are polymorphic (we just have \texttt{I}
for the input type, and \texttt{T} for the output type). The alternative parser
builds a new parser out of two existing parser combinator \texttt{p} and \texttt{q}.
Both need to be able to process input of type \texttt{I} and return the same
output type \texttt{Set[(T, I)]}. (There is an interesting detail of Scala, namely the
\texttt{=>} in front of the types of \texttt{p} and \texttt{q}. They will prevent the
evaluation of the arguments before they are used. This is often called
\emph{lazy evaluation} of the arguments.) The alternative parser should run
the input with the first parser \texttt{p} (producing a set of outputs) and then
run the same input with \texttt{q}. The result should be then just the union
of both sets, which is the operation \texttt{++} in Scala.
This parser combinator already allows us to construct a parser that either
a character \texttt{a} or \texttt{b}, as
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
new AltParser(CharParser('a'), CharParser('b'))
\end{lstlisting}
\end{center}
\noindent
Scala allows us to introduce some more readable shorthand notation for this, like \texttt{'a' || 'b'}.
We can call this parser combinator with the strings
\begin{center}
\begin{tabular}{rcl}
input string & & output\medskip\\
\texttt{\Grid{ac}} & $\rightarrow$ & $\left\{(\texttt{\Grid{a}}, \texttt{\Grid{c}})\right\}$\\
\texttt{\Grid{bc}} & $\rightarrow$ & $\left\{(\texttt{\Grid{b}}, \texttt{\Grid{c}})\right\}$\\
\texttt{\Grid{cc}} & $\rightarrow$ & $\varnothing$
\end{tabular}
\end{center}
\noindent
We receive in the first two cases a successful output (that is a non-empty set).
A bit more interesting is the \emph{sequence parser combinator} implemented in
Scala as follows:
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
class SeqParser[I, T, S]
(p: => Parser[I, T],
q: => Parser[I, S]) extends Parser[I, (T, S)] {
def parse(sb: I) =
for ((head1, tail1) <- p.parse(sb);
(head2, tail2) <- q.parse(tail1))
yield ((head1, head2), tail2)
}
\end{lstlisting}
\end{center}
\noindent
This parser takes as input two parsers, \texttt{p} and \texttt{q}. It implements \texttt{parse}
as follows: let first run the parser \texttt{p} on the input producing a set of pairs (\texttt{head1}, \texttt{tail1}).
The \texttt{tail1} stands for the unprocessed parts left over by \texttt{p}.
Let \texttt{q} run on these unprocessed parts
producing again a set of pairs. The output of the sequence parser combinator is then a set
containing pairs where the first components are again pairs, namely what the first parser could parse
together with what the second parser could parse; the second component is the unprocessed
part left over after running the second parser \texttt{q}. Therefore the input type of
the sequence parser combinator is as usual \texttt{I}, but the output type is
\begin{center}
\texttt{Set[((T, S), I)]}
\end{center}
Scala allows us to provide some
shorthand notation for the sequence parser combinator. So we can write for
example \texttt{'a' $\sim$ 'b'}, which is the
parser combinator that first consumes the character \texttt{a} from a string and then \texttt{b}.
Calling this parser combinator with the strings
\begin{center}
\begin{tabular}{rcl}
input string & & output\medskip\\
\texttt{\Grid{abc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{a}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\
\texttt{\Grid{bac}} & $\rightarrow$ & $\varnothing$\\
\texttt{\Grid{ccc}} & $\rightarrow$ & $\varnothing$
\end{tabular}
\end{center}
\noindent
A slightly more complicated parser is \texttt{('a' || 'b') $\sim$ 'b'} which parses as first character either
an \texttt{a} or \texttt{b} followed by a \texttt{b}. This parser produces the following results.
\begin{center}
\begin{tabular}{rcl}
input string & & output\medskip\\
\texttt{\Grid{abc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{a}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\
\texttt{\Grid{bbc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{b}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\
\texttt{\Grid{aac}} & $\rightarrow$ & $\varnothing$
\end{tabular}
\end{center}
Note carefully that constructing the parser \texttt{'a' || ('a' $\sim$ 'b')} will result in a tying error.
The first parser has as output type a single character (recall the type of \texttt{CharParser}),
but the second parser produces a pair of characters as output. The alternative parser is however
required to have both component parsers to have the same type. We will see later how we can
build this parser without the typing error.
The next parser combinator does not actually combine smaller parsers, but applies
a function to the result of the parser. It is implemented in Scala as follows
\begin{center}
\begin{lstlisting}[language=Scala,basicstyle=\small\ttfamily, numbers=none]
class FunParser[I, T, S]
(p: => Parser[I, T],
f: T => S) extends Parser[I, S] {
def parse(sb: I) =
for ((head, tail) <- p.parse(sb)) yield (f(head), tail)
}
\end{lstlisting}
\end{center}
\noindent
This parser combinator takes a parser \texttt{p} with output type \texttt{T} as
input as well as a function \texttt{f} with type \texttt{T => S}. The parser \texttt{p}
produces sets of type \texttt{(T, I)}. The \texttt{FunParser} combinator then
applies the function \texttt{f} to all the parer outputs. Since this function
is of type \texttt{T => S}, we obtain a parser with output type \texttt{S}.
Again Scala lets us introduce some shorthand notation for this parser combinator.
Therefore we will write \texttt{p ==> f} for it.
%\bigskip
%takes advantage of the full generality---have a look
%what it produces if we call it with the string \texttt{abc}
%
%\begin{center}
%\begin{tabular}{rcl}
%input string & & output\medskip\\
%\texttt{\Grid{abc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{a}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\
%\texttt{\Grid{bbc}} & $\rightarrow$ & $\left\{((\texttt{\Grid{b}}, \texttt{\Grid{b}}), \texttt{\Grid{c}})\right\}$\\
%\texttt{\Grid{aac}} & $\rightarrow$ & $\varnothing$
%\end{tabular}
%\end{center}
\end{document}
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