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\section*{Homework 6}

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\begin{enumerate}

\item (i) Give the regular expressions for lexing a language

      consisting of whitespaces, identifiers (some letters

      followed by digits), numbers, operations \texttt{=},

      \texttt{<} and \texttt{>}, and the keywords \texttt{if},

      \texttt{then} and \texttt{else}. (ii) Decide whether the

      following strings can be lexed in this language?

\begin{enumerate}

\item \texttt{"if y4 = 3 then 1 else 3"}

\item \texttt{"if33 ifif then then23 else else 32"}

\item \texttt{"if x4x < 33 then 1 else 3"}

\end{enumerate}

In case they can, give the corresponding token sequences. (Hint: 

Observe the maximal munch rule and priorities of your regular

expressions that make the process of lexing unambiguous.)

\solution{

  a and b are ok; c is not matched because x4x is not an identifier according to the rules.

}

\item Suppose the grammar

\begin{plstx}[margin=1cm]    

:\meta{E} ::= \meta{F} \;|\; \meta{F} \cdot \texttt{*} \cdot \meta{F} \;|\; \meta{F} \cdot \backslash \cdot \meta{F}\\

:\meta{F} ::= \meta{T} \;|\; \meta{T} \cdot \texttt{+} \cdot \meta{T} \;|\; \meta{T} \cdot \texttt{-} \cdot \meta{T}\\

:\meta{T} ::= $num$ \;|\; \texttt{(} \cdot E \cdot \texttt{)}\\

\end{plstx}

where \meta{E}, \meta{F} and \meta{T} are non-terminals, \meta{E} is

the starting symbol of the grammar, and $num$ stands for a number

token. Give a parse tree for the string \texttt{(3+3)+(2*3)}. Note that

\meta{F} and \meta{T} are ``exchanged'' in this grammar in comparison with

the usual grammar for arithmetic expressions. What does this mean in terms

of precedences of the operators?

\solution{

  \begin{tikzpicture}[level distance=10mm, black,sibling distance=10mm]

  \node {\meta{E}}

  child[sibling distance=20mm] {node {\meta{F}} 

    child {node {\meta{T}}

      child[sibling distance=5mm] {node {(}}

      child[sibling distance=5mm]  {node {\meta{E}}

        child {node {\meta{F}}

          child {node {\meta{T}} child {node {3}}}

          child {node {+}}

          child {node {\meta{T}} child {node {3}}}

        }  

      }

      child[sibling distance=5mm]  {node {)}}

    }

    child {node {+}}

    child {node {\meta{T}}

      child[sibling distance=5mm] {node {(}}

      child[sibling distance=5mm] {node {\meta{E}}

        child {node {\meta{F}}

          child {node {\meta{T}} child {node {2}}}

        }

        child {node {*}}

        child {node {\meta{F}}

          child {node {\meta{T}} child {node {3}}}

        }

        }

      child[sibling distance=5mm] {node {)}}

    }

    };

\end{tikzpicture}

  For the second part "1+2*3" will be parsed as (1+2)*3, meaning + and - bind

  tighter than * and $\backslash$

}

\item Define what it means for a grammar to be ambiguous. Give an example of

an ambiguous grammar.

\solution{

  Already the grammar

  \begin{plstx}[margin=1cm]

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

  \end{plstx}

  is ambiguous because a string like "1+2+3" can be parsed

  as (1+2)+3 or 1+(2+3).

  }

\item Suppose boolean expressions are built up from

\begin{center}

\begin{tabular}{ll}

1.) & tokens for \texttt{true} and \texttt{false},\\

2.) & the infix operations \texttt{$\wedge$} and \texttt{$\vee$},\\

3.) & the prefix operation $\neg$, and\\

4.) & can be enclosed in parentheses.

\end{tabular}

\end{center}

(i) Give a grammar that can recognise such boolean expressions

and (ii) give a sample string involving all rules given in 1.-4.~that 

can be parsed by this grammar.

\solution{

The simplest grammar is

\begin{plstx}[margin=1cm]    

  :\meta{B} ::= true \;|\; false \;|\;\meta{B} \cdot $\wedge$ \cdot \meta{B} \;|\; \meta{B} \cdot $\vee$ \cdot \meta{B}

  \;|\; $\neg$\meta{B} \;|\; (\cdot\meta{B}\cdot)\\

\end{plstx}

This is unfortunately a left-recursive grammar. Giving a not left-recursive one is a bit more involved.

  }

\item Parsing combinators consist of atomic parsers, alternative

  parsers, sequence parsers and semantic actions.  What is the purpose

  of (1) atomic parsers and of (2) map-parsers (also called semantic actions)?

  \solution{

    Atomic parsers look at a concrete prefix of the input, like num-tokens or identifiers.

    Map-parsers apply a function to the output of a parser. In this way you can transform

    the output from, for example, a string to an integer.

  }

\item Parser combinators can directly be given a string as

      input, without the need of a lexer. What are the

      advantages to first lex a string and then feed a

      sequence of tokens as input to the parser?

      \solution{ Reason 1 you can filter out whitespaces and comments,

        which makes the grammar rules simpler. If you have to make

        sure that a whitespace comes after a variable say, then your

        parser rule for variables gets more complicated. Same with

        comments which do not contribute anything to the parse tree.

        Reason 2) The lexer can already classify tokens, for example

        as numbers, keywords or identifiers. This again makes the grammar

        rules more deterministic and as a result faster to parse.

        The point is that parser combinators can be used to process

        strings, but in case of compilers where whitespaces and

        comments need to be filtered out, the lexing phase is actually

        quite useful.  }

\item The injection function for sequence regular expressions is defined

      by three clauses:

\begin{center}

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

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

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

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

\end{tabular}

\end{center}

     Explain why there are three cases in the injection function for sequence

     regular expressions.

     \solution{

       This is because the derivative of sequences can be of the form

       \begin{itemize}

       \item $(der\,c\,r_1)\cdot r_2$

       \item $(der\,c\,r_1)\cdot r_2 \;+\; der\,c\,r_2$  

       \end{itemize}

       In the first case the value needs to be of the form $Seq$, in the second case $\Left$ or $Right$.

       Therefore 3 cases.

       }

\item \POSTSCRIPT        

\end{enumerate}

\end{document}

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