author | Christian Urban <christian dot urban at kcl dot ac dot uk> |
Sat, 29 Oct 2016 21:47:22 +0100 | |
changeset 468 | bab72d4ac90e |
parent 459 | 780486571e38 |
child 525 | a2ee4b11c976 |
permissions | -rw-r--r-- |
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\documentclass{article} |
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\usepackage{../style} |
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\usepackage{../graphics} |
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\usepackage{../langs} |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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\begin{document} |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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\section*{Homework 9} |
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\HEADER |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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\begin{enumerate} |
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\item Describe what is meant by \emph{eliminating tail |
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recursion}? When can this optimization be applied and |
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why is it of benefit? |
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\item A programming language has arithmetic expression. For an |
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arithmetic expression the compiler of this language produces the |
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following snippet of JVM code. |
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\begin{lstlisting}[language=JVMIS,numbers=none] |
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ldc 1 |
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ldc 2 |
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ldc 3 |
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imul |
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ldc 4 |
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ldc 3 |
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isub |
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iadd |
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iadd |
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\end{lstlisting} |
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Give the arithmetic expression that produced this code. Make sure |
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you give all necessary parentheses. |
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\item Describe what the following two JVM instructions do! |
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\begin{lstlisting}[language=JVMIS,numbers=none] |
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iload 3 |
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istore 1 |
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\end{lstlisting} |
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\item \POSTSCRIPT |
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% \item It is true (I confirmed it) that |
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% |
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% \begin{center} if $\varnothing$ does not occur in $r$ |
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% \;\;then\;\;$L(r) \not= \{\}$ |
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% \end{center} |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \noindent |
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% holds, or equivalently |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% |
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% \begin{center} |
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% $L(r) = \{\}$ \;\;implies\;\; $\varnothing$ occurs in $r$. |
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% \end{center} |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \noindent |
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% You can prove either version by induction on $r$. The best way to |
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% make more formal what is meant by `$\varnothing$ occurs in $r$', you can define |
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% the following function: |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \begin{center} |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}} |
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% $occurs(\varnothing)$ & $\dn$ & $true$\\ |
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% $occurs(\epsilon)$ & $\dn$ & $f\!alse$\\ |
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% $occurs (c)$ & $\dn$ & $f\!alse$\\ |
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% $occurs (r_1 + r_2)$ & $\dn$ & $occurs(r_1) \vee occurs(r_2)$\\ |
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% $occurs (r_1 \cdot r_2)$ & $\dn$ & $occurs(r_1) \vee occurs(r_2)$\\ |
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% $occurs (r^*)$ & $\dn$ & $occurs(r)$ \\ |
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% \end{tabular} |
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% \end{center} |
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% |
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% \noindent |
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% Now you can prove |
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% |
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% \begin{center} |
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% $L(r) = \{\}$ \;\;implies\;\; $occurs(r)$. |
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% \end{center} |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \noindent |
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% The interesting cases are $r_1 + r_2$ and $r^*$. |
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% The other direction is not true, that is if $occurs(r)$ then $L(r) = \{\}$. A counter example |
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% is $\varnothing + a$: although $\varnothing$ occurs in this regular expression, the corresponding |
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% language is not empty. The obvious extension to include the not-regular expression, $\sim r$, |
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% also leads to an incorrect statement. Suppose we add the clause |
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% |
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% \begin{center} |
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% \begin{tabular}{@ {}l@ {\hspace{2mm}}c@ {\hspace{2mm}}l@ {}} |
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% $occurs(\sim r)$ & $\dn$ & $occurs(r)$ |
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% \end{tabular} |
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% \end{center} |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \noindent |
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% to the definition above, then it will not be true that |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \begin{center} |
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% $L(r) = \{\}$ \;\;implies\;\; $occurs(r)$. |
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% \end{center} |
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% |
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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% \noindent |
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% Assume the alphabet contains just $a$ and $b$, find a counter example to this |
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% property. |
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|
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Christian Urban <christian dot urban at kcl dot ac dot uk>
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\end{enumerate} |
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|
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\end{document} |
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%%% Local Variables: |
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%%% mode: latex |
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%%% TeX-master: t |
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%%% End: |