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diff --git a/exercices2.tex b/exercices2.tex index 10cd7d7..70d030b 100644 --- a/exercices2.tex +++ b/exercices2.tex @@ -87,6 +87,105 @@ Git: \input{vcline.tex} \tolerance=50000 + +% +% +% + +\exercice + +Considérons le fragment simplifié suivant de la grammaire d'un langage +de programmation hypothétique : +\[ +\begin{aligned} +\mathit{Instruction} &\rightarrow \mathtt{foo} \;|\; \mathtt{bar} \;|\; \mathtt{qux} \;|\; \mathit{Conditional}\\ +|&\phantom{\rightarrow} \mathtt{begin}\ \mathit{InsList}\ \mathtt{end}\\ +\mathit{Conditional} &\rightarrow \mathtt{if}\ \mathit{Expression}\ \mathtt{then}\ \mathit{Instruction}\\ +|&\phantom{\rightarrow} \mathtt{if}\ \mathit{Expression}\ \mathtt{then}\ \mathit{Instruction}\ \mathtt{else}\ \mathit{Instruction}\\ +\mathit{InsList} &\rightarrow \mathit{Instruction} \;|\; \mathit{Instruction}\ \mathit{InsList}\\ +\mathit{Expression} &\rightarrow \mathtt{true} \;|\; \mathtt{false} \;|\; \mathtt{happy} \;|\; \mathtt{trippy}\\ +\end{aligned} +\] +(Ici, les « lettres » ou tokens ont été écrits comme des mots, par +exemple $\mathtt{foo}$ est une « lettre » : les terminaux sont écrits +en police à espacement fixe tandis que les nonterminaux sont en +italique et commencent par une majuscule. On prendra +$\mathit{Instruction}$ pour axiome.) + +(1) Donner l'arbre d'analyse de : +$\mathtt{if}\penalty0\ \mathtt{happy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{if}\penalty0\ \mathtt{trippy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{foo}\penalty0\ \mathtt{else}\penalty0\ \mathtt{bar}\penalty0\ \mathtt{else}\penalty0\ \mathtt{qux}$ ; +expliquer brièvement pourquoi il n'y en a qu'un. + +(2) Donner deux arbres d'analyse distincts de : +$\mathtt{if}\penalty0\ \mathtt{happy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{if}\penalty0\ \mathtt{trippy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{foo}\penalty0\ \mathtt{else}\penalty0\ \mathtt{bar}$. +Que peut-on dire de la grammaire présentée ? + +\begin{corrige} +(1) L'arbre d'analyse de + $\mathtt{if}\penalty0\ \mathtt{happy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{if}\penalty0\ \mathtt{trippy}\penalty0\ \mathtt{then}\penalty0\ \mathtt{foo}\penalty0\ \mathtt{else}\penalty0\ \mathtt{bar}\penalty0\ \mathtt{else}\penalty0\ \mathtt{qux}$ + est le suivant (en notant $I$, $C$ et $E$ pour + $\mathit{Instruction}$, $\mathit{Condition}$ et + $\mathit{Expression}$ respectivement) : +\begin{center} +\tikzstyle{automaton}=[scale=0.5] +%%% begin ex2p1 %%% + +\begin{tikzpicture}[>=latex,line join=bevel,automaton] +%% +\node (foo) at (279bp,18bp) [draw,draw=none] {$\mathtt{foo}$}; + \node (then1) at (207bp,90bp) [draw,draw=none] {$\mathtt{then}$}; + \node (if0) at (27bp,234bp) [draw,draw=none] {$\mathtt{if}$}; + \node (if1) at (63bp,90bp) [draw,draw=none] {$\mathtt{if}$}; + \node (bar) at (423bp,18bp) [draw,draw=none] {$\mathtt{bar}$}; + \node (I1) at (243bp,234bp) [draw,draw=none] {$I$}; + \node (I0) at (207bp,378bp) [draw,draw=none] {$I$}; + \node (I3) at (423bp,90bp) [draw,draw=none] {$I$}; + \node (I2) at (279bp,90bp) [draw,draw=none] {$I$}; + \node (I4) at (387bp,234bp) [draw,draw=none] {$I$}; + \node (trippy) at (135bp,18bp) [draw,draw=none] {$\mathtt{trippy}$}; + \node (else0) at (315bp,234bp) [draw,draw=none] {$\mathtt{else}$}; + \node (else1) at (351bp,90bp) [draw,draw=none] {$\mathtt{else}$}; + \node (qux) at (387bp,162bp) [draw,draw=none] {$\mathtt{qux}$}; + \node (C1) at (243bp,162bp) [draw,draw=none] {$C$}; + \node (then0) at (171bp,234bp) [draw,draw=none] {$\mathtt{then}$}; + \node (C0) at (207bp,306bp) [draw,draw=none] {$C$}; + \node (E1) at (135bp,90bp) [draw,draw=none] {$E$}; + \node (E0) at (99bp,234bp) [draw,draw=none] {$E$}; + \node (happy) at (99bp,162bp) [draw,draw=none] {$\mathtt{happy}$}; + \draw [] (I2) ..controls (279bp,60.846bp) and (279bp,46.917bp) .. (foo); + \draw [] (C0) ..controls (192.52bp,276.85bp) and (185.36bp,262.92bp) .. (then0); + \draw [] (I0) ..controls (207bp,348.85bp) and (207bp,334.92bp) .. (C0); + \draw [] (C0) ..controls (250.16bp,277.03bp) and (271.72bp,263.05bp) .. (else0); + \draw [] (C1) ..controls (286.16bp,133.03bp) and (307.72bp,119.05bp) .. (else1); + \draw [] (C1) ..controls (257.48bp,132.85bp) and (264.64bp,118.92bp) .. (I2); + \draw [] (E0) ..controls (99bp,204.85bp) and (99bp,190.92bp) .. (happy); + \draw [] (C1) ..controls (199.84bp,133.03bp) and (178.28bp,119.05bp) .. (E1); + \draw [] (C1) ..controls (228.52bp,132.85bp) and (221.36bp,118.92bp) .. (then1); + \draw [] (C0) ..controls (263.23bp,285.79bp) and (310.83bp,268.84bp) .. (351bp,252bp) .. controls (353.93bp,250.77bp) and (356.97bp,249.43bp) .. (I4); + \draw [] (C1) ..controls (186.77bp,141.79bp) and (139.17bp,124.84bp) .. (99bp,108bp) .. controls (96.068bp,106.77bp) and (93.027bp,105.43bp) .. (if1); + \draw [] (C0) ..controls (150.77bp,285.79bp) and (103.17bp,268.84bp) .. (63bp,252bp) .. controls (60.068bp,250.77bp) and (57.027bp,249.43bp) .. (if0); + \draw [] (E1) ..controls (135bp,60.846bp) and (135bp,46.917bp) .. (trippy); + \draw [] (C1) ..controls (299.23bp,141.79bp) and (346.83bp,124.84bp) .. (387bp,108bp) .. controls (389.93bp,106.77bp) and (392.97bp,105.43bp) .. (I3); + \draw [] (C0) ..controls (221.48bp,276.85bp) and (228.64bp,262.92bp) .. (I1); + \draw [] (I4) ..controls (387bp,204.85bp) and (387bp,190.92bp) .. (qux); + \draw [] (I1) ..controls (243bp,204.85bp) and (243bp,190.92bp) .. (C1); + \draw [] (I3) ..controls (423bp,60.846bp) and (423bp,46.917bp) .. (bar); + \draw [] (C0) ..controls (163.84bp,277.03bp) and (142.28bp,263.05bp) .. (E0); +% +\end{tikzpicture} + +%%% end ex2p1 %%% +\end{center} + +Il est le seul possible car une fois acquis que les deux $\mathtt{if}$ +comportent chacun un $\mathtt{else}$, il se construit ensuite en +descendant de façon unique (l'instruction est forcément une condition, +qui s'analyse en $\mathtt{if}\ E\ \mathtt{then}\ I\ \mathtt{else}\ I$ +de façon unique, et chacun des morceaux s'analyse de nouveau de façon +unique). +\end{corrige} + + % % % @@ -186,6 +285,10 @@ $x_1\cdots x_{2i-1}$ et $x_{2i}\cdots x_{2n}$.) \end{corrige} +% +% +% + \exercice\label{square-words-not-algebraic} Soit $\Sigma = \{a,b\}$. Montrer que le langage $Q := \{ww : |