Differences

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--- lfa:2024:lab11 [2025/01/08 23:56]
cata_chiru [11.2. Closed under CFLs]
+++ lfa:2024:lab11 [2026/01/18 17:55] (current)
cata_chiru
@@ Line 135: / Line 135: @@
 <hidden Solution 11.2.4>
+__**Intuition**__
 Reverse each production rule in the grammar for L, with $ L \in CFL $.
-</hidden>
+__**Formal solution - induction**__
-<hidden Debug visual Solution 11.2.4>
+If \( G \) is a grammar, let \( H \)  be its reverse, so for production   \( A \to w \) in \( G \) we have  \( A \to w^R \)  in \( H \).
-If \( G \) is a grammar, let \( H \) be its reverse, so for production \( A \to w \) in \( G \) we have \( A \to w^R \) in \( H \).
-Then by induction we show that the original grammar \( G \) generates a string iff the reverse grammar \( H \) generates the reverse of the string. Formally:
+Then by induction we show that the original grammar  \( G \) generates a string iff the reverse grammar \( H \) generates the reverse of the string. Formally:
-\( A \xRightarrow{*}_G w \iff A \Rightarrow{*}_H w^R \).
+ \( A \to^{*}_G w \iff A \to^{*}_H w^R \).
 . **Basis**:
-In zero steps, we have $ \( A \Rightarrow{0}_G A \iff A \Rightarrow{0}_H A \) $.
+In zero steps, we have \( A \to^{0}_G A \iff A \to^{0}_H A \).
 . **Induction**:
-Assuming $ \( A \Rightarrow{*}_G w_1 B w_2 \iff A \Rightarrow{*}_H w_2^R B w_1^R \) $,
+Assuming \( A \to^{*}_G w_1 B w_2 \iff A \to^{*}_H w_2^R B w_1^R \),
-we can apply any production $ \( B \to u \) $ in \( G \) (and in \( H \) in reverse) and obtain:
+we can apply any production \( B \to u \) in \( G \) (and in \( H \) in reverse) and obtain:
-$ \( A \rightarrow^{*}_G w_1 u w_2 \) $
+\( A \to^{*}_G w_1 u w_2 \)
+\( A \to^{*}_H w_2^R u^R w_1^R \),   where indeed \( w_2^R u^R w_1^R \) is the reverse of \( w_1 u w_2 \).
-$ \( A \rightarrow^{*}_H w_2^R u^R w_1^R \) $,   where indeed $ \( w_2^R u^R w_1^R \) $ is the reverse of $ \( w_1 u w_2 \) $.
 </hidden>
@@ Line 164: / Line 166: @@
 **11.2.5.** Intersection with a regular language is a closure property.
+<hidden Solution 11.2.5.>
+Let \( L_1 \) be a context-free language and \( P = (K_1, \Sigma, \Gamma, \Delta_1, q^{1}_{0}, F_1) \)
+be its respective PDA.
+Let \( L_2 \) be a regular language and \( A = (K_2, \Sigma, \delta, q_{20}, F_2) \) be its respective DFA.
+We build the following PDA \( (K, \Sigma, \Gamma, \Delta, q_0, F) \) where:
+- \( K = K_1 \times K_2 \)
+- \( q_0 = (q^{1}_{0}, q^{2}_{0}) \)
+Transition rules for \( \Delta \):
+. \( ((q_1, q_2), c, \alpha, (q_1', q_2'), \beta) \in \Delta \)
+   iff \( (q_1, c, \alpha, q_1', \beta) \in \Delta_1 \) and \( \delta(q_2, c) = q_2' \).
+. \( ((q_1, q_2), \epsilon, \alpha, (q_1', q_2), \beta) \in \Delta \)
+   iff \( (q_1, \epsilon, \alpha, q_1', \beta) \in \Delta_1 \).
+- \( F = F_1 \times F_2 \)
+The PDA accepts \( L(P) \cap L(A) \).
+</hidden>
 **11.2.6.** Difference with a regular language is a closure property.
 <hidden Solution 11.2.6.>
-Let \( A \) be a context-free language and \( B \) a regular language. $ \overline{B} $ is regular, as complement is a closure property for regular languages. From 11.2.5, we know that the intersection is closed between CFLs and Regular Languages, and write A \ B as A \cup $ \overline{B} $.
+Let \( L_1 \) be a context-free language and \( L_2 \) a regular language.
+$ \overline{L_2} $ is regular, as complement is a closure property for regular languages.
+From 11.2.5, we know that the intersection is closed between CFLs and Regular Languages, and write $ L_1 \setminus L_2 = L_1 \cap \overline{L_2} $.
 </hidden>