Differences

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--- lfa:2024:lab11 [2024/12/16 21:43]
cata_chiru [11.2. Closed under CFLs]
+++ lfa:2024:lab11 [2025/01/09 00:14] (current)
cata_chiru [11.2. Closed under CFLs]
@@ Line 97: / Line 97: @@
 **11.2.1.** Concat is a closure property.
+<hidden Solution 11.2.1.>
+Let \( A \) and \( B \) be context-free languages over an alphabet \( \Sigma \), and let \( G_A = (V_A, \Sigma, R_A, S_A) \) and \( G_B = (V_B, \Sigma, R_B, S_B) \) be context-free grammars that generate \( A \) and \( B \) respectively. By renaming the variables if necessary, we assume that \( V_A \) is disjoint from \( V_B \), and that neither variable set contains the variable \( S \). We now construct a new CFG by writing:
+\[
+G = (V_A \cup V_B \cup \{S\}, \Sigma, R_A \cup R_B \cup \{S \to S_AS_B\}, S)
+\]
+</hidden>
 **11.2.2.** Union is a closure property.
+<hidden Solution 11.2.2.>
+Let \( A \) and \( B \) be context-free languages over an alphabet \( \Sigma \), and let \( G_A = (V_A, \Sigma, R_A, S_A) \) and \( G_B = (V_B, \Sigma, R_B, S_B) \) be context-free grammars that generate \( A \) and \( B \) respectively. By renaming the variables if necessary, we assume that \( V_A \) is disjoint from \( V_B \), and that neither variable set contains the variable \( S \). We now construct a new CFG by writing:
+\[
+G = (V_A \cup V_B \cup \{S\}, \Sigma, R_A \cup R_B \cup \{S \to S_A | S_B\}, S)
+\]
+</hidden>
 **11.2.3.** Kleene Star is a closure property.
+<hidden Solution 11.2.3.>
+Let \( A \) be a context-free languages over an alphabet \( \Sigma \), and let \( G_A = (V_A, \Sigma, R_A, S_A) \) be the context-free grammar that generate \( A \). We now construct a new CFG by writing:
+\[
+G = (V_A \cup \{S\}, \Sigma, R_A \cup \{S \to SS | S_A | \epsilon\}, S)
+\]
+</hidden>
 **11.2.4.** Reverse is a closure property.
 <hidden Solution 11.2.4>
+Reverse each production rule in the grammar for L, with $ L \in CFL $.
+</hidden>
+<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 \( H \).
 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:
-\( A \Rightarrow{*}_G w \iff A \Rightarrow{*}_H w^R \).
+\( A \xRightarrow{*}_G w \iff A \Rightarrow{*}_H w^R \).
 . **Basis**:
-   In zero steps, we have \( A \Rightarrow{0}_G A \iff A \Rightarrow{0}_H A \).
+In zero steps, we have $ \( A \Rightarrow{0}_G A \iff A \Rightarrow{0}_H A \) $.
 . **Induction**:
-   Assuming \( A \Rightarrow{*}_G w_1 B w_2 \iff A \Rightarrow{*}_H w_2^R B w_1^R \),
+Assuming $ \( A \Rightarrow{*}_G w_1 B w_2 \iff A \Rightarrow{*}_H w_2^R B w_1^R \) $,
-   we can apply any production \( B \to u \) in \( G \) (and in \( H \) in reverse) and obtain:
-   \( A \Rightarrow{*}_G w_1 u w_2 \)
-   \( A \Rightarrow{*}_H w_2^R u^R w_1^R \),
-   where indeed \( w_2^R u^R w_1^R \) is the reverse of \( w_1 u w_2 \).
+we can apply any production $ \( B \to u \) $ in \( G \) (and in \( H \) in reverse) and obtain:
+$ \( A \rightarrow^{*}_G w_1 u w_2 \) $
+$ \( A \rightarrow^{*}_H w_2^R u^R w_1^R \) $,   where indeed $ \( w_2^R u^R w_1^R \) $ is the reverse of $ \( w_1 u w_2 \) $.
 </hidden>
 **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 \( L_1 \) 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>