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lfa:2024:lab11 [2025/01/08 23:46] cata_chiru [11.2. Closed under CFLs] |
lfa:2024:lab11 [2025/01/09 00:14] (current) cata_chiru [11.2. Closed under CFLs] |
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**11.2.5.** Intersection with a regular language is a closure property. | **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 \( \Delta \): | ||
+ | |||
+ | 1. \( ((q_1, q_2), c, \alpha, (q_1', q_2'), \beta) \in \Delta \) | ||
+ | iff \( (q_1, c, \alpha, q_1', \beta) \in \Delta_1 \) and \( \delta(q_2, c) = q_2' \). | ||
+ | |||
+ | 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. | **11.2.6.** Difference with a regular language is a closure property. | ||
<hidden Solution 11.2.6.> | <hidden Solution 11.2.6.> | ||
- | Let \( A \) be a context-free language and \( B \) a regular language. \hat{B} is regular, as complement is closed under regular languages. From 11.2.5 we know that the intersection is closed betwwen CFLs and Regular Languages, and write A \ B as A \cup \hat{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> |