Differences
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lfa:2025:lab10 [2026/01/18 17:57] cata_chiru |
lfa:2025:lab10 [2026/01/20 11:49] (current) cata_chiru |
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| - | ====== 11. Closure properties for Context-Free Languages ====== | + | ====== 10. Closure properties for Context-Free Languages ====== |
| - | Analogy with Subtraction between $ \mathbb{Z} $ and $ \mathbb{N} $. | + | Analogy with Subtraction, Division between $ \mathbb{Z} $ and $ \mathbb{N} $. |
| - | ===== 11.0. Lexer Spec ===== | + | ===== 10.0. Lexer Spec ===== |
| Given the following specs, construct the lexer DFA as presented in Lecture 14: | Given the following specs, construct the lexer DFA as presented in Lecture 14: | ||
| * PAIRS: $ (10 | 01)* $ | * PAIRS: $ (10 | 01)* $ | ||
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| - | ===== 11.1. Not Closed under CFLs ===== | + | ===== 10.1. Not Closed under CFLs ===== |
| - | **11.1.1.** Intersection is not a closure property. | + | **10.1.1.** Intersection is not a closure property. |
| <hidden Solution 11.1.1 > | <hidden Solution 11.1.1 > | ||
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| </hidden> | </hidden> | ||
| - | **11.1.2.** Complement is not a closure property. | + | **10.1.2.** Complement is not a closure property. |
| - | <hidden Solution 11.1.2 > | + | <hidden Solution 10.1.2 > |
| Suppose Complement is a closure property $\Rightarrow$ complements of CFLs are CFLs, we will prove in 11.2.2 that Union is a closure property. | Suppose Complement is a closure property $\Rightarrow$ complements of CFLs are CFLs, we will prove in 11.2.2 that Union is a closure property. | ||
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| </hidden> | </hidden> | ||
| - | **11.1.3.** Difference is not a closure property. | + | **10.1.3.** Difference is not a closure property. |
| - | <hidden Solution 11.1.3 > | + | <hidden Solution 10.1.3 > |
| Suppose Difference is a closure property. | Suppose Difference is a closure property. | ||
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| - | ===== 11.2. Closed under CFLs ===== | + | ===== 10.2. Closed under CFLs ===== |
| - | **11.2.1.** Concat is a closure property. | + | **10.2.1.** Concat is a closure property. |
| - | <hidden Solution 11.2.1.> | + | <hidden Solution 10.2.1.> |
| 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: | 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: | ||
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| - | **11.2.2.** Union is a closure property. | + | **10.2.2.** Union is a closure property. |
| - | <hidden Solution 11.2.2.> | + | <hidden Solution 10.2.2.> |
| 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: | 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: | ||
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| </hidden> | </hidden> | ||
| - | **11.2.3.** Kleene Star is a closure property. | + | **10.2.3.** Kleene Star is a closure property. |
| - | <hidden Solution 11.2.3.> | + | <hidden Solution 10.2.3.> |
| 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: | 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: | ||
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| - | **11.2.4.** Reverse is a closure property. | + | **10.2.4.** Reverse is a closure property. |
| - | <hidden Solution 11.2.4> | + | <hidden Solution 10.2.4> |
| __**Intuition**__ | __**Intuition**__ | ||
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| </hidden> | </hidden> | ||
| - | **11.2.5.** Intersection with a regular language is a closure property. | + | **10.2.5.** Intersection with a regular language is a closure property. |
| - | <hidden Solution 11.2.5.> | + | <hidden Solution 10.2.5.> |
| Let \( L_1 \) be a context-free language and \( P = (K_1, \Sigma, \Gamma, \Delta_1, q^{1}_{0}, F_1) \) | 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. | be its respective PDA. | ||
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| </hidden> | </hidden> | ||
| - | **11.2.6.** Difference with a regular language is a closure property. | + | **10.2.6.** Difference with a regular language is a closure property. |
| - | <hidden Solution 11.2.6.> | + | <hidden Solution 10.2.6.> |
| Let \( L_1 \) be a context-free language and \( L_2 \) a regular language. | 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. | $ \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} $. | + | From 10.2.5, 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> | </hidden> | ||