11. Closure properties for Context-Free Languages

11. Closure properties for Context-Free Languages

11.0. Lexer Spec

Given the following specs, construct the lexer DFA as presented in Lecture 14:

PAIRS: $ (10 | 01)* $
ONES: $ 1+ $
NO_CONSEC_ONE: $ (1 | \epsilon)(01 | 0)* $

Separate the following input strings into lexemes:

a) 010101

Solution a)

Although the entire string is matched by PAIRS and NO_CONSEC_ONE, PAIRS is defined first, thus it will be the first picked.
PAIRS “010101”

b) 1010101011

Solution b)

First we have a maximal match on “101010101” for regex NO_CONSEC_ONE. The remaining string, “1”, is matched by both ONES and NO_CONSEC_ONE, but ONES is defined first.
NO_CONSEC_ONE “101010101”
ONES “1”

c) 01110101001

Solution c)

PAIRS “01”
ONES “11”
NO_CONSEC_ONES “0101001”

d) 01010111111001010

Solution d)

PAIRS “010101”
ONES “11111”
NO_CONSEC_ONES “001010”

e) 1101101001111100001010011001

Solution e)

ONES “11” PAIRS “01101001”
ONES “1111”
NO_CONSEC_ONES “0000101001”
PAIRS “1001”

11.1. Not Closed under CFLs

11.1.1. Intersection is not a closure property.

Solution 11.1.1

Take:

$ L_1 = \{ a^{n} b^{n} c^{m} \ | \: n, m \in \mathbb{N} \} $

$ L_2 = \{ a^{m} b^{n} c^{n} \ | \: n, m \in \mathbb{N} \} $

$ \Rightarrow L_1 \bigcap L_2 = \{ a^{n} b^{n} c^{n} \ | \: n \in \mathbb{N} \} $ is not context free

11.1.2. Complement is not a closure property.

Solution 11.1.2

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.

Take, $ L_1, L_2 \in CFL \Rightarrow \overline{\overline{L_1} \bigcup \overline{L_2}} = L_1 \bigcap L_2 \in CFL $

This would imply $ L_1 \bigcap L_2 $ is a CFL, and at 11.1.1 we have proven that intersection is not a closure property.

Contradiction!

11.1.3. Difference is not a closure property.

Solution 11.1.3

Suppose Difference is a closure property.

Take, $ \Sigma^*, L_1 \in CFL \Rightarrow \Sigma^* \backslash L_1 = \overline{L_1}\in CFL $

This would imply $ \overline{L_1} $ is a CFL, and at 11.2.1 we have proven that intersection is not a closure property.

Contradiction!

11.2. Closed under CFLs

11.2.1. Concat is a closure property.

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:

\[ G = (V_A \cup V_B \cup \{S\}, \Sigma, R_A \cup R_B \cup \{S \to S_AS_B\}, S) \]

11.2.2. Union is a closure property.

Solution 11.2.2.

\[ G = (V_A \cup V_B \cup \{S\}, \Sigma, R_A \cup R_B \cup \{S \to S_A | S_B\}, S) \]

11.2.3. Kleene Star is a closure property.

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:

\[ G = (V_A \cup \{S\}, \Sigma, R_A \cup \{S \to SS | S_A | \epsilon\}, S) \]

11.2.4. Reverse is a closure property.

Solution 11.2.4

Reverse each production rule in the grammar for L, with $ L \in CFL $.

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 \xRightarrow{*}_G w \iff A \Rightarrow{*}_H w^R $.

1. Basis:

In zero steps, we have $ $ A \Rightarrow{0}_G A \iff A \Rightarrow{0}_H A $ $.

2. Induction: 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 $ $.

11.2.5. Intersection with a regular language is a closure property.

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) $.

11.2.6. Difference with a regular language is a closure property.

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} $.

Table of Contents

11. Closure properties for Context-Free Languages

11.0. Lexer Spec

11.1. Not Closed under CFLs

11.2. Closed under CFLs