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Question

Context free grammar is not closed under:

The correct answer is

Complementation

Let's explore the closure properties of Context-Free Grammars (CFGs). A class of languages is said to be closed under an operation if applying that operation to languages within the class always produces a language that is also within that class. Context-Free Languages (CFLs), which are generated by CFGs, have certain closure properties.

Understanding CFG Closure Properties

We need to determine which of the given operations does not guarantee that the resulting language is still a Context-Free Language when applied to one or more CFLs.

Closure Under Union

If $L_1$ and $L_2$ are two Context-Free Languages, their union $L_1 \cup L_2$ is also a Context-Free Language. This property holds true. If $G_1$ generates $L_1$ and $G_2$ generates $L_2$, we can construct a new CFG $G$ that generates $L_1 \cup L_2$. The new grammar combines the rules of $G_1$ and $G_2$ under a new start symbol.

  • Operation: Union ($\cup$)
  • Closed: Yes
  • Explanation: Given grammars $G_1=(V_1, \Sigma, P_1, S_1)$ and $G_2=(V_2, \Sigma, P_2, S_2)$ generating $L_1$ and $L_2$ respectively (assume $V_1 \cap V_2 = \emptyset$), we can form a new grammar $G=(V_1 \cup V_2 \cup \{S'\}, \Sigma, P_1 \cup P_2 \cup \{S' \to S_1, S' \to S_2\}, S')$. This new grammar generates $L_1 \cup L_2$.

Closure Under Concatenation

If $L_1$ and $L_2$ are two Context-Free Languages, their concatenation $L_1L_2$ is also a Context-Free Language. This property also holds. If $G_1$ generates $L_1$ and $G_2$ generates $L_2$, a new CFG can be constructed whose start symbol derives the start symbol of $G_1$ followed by the start symbol of $G_2$.

  • Operation: Concatenation
  • Closed: Yes
  • Explanation: Given grammars $G_1=(V_1, \Sigma, P_1, S_1)$ and $G_2=(V_2, \Sigma, P_2, S_2)$ generating $L_1$ and $L_2$ respectively ($V_1 \cap V_2 = \emptyset$), we can form a new grammar $G=(V_1 \cup V_2 \cup \{S'\}, \Sigma, P_1 \cup P_2 \cup \{S' \to S_1 S_2\}, S')$. This new grammar generates $L_1L_2$.

Closure Under Kleene Star

If $L$ is a Context-Free Language, its Kleene Star $L^*$ is also a Context-Free Language. This property holds. If $G$ generates $L$, a new CFG can be constructed to generate zero or more concatenations of strings from $L$.

  • Operation: Kleene Star ($*$)
  • Closed: Yes
  • Explanation: Given grammar $G=(V, \Sigma, P, S)$ generating $L$, we can form a new grammar $G'=(V \cup \{S'\}, \Sigma, P \cup \{S' \to S S', S' \to \epsilon\}, S')$. This new grammar generates $L^*$. (Some definitions use $\lambda$ instead of $\epsilon$ for the empty string).

Closure Under Complementation

If $L$ is a Context-Free Language, is its complement $\overline{L}$ (all strings over the alphabet $\Sigma$ that are *not* in $L$) also guaranteed to be a Context-Free Language? No, the class of Context-Free Languages is not closed under complementation.

  • Operation: Complementation
  • Closed: No
  • Explanation: The most common way to demonstrate this is by using the property of intersection. Context-Free Languages are not closed under intersection. If CFLs were closed under complementation and union, they would also be closed under intersection, because the intersection of two languages $L_1$ and $L_2$ can be expressed using complementation and union: $L_1 \cap L_2 = \overline{\overline{L_1} \cup \overline{L_2}}$. Since CFLs are closed under union but not intersection, it must be the case that they are not closed under complementation. A classic example showing non-closure under intersection is $L_1 = \{a^n b^n c^m \mid n, m \ge 0\}$ (a CFL) and $L_2 = \{a^k b^m c^m \mid k, m \ge 0\}$ (also a CFL). Their intersection $L_1 \cap L_2 = \{a^n b^n c^n \mid n \ge 0\}$ which is known to be a language that is not context-free. Therefore, CFLs are not closed under intersection, and consequently, not closed under complementation.

Based on this analysis, Context Free Grammars (and the languages they generate) are not closed under Complementation.

Operation Is CFG/CFL Closed Under This Operation?
Union Yes
Concatenation Yes
Kleene Star Yes
Complementation No

Revision Table: CFG Closure Properties

Property Closed (Yes/No)
Union Yes
Concatenation Yes
Kleene Star Yes
Intersection No
Complementation No
Difference No
Reverse Yes
Homomorphism Yes
Inverse Homomorphism Yes

Additional Information on Formal Language Closure Properties

Understanding closure properties is crucial in formal language theory and automata theory. Different classes of languages (like Regular Languages, Context-Free Languages, Context-Sensitive Languages, and Recursively Enumerable Languages) have different sets of closure properties. This helps classify and distinguish between the generative power of different grammar types or the recognition power of different automata types.

  • Regular Languages: Closed under Union, Concatenation, Kleene Star, Complementation, Intersection, Difference, Reverse, Homomorphism, Inverse Homomorphism. They are closed under all standard set operations.
  • Context-Free Languages: Closed under Union, Concatenation, Kleene Star, Reverse, Homomorphism, Inverse Homomorphism. Not closed under Intersection, Complementation, Difference.
  • Context-Sensitive Languages: Closed under Union, Concatenation, Kleene Star, Complementation, Intersection, Reverse. Not closed under Homomorphism (in general), Inverse Homomorphism.
  • Recursively Enumerable Languages: Closed under Union, Concatenation, Kleene Star, Intersection, Reverse, Homomorphism. Not closed under Complementation, Difference, Inverse Homomorphism.

The fact that CFLs are not closed under complementation means that for every CFL $L$, its complement $\overline{L}$ is not necessarily context-free. There exist CFLs whose complements are not CFLs.

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Important Questions from Context Free Languages - Teaching

  1. Consider the following statements about Context Free Language (CFL):

    Statement I: CFL is closed under homomorphism.

    Statement II: CFL is closed under complement.

    Which of the following is correct?

  2. Match List-I with List-II:

    Where L1: Regular language

    L2: Context-free language

    L3: Recursive language

    L4: Recursively enumerable language

    List-1

    List-2

    (a) L' 3 U L 4

    (i) Context-free language

    (b) L' 2 U L 3

    (ii) Recursively enumerable language

    (c) L 1* ∩ L 2

    (iii) Recursive language

    Choose the correct from those given below: 

  3. How can the decision algorithm be constructed for deciding whether context-free language L is finite?

    (a) By Constructing redundant CFG in CNF generating language L

    (b) By constructing non-redundant CFG G in CNF generating language L

    (c) By constructing non-redundant CFG in CNF generating language L-{∧} (∧ stands for null)

    Which of the following is correct?

  4. Which of the following statements is true ?

  5. Consider the following statements with respect to the language L = {a nb n|n≥ 0}

    S 1: L 2is context free language

    S 2: L kis context-free language for any given k ≥ 1

    S 3: L̅ and L* are context free languages

    Which one of the following is correct?

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