Question:medium

Consider the following languages: \[ L_1 = \{a^n w a^n | w \in \{a, b\}^*\} \] \[ L_2 = \{ w x w^R | w, x \in \{a, b\}^*, |w|, |x| > 0 \} \] Note that \( w^R \) is the reversal of the string \( w \). Which of the following is/are TRUE?

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Context-free languages are more powerful than regular languages and can handle patterns like matching numbers of symbols or reversals.
Updated On: Jan 30, 2026
  • \( L_1 \) and \( L_2 \) are regular.
  • \( L_1 \) and \( L_2 \) are context-free.
  • \( L_1 \) is regular and \( L_2 \) is context-free.
  • \( L_1 \) and \( L_2 \) are context-free but not regular.
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The Correct Option is A, B, C

Solution and Explanation

(A) False.
\(L_1\) is not a regular language. Any language that requires comparing or matching counts across different parts of the string (such as ensuring equality or symmetry) cannot be accepted by a finite automaton. Finite automata have no memory to store and compare such information, hence \(L_1\) is non-regular. Therefore, the statement in option (A) is false.

(B) True.
\(L_1\) is a context-free language since a pushdown automaton can use its stack to keep track of the number of symbols in one part of the string and match them later. Similarly, \(L_2\) is context-free because pushdown automata can recognize patterns involving a string and its reverse by pushing symbols onto the stack and popping them in reverse order. Hence, both languages are context-free, making option (B) true.

(C) True.
\(L_1\) is regular because it follows a simple repetitive structure that can be recognized without additional memory. \(L_2\) is context-free since recognizing a string followed by its reverse requires stack-based memory, which is provided by pushdown automata. Thus, the statement in option (C) is true.

(D) True.
Both \(L_1\) and \(L_2\) are context-free languages as they can be generated by context-free grammars. However, neither language is regular due to the need for matching conditions and reversal properties, which cannot be handled by finite automata. Hence, option (D) is true.
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