Question

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?

• A. \( L_1 \) and \( L_2 \) are regular.
• B. \( L_1 \) and \( L_2 \) are context-free.
• C. \( L_1 \) is regular and \( L_2 \) is context-free.
• D. \( L_1 \) and \( L_2 \) are context-free but not regular.

Hint:

Context-free languages are more powerful than regular languages and can handle patterns like matching numbers of symbols or reversals.

Answer 1

The Correct Option is A, B, C

(A) False. \( L_1 \) is not regular, as it involves matching the number of \( a \)'s on both sides of the string, which is a non-regular operation.

(B) True. \( L_1 \) is context-free because it can be generated by a context-free grammar that matches \( a^n \) on both sides. \( L_2 \) is context-free because it involves two segments with the string and its reversal, which can be described by a context-free grammar.

(C) True. \( L_1 \) is regular, as it is a language of balanced strings with \( a^n \), and \( L_2 \) is context-free because it involves a string and its reversal, which can be described by a context-free grammar.

(D) True. \( L_1 \) and \( L_2 \) are context-free because they can be generated by context-free grammars, but they are not regular due to the matching conditions in \( L_1 \) and the reversal in \( L_2 \).