Question

Consider the following languages: 

\( L_1 = \{ ww \mid w \in \{a,b\}^* \} \) 
\( L_2 = \{ a^n b^n c^m \mid m, n \geq 0 \} \) 
\( L_3 = \{ a^m b^n c^n \mid m, n \geq 0 \} \) 

Which of the following statements is/are FALSE?

• A. \( L_1 \) is not context-free but \( L_2 \) and \( L_3 \) are deterministic context-free.
• B. Neither \( L_1 \) nor \( L_2 \) is context-free.
• C. \( L_2, L_3 \) and \( L_2 \cap L_3 \) are all context-free.
• D. Neither \( L_1 \) nor its complement is context-free.

Hint:

Context-free languages are not closed under intersection and complement, which makes problems like checking the complement non-trivial.

Answer 1

The Correct Option is B, C, D

Let's go over the properties of the languages \(L_1\), \(L_2\), and \(L_3\):

Option (A) - False:
- \(L_1 = \{ ww \mid w \in \{a,b\}^ \}\) is not context-free, and it is not deterministic context-free. The language \( L_1 \) represents strings that consist of two identical halves, which requires a machine with more than a single stack, and hence cannot be parsed by a pushdown automaton. Thus, \(L_1\) is not context-free.
- However, both \(L_2\) and \(L_3\) are context-free and also deterministic context-free languages. Both languages can be parsed by a deterministic pushdown automaton because they can be represented by context-free grammars and don't require non-determinism.

Thus, option (A) is false.

Option (B) - True:
- \(L_1\) is not context-free (as mentioned above).
- \(L_2 = \{ a^m b^m c^n \mid m, n \geq 0 \}\) is context-free because it can be generated by a context-free grammar, as it has two different components: one for \( a^m b^m \) and one for \( c^n \). However, it is not deterministic context-free, because it requires the pushdown automaton to handle two different strings of different lengths, making it non-deterministic. Hence, the statement in option (B) is true.

Option (C) - False:
- \(L_2 = \{ a^m b^m c^n \mid m, n \geq 0 \}\) is context-free.
- \(L_3 = \{ a^m b^n c^n \mid m, n \geq 0 \}\) is context-free.
- However, the intersection of \(L_2\) and \(L_3\), \(L_2 \cap L_3\), is not context-free. The intersection would require matching both the number of \( b \)'s and \( c \)'s, which requires non-context-free properties. Therefore, this statement is false.

Option (D) - False:
- \(L_1\) is not context-free.
- However, the complement of \(L_1\) is also not context-free. This is a result of the closure properties of context-free languages, which do not allow the complement of a non-context-free language to be context-free. Hence, the statement is false.

Thus, the correct answers are (B), (C), and (D).