Question:
Consider the following two languages over the alphabet \( \{a, b, c\} \), where \( m \) and \( n \) are natural numbers.
\[
L_1 = \{ a^m b^{m+n} c^{m+n} \mid m, n \geq 1 \}
\]
\[
L_2 = \{ a^m b^n c^{m+n} \mid m, n \geq 1 \}
\]
Which ONE of the following statements is CORRECT?
Consider the following two languages over the alphabet \( \{a, b, c\} \), where \( m \) and \( n \) are natural numbers.
\[
L_1 = \{ a^m b^{m+n} c^{m+n} \mid m, n \geq 1 \}
\]
\[
L_2 = \{ a^m b^n c^{m+n} \mid m, n \geq 1 \}
\]
Which ONE of the following statements is CORRECT?
Show Hint
When analyzing whether a language is context-free, look for dependencies between symbols that involve complex relations, as these often lead to non-context-free languages. Simple counting relationships, however, can typically be expressed using context-free grammars.
Updated On: Jan 30, 2026
- Both \( L_1 \) and \( L_2 \) are context-free languages.
- \( L_1 \) is a context-free language but \( L_2 \) is not a context-free language.
- \( L_1 \) is not a context-free language but \( L_2 \) is a context-free language.
- Neither \( L_1 \) nor \( L_2 \) are context-free languages.
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The Correct Option is C
Solution and Explanation
Step 1: Analyze \( L_1 \)
The language \( L_1 = \{ a^m b^{m+n} c^{m+n} \mid m, n \geq 1 \} \) defines a relationship between the number of \( b \)'s and \( c \)'s, where the number of \( b \)'s and \( c \)'s depend on both \( m \) and \( n \). This dependency makes the language non-context-free. Specifically, this kind of relationship between different symbols typically cannot be captured by a context-free grammar.
Step 2: Analyze \( L_2 \)
The language \( L_2 = \{ a^m b^n c^{m+n} \mid m, n \geq 1 \} \) defines a more straightforward relationship, where the number of \( c \)'s is the sum of the number of \( a \)'s and \( b \)'s. This can be expressed by a context-free grammar, as it is a simple counting relationship that a context-free grammar can handle. The relationship between \( b \) and \( c \) can be captured by generating \( a^m b^n \) first, and then ensuring that the number of \( c \)'s is the sum \( m+n \).
Step 3: Conclusion
\( L_1 \) is not context-free due to the more complex dependency between \( b \) and \( c \).
\( L_2 \) is context-free because it represents a simpler relationship between \( a \), \( b \), and \( c \).
Thus, the correct answer is (C) \( L_1 \) is not a context-free language but \( L_2 \) is a context-free language.
The language \( L_1 = \{ a^m b^{m+n} c^{m+n} \mid m, n \geq 1 \} \) defines a relationship between the number of \( b \)'s and \( c \)'s, where the number of \( b \)'s and \( c \)'s depend on both \( m \) and \( n \). This dependency makes the language non-context-free. Specifically, this kind of relationship between different symbols typically cannot be captured by a context-free grammar.
Step 2: Analyze \( L_2 \)
The language \( L_2 = \{ a^m b^n c^{m+n} \mid m, n \geq 1 \} \) defines a more straightforward relationship, where the number of \( c \)'s is the sum of the number of \( a \)'s and \( b \)'s. This can be expressed by a context-free grammar, as it is a simple counting relationship that a context-free grammar can handle. The relationship between \( b \) and \( c \) can be captured by generating \( a^m b^n \) first, and then ensuring that the number of \( c \)'s is the sum \( m+n \).
Step 3: Conclusion
\( L_1 \) is not context-free due to the more complex dependency between \( b \) and \( c \).
\( L_2 \) is context-free because it represents a simpler relationship between \( a \), \( b \), and \( c \).
Thus, the correct answer is (C) \( L_1 \) is not a context-free language but \( L_2 \) is a context-free language.
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