GATE2025_CS1_Q45

Consider the following two languages over the alphabet {a, b, c}, where m and n are natural numbers.

L₁ = {a^mb^mc^m+n | m, n ≥ 1}

L₂ = {a^mbⁿc^m+n | m, n ≥ 1}

Which ONE of the following statements is CORRECT?

✅ Final Answer:

The correct statement is option (C) L₁ is not a context-free language but L₂ is a context-free language.

🔹 Short Answer:

L₁ is not Context-Free: The language L₁ = {a^mb^mc^m+n} can be written as {a^mb^mc^mcⁿ}. A pushdown automaton (PDA) needs to verify two separate dependencies: that the number of 'a's equals the number of 'b's, AND that it also equals the first part of the 'c's. A single stack cannot handle these two independent comparisons simultaneously.
L₂ is Context-Free: The language L₂ = {a^mbⁿc^m+n} requires that the number of 'c's equals the sum of 'a's and 'b's. A PDA can easily recognize this by pushing a symbol onto the stack for every 'a' and every 'b' it reads, and then popping one symbol for every 'c'. If the stack is empty after all characters are read, the string is accepted.

🔸 Long Answer:

To determine if these languages are context-free, we must check if a Pushdown Automaton (PDA) can be constructed to recognize them. A PDA uses a single stack, which limits its memory and comparison capabilities.

Analysis of L₁ = {a^mb^mc^m+n | m, n ≥ 1}

Let's rewrite the language as L₁ = {a^mb^mc^mcⁿ | m, n ≥ 1}. To recognize a string from this language, a machine must verify two conditions:

1. The number of 'a's is equal to the number of 'b's.
2. The number of 'a's (or 'b's) is also equal to the number of 'c's in the first part of the c-block.

Let's trace the logic for a PDA:

• To verify condition 1, the PDA could push m symbols onto the stack while reading the 'a's.
• Then, it would pop one symbol for each 'b' it reads. After reading all m 'b's, the stack would be empty.
• Now, the PDA has to verify that the number of 'c's is m+n. The first part of this requires matching m 'c's. But since the stack is now empty, the PDA has lost all information about the value of m. It cannot perform this second check.

Because a single stack cannot remember and verify two independent counts, L₁ is not a Context-Free Language. It is a well-known example of a Context-Sensitive Language, provable using the Pumping Lemma for CFLs.

Analysis of L₂ = {a^mbⁿc^m+n | m, n ≥ 1}

This language requires that the total count of 'c's matches the combined count of 'a's and 'b's. A PDA is perfectly suited for this task.

A PDA can be designed with the following logic:

1. For every 'a' read, push a symbol (e.g., 'X') onto the stack.
2. For every 'b' read, also push a symbol ('X') onto the stack.
3. After all 'a's and 'b's are read, the stack will contain m + n symbols.
4. For every 'c' read, pop one symbol from the stack.
5. If the input is fully read and the stack becomes empty at the exact same time, the string is accepted. Otherwise, it is rejected.

Since a PDA can be constructed for L₂, L₂ is a Context-Free Language.

Conclusion

Based on our analysis:

• L₁ is NOT context-free.
• L₂ IS context-free.

This matches option (C).

Bibliography:
- Hopcroft, J. E., Motwani, R., & Ullman, J. D. (2006). Introduction to Automata Theory, Languages, and Computation. Addison-Wesley. (Chapters 5 & 6).
- Sipser, M. (2012). Introduction to the Theory of Computation. Cengage Learning. (Chapter 2: Context-Free Languages).

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