GATE2025_CS1_Q44

Consider the following two languages over the alphabet {a, b}:

L₁ = {αβα | α ∈ {a,b}⁺ AND β ∈ {a,b}⁺ }

L₂ = {αβα | α ∈ {a}⁺ AND β ∈ {a,b}⁺ }

Which ONE of the following statements is CORRECT?

✅ Final Answer:

The correct statement is option (C) L₁ is not a regular language but L₂ is a regular language.

🔹 Short Answer:

L₁ is not regular because it requires matching an arbitrarily long prefix α with the suffix, which cannot be done with finite memory. It is a context-sensitive language.
L₂ is regular because the constraint on α (it must be a string of one or more 'a's) allows us to simplify the language. Any string that starts with 'a', ends with 'a', and has at least one character in between can be represented by choosing α = a. This makes L₂ equivalent to the regular language described by the expression a(a|b)⁺a.

🔸 Long Answer:

This is a tricky question that hinges on a subtle difference between the definitions of L₁ and L₂. The key is the existential nature of set-builder notation: a string `w` belongs to the language if there *exists* at least one way to decompose it according to the rule.

Analysis of Language L₂

L₂ = {αβα | α ∈ {a}⁺ AND β ∈ {a,b}⁺ }

Let's determine the set of strings that belong to L₂. A string w is in L₂ if it can be broken down into `αβα` where `α` is a non-empty sequence of 'a's and `β` is a non-empty sequence of 'a's and/or 'b's.

Consider the language of strings that start with 'a', end with 'a', and have a length of at least 3. A regular expression for this is L' = a(a|b)⁺a. Let's see if L₂ and L' are the same.

• Is L' ⊆ L₂? Let's take any string w from L'. We can write w = aua, where u ∈ {a,b}⁺. Can we find a valid decomposition for w to be in L₂? Yes. We can choose:
  • α = a. This satisfies `α ∈ {a}⁺`.
  • β = u. This satisfies `β ∈ {a,b}⁺`.
  With this choice, αβα = aua = w. So, every string in L' is also in L₂.
• Is L₂ ⊆ L'? Let's take any string s from L₂. By definition, s = αβα where `α = aⁿ` (for n ≥ 1) and `β` is a non-empty string.
  • The string s starts with 'a' (since `α` does).
  • The string s ends with 'a' (since `α` does).
  • The length of s is `2n + |β|`. Since `n ≥ 1` and `|β| ≥ 1`, the minimum length is `2(1) + 1 = 3`.
  So, every string in L₂ starts and ends with 'a' and has length at least 3. Thus, every string in L₂ is also in L'.

Since L' ⊆ L₂ and L₂ ⊆ L', the languages are identical. Because L' can be described by the regular expression a(a|b)⁺a, **L₂ is a regular language.**

Analysis of Language L₁

L₁ = {αβα | α ∈ {a,b}⁺ AND β ∈ {a,b}⁺ }

Here, `α` can be any non-empty string of 'a's and 'b's. The crucial part is that the prefix `α` must be identical to the suffix `α`. A finite automaton (DFA or NFA) has no memory mechanism (like a stack) to store an arbitrarily long `α` and compare it with the end of the string.

For example, consider the strings in L₁:

• If `α = ab`, `β = a`, the string is ab a ab.
• If `α = abba`, `β = b`, the string is abba b abba.

Recognizing this pattern requires unbounded memory. This is a classic example of a language that is not regular. It is, in fact, not even context-free and belongs to the class of context-sensitive languages. We could formally prove this using the Pumping Lemma for regular languages.

Conclusion

L₁ is not regular due to the requirement of matching an arbitrary-length prefix and suffix. L₂ is regular because the constraint on `α` is loose enough that we can always find a simple decomposition (`α=a`) that fits the regular expression a(a|b)⁺a.

Bibliography:
- Sipser, M. (2012). Introduction to the Theory of Computation. Cengage Learning. (Chapter 1: Regular Languages, Pumping Lemma).
- Hopcroft, J. E., Motwani, R., & Ullman, J. D. (2006). Introduction to Automata Theory, Languages, and Computation. Addison-Wesley.

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