GATE | Gate IT 2008 | Question 31

Consider the following languages.

L₁ = {aⁱ b^j c^k | i = j, k ≥ 1}

L₁ = {aⁱ b^j | j = 2i, i ≥ 0}

Which of the following is true?
(A) L₁ is not a CFL but L₂ is
(B) L₁ ∩ L₂ = ∅ and L₁ is non-regular
(C) L₁ ∪ L₂ is not a CFL but L₂ is
(D) There is a 4-state PDA that accepts L₁, but there is no DPDA that accepts L₂


Answer: (B)

Explanation: A: Both L₁ and L₂ are CFL

B: L₁ ∩ L₂ = ∅ is true

L₁is not regular => true

=> B is true

C: L₁ ∪ L₂ is not a CFL nut L₂ is CFL is closed under Union

=> False

D: Both L₁ and L₂ accepted by DPDA

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