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GATE | Gate IT 2008 | Question 31

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  • Last Updated : 28 Jun, 2021
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Consider the following languages.

L1 = {ai bj ck | i = j, k ≥ 1}

L1 = {ai bj | 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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