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GATE | GATE-CS-2014-(Set-1) | Question 24
  • Last Updated : 18 Apr, 2014

Which one of the following is TRUE?

(A) A
(B) B
(C) C
(D) D

Answer: (C)

Explanation: (A) L = {a n b n |n >= 0} is not regular because there does not exists a finite automaton that can
derive this grammar. Intuitively, finite automaton has finite memory, hence it can’t track
number of as. It is a standard CFL though.

(B) L = {a n b n |n is prime} is again not regular because there is no way to remember/check if
current n is prime or not. Hence, no finite automaton exists to derive this grammar, thus it
is not regular.

(C) L = {w|w has 3k+1 bs} is a regular language because k is a fixed constant and we can easily
emulate L as a ∗ ba ∗ … ∗ such that there are exactly 3k + 1 bs and a ∗ s surrounding each b in
the grammar.


(D) L = {ww| w ∈ Σ ∗ } is again not a regular grammar, infact it is not even a CFG. There is no
way to remember and derive double word using finite automaton.

Hence, correct answer would be (C).
This solution is contributed by vineet purswani.

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