Test: Pumping Lemma - Computer Science Engineering (CSE) MCQ

The pumping lemma is often used to prove that a particular language is non-regular. Pumping lemma for regular language is generally used for proving a given grammar is not regular.
Hence the correct answer is a given grammar is not regular.

The pumping lemma is a necessary condition for language to be regular,
In other words, Not satisfying the pumping lemma is a sufficient condition for language to be not regular.
So statements 2 and 3 are false.

I and III are regular languages while II is a context-sensitive language.
I is a regular language because 'n' is bounded and III is regular because a finite automaton is capable of modular counting but not unbounded counting.
The language generated by II is {a, a⁴, a⁹, a¹⁶,.....}. The power of the series is not in arithmetic progression due to which a loop cannot be formed. Therefore, II is not a regular language because we cannot find a substring which is repeated arbitrarily.
Therefore, both I and III are regular languages.

Regular language L = {x|x = a^2+3k or x = b^10+12k k ≥ 0}.
L = {a², a⁵, a⁸, a¹¹ … ∪ b¹⁰, b²², b³⁴…}
Pumping Lemma:
Let L be an infinite regular language. Then there exists some positive integer m such that any w ϵ L with |w|≥ m can be decomposed as
w = xyz
with |xy|≤ m such that wi = xyⁱz
is also L for all i = 0, 1, 2, …
Therefore, minimum Pumping Length should be 11, because string with length 10 (i.e., w = b¹⁰) does not repeat anything, but string with length 11 (i.e., w = b11) will repeat states.
Hence option 1, 2, and 3 are eliminated.
Therefore 24 can be the pumping lemma length.

Concept:
Pigeon-hole principle: If there are total p pigeons seating into q holes then what pigeon-hole principle says that at least one hole must have more than one pigeon.
Pumping lemma: There is an infinite number of input sequences possible and for every finite automaton there is a finite number of states.
Therefore, the logic pumping lemma is a good example of the pigeon-hole principle