Q. Let w= xyz and y refers to the middle portion and |y|>0.What do we call the process of repeating y 0 or more times before checking that they still belong to the language L or not?

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Q. There exists a language L. We define a string w such that w∈L and w=xyz and |w| >=n for some constant integer n.What can be the maximum length of the substring xy i.e. |xy|<=?

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Q. Fill in the blank in terms of p, where p is the maximum string length in L.
Statement: Finite languages trivially satisfy the pumping lemma by having n = ______

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Q. Answer in accordance to the third and last statement in pumping lemma:
For all _______ xyiz ∈L

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Q. Let w be a string and fragmented by three variable x, y, and z as per pumping lemma. What does these variables represent?

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Q. Which of the following one can relate to the given statement:
Statement: If n items are put into m containers, with n>m, then atleast one container must contain more than one item.

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Q. Which kind of proof is used to prove the regularity of a language?

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Q. The language of balanced paranthesis is

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Q. State true or false:
Statement: Pumping lemma gives a necessary but not sufficient condition for a language to be regular.

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Q. Which of the following is/are an example of pigeon hole principle?

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Q. Pigeonhole principle can be applied in the following computer science algorithms:

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Q. If n objects are distributed over m places, and n < m, then some of the places receive:

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Q. Which of the following fields may have pigeonhole principle violated?

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Q. Which of the following is not an application of Pumping Lemma?

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Q. Which of the following can refer a language to be non regular?

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Q. Which of the following is not an example of counting argument?

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Q. If L1, L2 are regular and op(L1, L2) is also regular, then L1 and L2 are said to be ____________ under an operation op.

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Q. Suppose a regular language L is closed under the operation halving, then the result would be:

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Q. If L1′ and L2′ are regular languages, then L1.L2 will be

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Q. If L1 and L2′ are regular languages, L1 ∩ (L2′ U L1′)’ will be

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