Question

Is the following language regular? If so, give an NFA that accepts this language and if not, prove why it is not using the pumping lemma L= {an2∣n≥1}.

Answer 1

The language L = a^n2 is not regular. Using the pumping lemma, we proved by contradiction that no NFA can accept this language since it does not satisfy the necessary conditions defined by the lemma.

Step-by-step explanation:

The question is whether the language L = n ≥1 is regular. To determine this, we can use the pumping lemma, a fundamental tool in formal language theory which is often used to prove that certain languages are not regular.

Assume for the sake of contradiction that L is regular. The pumping lemma states that for a regular language, there exists some integer p (the pumping length), such that any string s in the language of length at least p can be divided into three parts, xyz, satisfying the following conditions:

• |xy| ≤ p
• |y| > 0
• For all i ≥0, the string xyiz is also in the language.

If L were regular, we could pick a string s = ap2 from L, where p is the pumping length given by the lemma. However, no matter how we divide s into xyz with |y| > 0, pumping y (by repeating it i times) will never give us a string of the form an2, since (p+i)2 is not equal to p2 for any i other than 0. This means the third condition of the pumping lemma cannot be satisfied, and thus, L cannot be a regular language. Therefore, we conclude that L is not a regular language, and it is not possible to construct an NFA for it.