Let A = {omega elementof {a, b}* | omega contains more a\'s than b\'s}. Suppose

Question

Let A = {omega elementof {a, b}* | omega contains more a's than b's}. Suppose you are trying to prove that A is not regular using the pumping lemma. Your proof starts (correctly) like this: Suppose for a contradiction that A is regular. Let p be the pumping length given by the pumping lemma. Now you have to choose string s. For each choice of s below, state whether or not this choice of s can be used to finish the proof that A is not regular. If your answer is YES you do not have to explain anything. If your answer is NO. you should also briefly explain why it can not be used. (a) s = a^p+1 b. (b) s = a^p+1 b^P. (c) s = a^pb^p. (d) s = (ab)^p a. (e) s = baa. (f) s = b^pa^p+1.

Explanation / Answer

(a) No. If we divide this in to x,y,z. y will contain a's. So let y=a^k, k>0. So k can be even be 1. x(y^0)z will be a^(p+1-1) b. We know p>1 so x(y^0)z still belongs to the language.

(b) Yes.

(c) No. This string does not even belong to language A.

(d) Yes.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.