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2. Prove that {a6c |m,n0}is not a regular language. Answer:
3. Let L = { M M is a Turing machine and L(M) is empty), where L(M) is the language accepted by M. Prove L is undecidable by
4. a) Define the concept of NP-completeness b) If A is NP-complete, and A has a polynomial time algorithm, then a polynomial
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Answer #1

2. Proof:

Let's say ? is the constant of the pumping lemma, and let's consider ? = ??N?N? and |?| = 2?+1 ≥ ?. By the lemma, it can be written ? = ???, such that |??| ≤ ? with ?? such that for all ?≥0 it is ??k??.

Now ? must be composed of only one symbol (in this case, as |??|≤?, it has to be only ? or some ?s) as otherwise ??k? for ?>1 isn't of the right form (?s followed by ?s followed by ?s). Now ?=?, ?=? is impossible, because in that case we have ??0? = ?N?N?. But ? just ?s with ?≠1 gives a string with number of ?s and ?s different, thus not in ?. No division into ?,?,? works, the language fails the pumping lemma. Hence it can't be regular.

Hope this helped. Please do upvote and if there are any queries please ask in comments section.

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