Question

Prove that the following language is not regular: { w₁aw₂ | w₁,w₂ ∈ {a,b}* and |w₁| = |w₂| } In other words, L consists of strings of odd length over the alphabet {a, b} which have a as its middle symbol. SHOW ALL WORK, THANKS!

Answer 1

Answer: It is not defining finite automata. Because there can be infinite string which having 'a' on middle position and remaining element is either 'a or 'b'. Proof: No DFA has enough states to keep track the given language. The language LWaw2 wi,w2 E (a.b* and |wil|w2l 3 is not regular Suppose (for condition) Lea is recognized by DFA M-(Q. 10,1), 5, qo, F) Let w = {a}*. For any wł, w2 E W with w1 W2, δ"M(q0,w1)-δ,M(q0,W2). Let us observe that if the claim holds, then M has infinitely many states, and so is not a finite automaton, giving the desired contradiction. Claim: For any wl, w2 E W with wi W2, δ,M140,W1)-δ,M(Qo,W2). Proof claim: Suppose (for contradiction) there is wl and w2 such that δ,M(Qo,w1) δ, M (Qo,w2-{aj. Without loss of generality we can assume that wi-ai and w2-3, with accepts both a a bi and ai a bi or neither. Butalabie Lea- This the contradiction the assumption that M recognizes Leq