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Suppose M=(Q,2,5,5, A) is a DFA. For states p, q EQ (p can be same as q) argue that Lpq = {w | 8*(p,w)=q} is
Suppose M=(Q,2,5,5, A) is a DFA. For states p, q EQ (p can be same as q) argue that Lpq = {w | 8*(p,w)=q} is regular. Recall that PREFIX(L) = {w|wx L, X 2*} is the set of all prefixes of strings in L. Express PREFIX(L(M)) as UgezLs,q for a suitable set of states 2 C Q. Why does this prove that PREFIX(L(M)) is regular whenever L is regular? For a language L let MID(L) = {w|xwy EL,x,y 2*}. Prove that MID(L) is regular if L is regular 1. (a) Draw an NFA that accepts the language {w there is exactly one block of is of even length}. (A block of is is a maximal substring of is.) (b) i. Draw an NFA for the regular expression (010)* +(01)* +0%. ii. Now using the powerset construction (also called the subset construcion), design a DFA for the same language. Label the states of your DFA with names that are sets of states of your NFA. You should use the incremental construction so that you only generate the states that are reachable form the start state
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