Question: Consider L = { abwba : w in Sigma but does not contain the substring ba } over Sigma = { a ,
Consider L abwba : w in Sigma
but does not contain the substring ba over Sigma a b c d
Show that L is a regular language by drawing a DFA for it with the DFA having
as few states as you can.
Draw an NFA for L with as few states as you can.
Convert the NFA to a GNFA.
Convert the GNFA to an RE Regular Expression using the procedure discussed
in class. Show the steps. You may, but are not required to shorten intermediate
REs using identities like R and epsi R R and R cup R to reduce expression
swell. PLEASE DRAW THEM on paper if possible,im trying to correct my answers
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