Question: Consider L = { abwba: w in Sigma * but does not contain the substring ba } over Sigma { a , b
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
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