S => AB => AbA => Aba => aSba => aABba => aAbAba => aAbaba => aababa.

111001 cannot be the suffix of any string accepted by this DFA. Suppose s=w111001. No matter what state the DFA reaches after reading w, it will go to state D after reading “111”, then go to state B after reading “00” and finally reaches state C after reading “1”.

Every string in the language {011, 10, 110}* has to be formed from zero or more uses of the strings 011, 10, and 110. A string may be used more than once.

The states traversed are ABDBDABDAC, and the only edge not traversed C D.

The cycle of unit-productions S → A → D → S says that any pair involving only S, A, and D is a unit pair. Similarly, the cycle S → B → E → S tells us that any pair involving S, B, and E is a unit pair.

First, notice that A generates strings of the form 021, where n is 0 or more. Also, B gives zero or more 1’s, which is followed by one 3, and then A gives something. Since S generates something an A can generate followed by something a B can generate, the strings in L (G) are of the form 0 21 30 21.

S => a. A string of length 2 has only one derivation, e.g., S => SS => aS => ab.

We can build a DFA for as such: firstly we get the DFA for: Then, we copy all the states and transitions to the DFA for. However, we mark any state as a final state in if and only if is a final state in.

All generate all strings of 0’s and 1’s thus are these pairs are equivalent.

According to pumping lemma, is not a regular language. It is the language of the DFA with two states to achieve an even number of 0’s…