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Answer: d.

Obviously, the language regular, since the minimal automata for L with the states that represent A1 and A2 as accepting states, accept this language.

Now, take the language {epsilon, 1, 11}, it has 4 equivalence classes. The automata that accepts {epsilon, 1} has 3. Done.

Q 2

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Answer: d

This is a regular language.

Create an k+1-state NFA from the k-state DFA that accept L by adding a start state with epsilon transitions to every state that is accessible from q0. Convert the NFA to DFA using the algorithm we saw in class. This will result in a DFA with at most 2^(k+1) states.

Q 3

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Answer: c

Language is not finite (n is unbounded)

Language is not regular. (choose m = m = l, pump y to 0)

Language is CFG (push into stack for every d, pop for every c, if you get a c with an empty stack - fail. Count k <= 100 using 100 states).

Thank you. ]]>