Question

Can you check on this: https://dl.dropbox.com/u/25439537/finite%20automata.png

This is a checked homework, so don’t worry. I just want to clarify whether my answer is correct or not, because it is marked by my teacher as incorrect.

My answer is ((a+b)(a+b))*a
The first (a+b) signifies the upper arrows. The second (a+b) signifies the lower arrows. The last ‘a’ tells us that it should always end in ‘a’.

I just want to record evidences from a lot of experts so that I can give it to my teacher.

Answer 1

I believe your answer is correct.

Let’s consider the whole process as two parts: (1) start with start, and go back to start; and (2) go from start to end and accept. Obviously, the (1) part is a loop.

For (1), starting with start, either accept b or a. For b, it’s b(a+b) to go back. For a, it’s a(a+b) to go back. So (1) is b(a+b) + a(a+b) which is (a+b)(a+b).

For (2), it’s a‘.

So, the final result is (loop in (1))* (2) i.e. ( (a+b)(a+b) )* a.

Follow the description above, you can also come up with a proof of the equivalence between the two. Proof part (a) every sequence accepted by the automata is in the set ((a+b)(a+b))*a; part (b) every sequence in the set ((a+b)(a+b))*a is accepted by the automata.