将DFA排除为正则表达式

Question

I have a few questions regarding state elimination and the terminology.

In the example above is a DFA with a accepting state where you must start with symbol 0 and end with 1.

If I were to convert it to regular expression, the upper would be

and the bottom would be

Here is my problem, I have no idea how to add the top part and the bottom part into on single expression. I am also not totally sure how to eliminate q2 symbol 1 further.

Would it be 0(0*(0+1))1* ?

Thank you to anyone who could help!

Answer 1

There is a much more well known and comprehensible algorithm available to accomplish this task.

To convert a DFA G into a Regular expression we first convert G into a 'GNFA'. Let for example G be the following DFA (q is the start state):

The process of converting a DFA to a GNFA is as follows:

1. Add new start state with an epsilon transition to the original start state.
2. Add new accept state, add epsilon transitions from every original accept state to the newly added accept state, then make all the original accept states into normal states.

This is the resulting GNFA:

Then we remove each state in-between the new start state and the new accept state one at a time, adjusting the graph to maintain correctness. The process works as follows: Let x, y, and z be states in our DFA. Additionally, the transitions are as follows x->y on input a, y->y on input b and y->z on input c. Say we want to remove y. For every transition from some node n to y and for every transition from y to m, we must add a new transition n->m. The transition from n to m would be the content of the transition from n to y, followed by the content of transition y->y with a kleene star followed by the content of the transition from y->m. In this case x->y on a, y->y on b and y->z on c, after removing state y, there would be a transition from x->z on a(b*)c .

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Consider our DFA in the images. After removing state q we get:

After removing state r, we get:

Finally, after removing state s we are left with:

This is our complete regular expression. Using this process completely avoids any of the problems you are facing. However, I will also provide you with a direct answer to your question. For starters, the upper part would not be what you suggested. Instead, it would become: This simplifies into: This is our final regex, since the bottom part has no accept state and is thus irrelevant.

Answer 2

you start from (q0) state if u input (0) then you can reach to the final ; instead if u input (1) you couldnt reach final . so consider only states (q0) (q1) (q2) and apply elimination rule on these states

after elimination RE would be as follow

0(0)*1 . (1+0(0)*1)*

start from 0 and ending on 1