How to convert regular expressions to grammar and DFA

Question

Hi I am trying to figure out the regular grammar that represents the regular expression L = (a+ab)* as DFA and generates L.

The picture below shows the process of expressing the expression as NFA and then converting it to DFA.

So if you get regular grammar from DFA,

A-> aB | bC | e

B-> aB | bA | e

C-> aC | bC

But the problem is, when you get the regular expression with this grammar, you get a much more complex expression, not (a+b)*.

C = aC + bC = (a+b)*

B = aB + bA + e = a*(bA+e)

A = aB + bC + e = aa* bA + aa* + b(a+b) + e = (aa* b)* (aa* +b(a+b)*+e)

I wonder if there is a problem with my solution.

Answer 1

Actually X = (aa*b)*(aa*+b(a+b)*+e) can be simplified to (a+b)*. Here I tried to explain my steps for simplifying X to (a+b)*:

1. We will divide our simplification into three parts: a) empty string , b) every possible string starting with b , c) every possible string starting with a . If we can obtain these, it will mean that X=(a+b)*.
2. If you take the first part (aa*b)* as e and look at the second part, you will see that we can obtain a) empty string and b) every possible string starting with b . We only need to obtain c) every possible string starting with a .
3. This part was a struggle but I think I obtained it. Now we look only for strings starting with a. First of all, we have aa* in the second part obtaining all 'a's. Moreover, if we have 2 or more consecutive 'b's, the rest of the string will be accepted by the second part b(a+b)*. So the only concern is single 'b's. Finally, single 'b's are obtained by the first part.

Although it was complicated (and probably unnecessary), this is a proof that X=(a+b)*. As a result, your solution was correct.