自动机 - 正则表达式（Union Case）

Question

Automata 1) Recognizes strings with at least 2 a

Regular Expression = b*ab*a(a+b)*

Automata 2) Recognizes strings with at least 2 b

Regular Expression = a*ba*b(a+b)*

The regular expression obtained from A3 = A1 U A2 is equivalent to R3 = R1 + R2? Or it's not?

R3 = b*ab*a(a+b)* + a*ba*b(a+b)*

Answer 1

Regular expressions are not like finite state parsers and it's usually a mistake to try to incorporate them into complex parsing scenarios.

But also, they are marvelous tools for specific problems. After reading your descriptive requirements, there is a simple regular expression that accomplishes it, but in a way you might not expect. Your requirements:

1. strings with at least 2 a

2. strings with at least 2 b

3. The Union of the two, or strings withat least two a's or two b's

   ([ab]).*?\\1

This expression opens a capture group to capture either a or b. Then it allows zero or more 'any characters' followed by whatever was captured in the capture group (\\1).

Answer 2

There is neither "one" automaton nor "one" regular expression for any language; generally there many reasonable ones and many more (maybe infinitely many) unreasonable ones. In this sense, your question is not entirely well-posed: the regular expression corresponding to the union of two DFAs may or may not look like regular expressions for the original DFAs, +'ed together.

So, if you mean, can they look the same, the answer is likely yes. If you mean, must they look the same, answer is likely no. If you instead want to fix the algorithms for constructing the union machine and getting the regular expression, maybe we could show that a fixed method of doing it always gives the same answer.

In your specific case, applying the Cartesian Product Machine construction to get a DFA for the union of the original DFAs and then applying the construction from the proof of equivalence between DFAs and REs, we can see that the structure of the RE obtained by +'ing the original REs can't be achieved starting from a DFA; you'd have needed an NFA to get a + between the LHS and RHS, but DFAs can only + among individual symbols, not subexpressions. Of course, it might be possible the RE can be algebraically manipulated to derive the target RE, but that isn't exactly the same.

All of the above hold for the question of equality of REs. However, you asked about equivalence. Almost always, we say two REs are equivalent if they generate the same language. If this is what you meant, then yes, +ing the two REs will give an RE equivalent to the one obtained by constructing a union machine and deriving an RE from that. The REs will not look the same but will generate the same language, just as (ab + e)(abab)* and (ab)* generate the same language despite looking a bit different.