How to find the intersection of two NFA

Question

In DFA we can do the intersection of two automata by doing the cross product of the states of the two automata and accepting those states that are accepting in both the init

Answer 1

Probabaly a late answer, but since I had the similair problem today I felt like sharing it. Realise the meaning of intersection first. Here, it means that given the string e, e should be accepted by both automata.

Consider the folowing automata:

1. m1 accepting the language {w | w contains '11' as a substring}
2. m2 accepting the language {w | w contains '00' as a substring}

Intiuitivly m = m1 ∩ m2 is the automaton accepting the strings containing both '11' and '00' as a substring. The idea is to simulate both automata simultaneously.

Let's now formally define the intersection.

m = (Q, Σ, Δ, q0, F)

• Let's start by defining the states for m this is, as mentioned above the carthesian product of the states in m1 and m2. So, if we have a1, a2 as labels for the states in m1, and b1, b2 the states in m2, Q will consist out of following states: a1b1, a2b1, a1b2, a2b2. The meaning behind this is to keep track of where where are in both m1 and m2.

• Σ most likely remains the same, however in some cases they differ and we just take the union of alphabets in m1 and m2.

• q0 is now the state in Q containing both the start state of m1 and the start state of m2. (a1b1, to give an example.)

• F contains state s IF and only IF both states mentioned in s are accept states of m1, m2 respectively.

• Last, but not least Δ; we define delta again in terms of the carthesian product, as follows: Δ(a1b1, E) = Δ(m1)(a1, E) x Δ(m2)(b1, E) as also mentoined in one of the answers above if I am not mistaken. The intuitive idea behind this is to just tear a1b1 apart, and consider the states a1 and b1 in their original automaton. Now we 'iterate' each possible edge, lets pick E for example, and see where it brings us in the original automata. After that we glue these results together using the carthesian product. If (a1, E) is present in m1, but not Δ(b1, E) in m2 the edge will not exist in m, otherwise we'll have some kind of a union construction.