I have to construct a DFA for the language $$\{w \mid w \text{ has exactly two } a's \text{ and at leat two } b's\}$$
To construct it we have to construct first the DFA's for the languages $$\{ w \mid \text{ has exactly two } a's\}$$ and $$\{w \mid w \text{ has at least two } b's\}$$
right?? I tried to construct these two DFA's...
For the language $\{ w \mid \text{ has exactly two } a's\}$ I got the following:
But as it is right now it isn't a DFA, but a NFA...
What is $\delta (q_3, a)$ ??
For the language $\{w \mid w \text{ has at least two } b's\}$ I got the following:
Is this correct??
To construct now the DFA of the initial language do we take the intersection of the two DFA's ?? $$$$
EDIT:
$q_i$ are the state of the DFA of the language $\{ w \mid \text{ has exactly two } a's\}$ and $q_i'$ are the states of the DFA of the language $\{w \mid w \text{ has at least two } b's\}$.
To answer your first question, add a non-accepting state $q_4$ to your first automaton; $\delta(q_3,a)=q_4$, and $\delta(q_4,a)=\delta(q_4,b)=q_4$. (I think of this kind of state as a garbage or dump state; I’ve seen it called a trap state and a dead state.)
If you know how to combine two DFAs to build one that accepts the intersection of their languages, then yes, you can apply that algorithm to build the desired DFA: the language that you want is the intersection of the two simpler languages. If you have questions about that algorithm, let me know where you’re having trouble.
Added: Your combined DFA isn’t quite right. Notice that it doesn’t accept $aabb$: you end up at $\langle q_4,q_3'\rangle$. The problem is that you misdirected the $b$ transitions out of the $\langle q_3,q_k'\rangle$ states. If you check the small DFAs, you’ll see that you should have
$$\langle q_3,q_1'\rangle\overset{b}\longrightarrow\langle q_3,q_2'\rangle\overset{b}\longrightarrow\langle q_3,q_3'\rangle\;,$$
with a $b$ transition from $\langle q_3,q_3'\rangle$ to itself. Remember, each transition in this DFA should mimic transitions in the two small ones: $\langle q_i,q_j'\rangle$ goes to $\langle q_k,q_\ell'\rangle$ on input $x$ if your first DFA goes from $q_i$ to $q_k$ on input $x$ and your second from $q_j'$ to $q_\ell'$ on input $x$.
Otherwise it’s fine, though if you wished to do so, you could combine the three $q_4$ states at the bottom into a single trap state: anything that reaches any one of them eventually ends up trapped at $\langle q_4,q_3'\rangle$ without being accepted.