Making the difference of a Non-Finite Deterministic Automaton.

Question

I am unsure if I should ask this in this forum, but as I saw questions concerning DFAs and NFAs on here, I figured I could do the same. In case I shouldn't, just say so, I will delete it and ask it other place( would nice if you could advise me where).

I have to do this:

I started off by making a table of its entries and transitions, which gave me this: Here the "*" denotes an accept state, and "N/A" denotes a state that doesn't exist for a transition. So far I am comfortable with this( tell me if you find any errors)

Now for the complementary, I do two things: I change the status between accept and non-accept states, and then I create one additional state for those cases where a state didn't exist for some transition, which I simply call a "3" state, and add the transitions from him to himself. Which gives me this:

My question is whether I did everything the right way? Or did I miss something?

EDIT 1: Ass stated in one answer, I forgot the transition from $1 \rightarrow 0$ So here's the update:

And then, I have to modify the DFA of the complimentary NFA, which gives:

Is this correct now?

Answer 1

Your first transition table is missing an entry for $\{1\}$:

$$\{1\}\overset{a,b}\longrightarrow\{0\}\;.$$

Also, you want to build a DFA equivalent to the given NFA, so your N/A should actually be a state $\varnothing$, corresponding to the empty subset of the states of the original automaton, and you need transitions for it:

$$\varnothing\overset{a,b}\longrightarrow\varnothing\;.$$

Then your $\text{N/A}\cup\{0\}$, for instance, becomes $\varnothing\cup\{0\}=\{0\}$, and similarly your $\text{N/A}\cup\{2\}$ is just $\{2\}$.

This means that in the DFA for the complementary language the $b$-transition from $\{2\}$ should go to $\varnothing$, the $b$-transition from $\{1,2\}$ should go to $\{0\}$, the $b$-transition from $\{0,2\}$ should go to $\{2\}$, and the last line should have $\varnothing$ wherever it now has $\{3\}$. Finally, $\varnothing$ is a non-acceptor state in the DFA equivalent to the original NFA, so it becomes an acceptor state in the final DFA.