Converting DFA to Regular Expression (DFA final state is the empty set)

Question

So the original question is to get the complement of the regex 0*1*.
0*1* = {ε,0,1,00,01,11,000,001,011,111,...}, essentially any number of 0s followed by any number of 1s.

I made this DFA for 0*1*:

• Note: "↫" denotes a self-loop.

->((q1))↫0 ---1--> ((q2))↫1 ---0---> (∅)↫0,1

Then I did the complement by swapping the final states with the non final states, resulting in the empty set being the only final state:

-->(q1)↫0 ---1--> (q2)↫1 ---0---> ((∅))↫0,1

But now I'm confused, I've looked at countless examples online and I'm not sure how to convert this particular DFA to a regex, my main problem being that the final state is an empty set.

These are the computations I did:

For q1:

q1 = Ɛ + q1(0), because R=Q+RP is equal to R=QP*...

q1 = Ɛ(0*), because Ɛ(R)=R...

q1 = 0*

Then, for q2:

q2=q1(1)+q2(1), substitute q1 to get q2=0*(1)+q2(1), then...

q2=0*1(1)*

For ∅: (<-This is where I think things start to get shaky)

∅ = q2(0) + ∅(0) + ∅(1), can be reorganized ∅ = q2(0) + ∅(0 + 1), then substitute q2 and get ∅ = 0*1(1)*0 + ∅(0 + 1), then this can become:

∅ = 0*1(1)* 0(0 + 1)*

So I would consider the complement of the regex 0*1* to be 0*1(1)* 0(0 + 1)*. But that just doesn't seem right. Am I overcomplicating this? Please any help would be appreciated.

Edit 1: - Edited to fix typos

Answer 1

We start at state $A$.

$$\begin{array}{c|c|c} \text{State}&\text{Next}&\text{Goto}\\\hline A&0&A\\\hline A&1&B\\\hline B&0&C\\\hline B&1&B\\\hline C&0&C\\\hline C&1&C \end{array}$$

where states $A$ and $B$ are accepted.

---

In its complement, $C$ is accepted.

We have these paths from $A$ to $C$ forcing no repetition of states:

• $A \overset{1}{\longmapsto} B \overset{0}{\longmapsto} C$

There's only one path.

This corresponds to the regex 0*11*0[01]*, which can be shortened to 0*1+0.* assuming that . only matches 0 or 1.