Convert the following NFA to a reduced DFA(final state is marked by *)

The NFA that is made from the table is as follows:

Now, as {p} is initial state, it is added to the solution

We take the state transitions of all states for each input

$\delta({p},0)={p,q}$

Therefore, {p,q} is added to the solution

$\delta({p},1)={q}$

$\delta({q},0)={r}$

Therefore, {r} is added to the solution

$\delta({q},1)={r}$

$\delta({r},0)={s}$

Therefore, {s} is added to the solution

$\delta({r},1)=\Phi$

$\delta({s},0)={s}$

$\delta({s},1)={s}$

$\delta({p,q},0)={p,q,r}$

Therefore, {p,q,r} is added to the solution

$\delta({p,q},1)={q,r}$

Therefore, {q,r} is added to the solution

$\delta({p,q,r},0)={p,q,r,s}$

$\delta({p,q,r},1)={q,r}$

$\delta({q,r},0)={r,s}$

Therefore, {r,s} is added to the solution

$\delta({q,r},1)={r}$

$\delta({r,s},0)={s}$

$\delta({r,s},1)={s}$

Therefore, the solutions set S ={{p,q},{r},{s},{p,q,r},{q,r},{r,s},{q,r,s}}