Can final and non-final states having identical configurations be merged together in a minimal DFA?

Question

I have been practicing some questions on automata theory where I came across a question on minimal dfa at which I can't figure out where I have gone wrong.I am getting 4 states in the minimal dfa but my book says answer is 3.The question asks a given NFA to convert into a minimal DFA and count the number of states in the latter one.The given NFA(p and r are initial and final states respectively) is:

{p}---b-->{q}
{q}---a--->{r}
{q}---b--->{s}
{r}---a--->{r}
{r}---b--->{s}
{s}---a--->{r}
{s}---b--->{s}

I am getting 4 states:[r],[p],[q,s],[dead].Can the final [r] and the non-final state [q,s] be merged here since they lead to the similar configuration on receiving inputs a and b??I have learned that final and non-final states cannot be in the same equivalence class...

Answer 1

OK, let's start with all possible states for our DFA, there will be 17 (2^4 for the 4 symbols plus 1 for the empty set). These will be:

{}
{p}
{q}
{r} 
{s}
{p,q}
{p,r}
{p,s} 
{q,r} 
{q,s}
{r,q}
{r,s}
{p,q,r}
{p,q,s}
{p,r,s}
{q,r,s}
{p,q,r,s}

Now that we have all the possible sets, let's highlight all that are reachable from the start state p:

{}
{p} --- Start state. From here we can get to {q} only as defined by the transition {p}--b-->{q}
{q} --- from here, we get to r or s, so {r,s} as defined by {q}--a-->{r} and {q}--b-->{s}
{r} 
{s}
{p,q}
{p,r}
{p,s} 
{q,r} 
{q,s}
{r,q}
{r,s} --- from here, we can only get to r or s (the same state!), so we're at a dead end.
{p,q,r}
{p,q,s}
{p,r,s}
{q,r,s}
{p,q,r,s}

The three accessible states are thus {p},{q}, and {r,s}. The reason the "dead" state, or empty set is not reachable is that none of the accessible transitions lead to it. Hope this helps!