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I'm a newbee learning DFA minimization. And I found that(strangely) Brzozowki's algorithm cannot give me a minimized DFA on this example: enter image description here In this DFA, $S_0$ and $S_1$ are nondistinguishable and should be merged after minimization, but they are not. The process of minimization is like this: enter image description here I tried to find out why by reading @Hendrik Jan's proof here, Proof of Brzozowski's algorithm for DFA minimization?

However, I find that this case bypasses the proof, because that in the reversed DFA, the initial state $q_0$ cannot be reached by $q_0$ using any words (but initial state is always considered reachable and cannot be removed). So did I miss anything, or does Brzozowki's algorithm has any specification for this corner case?

(It bypasses the proof like this: in the second series of reversal-determinization, $S_0 = \{q_1\}$ and $S_1 = \{q_0, q_1\}$, for any $w$, $\delta(q_0, w)$ in $S_0$ iff $\delta(q_0, w)$ in $S1$, so we say $S_0 = S_1$, assuming all states can be reached by $\delta(q_0, w)$, but $\delta(q_0,w)$ will never be $q_0$)

the initial DFA I was working on: enter image description here

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    $\begingroup$ $S_0$ is not a final state and $S_1$ is. So they cannot be merged. $\endgroup$ – rici Apr 7 at 0:27
  • $\begingroup$ But if we merge $S_0$ with $S_1$ (and the merged one is also a final state), we get the same DFA with fewer states, does it mean that the original DFA (without $S_0$ and $S_1$ merged) is not a minimal DFA? $\endgroup$ – Changda Li Apr 9 at 2:37
  • $\begingroup$ The merged DFA accepts $\epsilon$. The original one does not. So they are not equivalent. $\endgroup$ – rici Apr 9 at 4:23
  • $\begingroup$ Oh, I finally find out my issue. Actually the example I gave above is my "DIY" version trying to make it simpler. And the initial one that I didn't get right is updated in the question. I found out that I used the approach mentioned in Udacity course to do the minimization using Brzozowki's algorithm, which is creating a new state as initial state in the reversed NFA, when there are 2 or more final states. And if two sets are different only on the "created" initial states, I treated them as two different ones, which is not true. I shouldn't add a new state at the begining. $\endgroup$ – Changda Li Apr 11 at 3:54
  • $\begingroup$ Thanks @rici for helping me out, and your explanation makes things clearer! $\endgroup$ – Changda Li Apr 11 at 3:56
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I have looked into some other corner cases and have a tentative answer: when we are doing the determinization after reversal, there is a corner case where in the DFA before reversal, some states cannot be reached by applying any word to the initial state (two possibilities: the initial state cannot go back to itself or we are adding a state as the new initial state, because when we are doing reversal, there are 2 or more final states). In these cases, two sets are considered the same set if the only differences between them are the "unreachable" states before reversal.

I think we can prove that by reversing the proof process in the link cited in the question.

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