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So, I was watching a video about the conversion of $\epsilon$-NFA to a DFA. In the resulted DFA, she didn't write the state 4 in any successor set of the sets containing the state 3, and her explanation was that this happens because the state 3 has no outgoing transitions that are labeled with letters from $\Sigma$ in the input NFA. See the following image.

enter image description here

So my question is, if the state 3 had an outgoing transition labeled with 0 in the input NFA (see the 2nd image below), then does the state 3 contribute the states 2,4,5, and 6 in the successor set upon reading the letter 0?

enter image description here

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  • $\begingroup$ What do you mean by take a state while inputing? Can you explain the question in more detail? $\endgroup$ – Bader Abu Radi Jan 3 at 9:36
  • $\begingroup$ @BaderAbuRadi I already linked the video at a specific time where she explains this but anyway I'll edit my question to explain in more detail. $\endgroup$ – Moki Jan 3 at 9:55
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What she actually describing is the following construction. For an $\epsilon$-NFA $ A = \langle Q, \Sigma, \delta, Q_0, F \rangle$, the intuitive idea is as follows. Consider a state $q$. If there is an $\epsilon$-transition from q to $s$, then whenever we reach q, we can also reach $s$. Yet, there may be $\epsilon$-transitions from $s$, so we also need to take them into account. For every state $q\in Q$, let $EC(q) \in 2^Q$ be the $\epsilon$-closure of $q$ which is formally defined as $EC(q) = \{s \in Q: \text{ $s$ is reachable from $q$ using only $\epsilon$-transitions}\}$. In particular, for every state $q\in Q$, $q\in EC(q)$ as a state is reachable from itself by a path of length 0.

Now Given the sets $EC(q)$ for every state $q \in Q$, we define the DFA B as follows. $B = \langle 2^Q, \Sigma, \kappa, \bigcup\limits_{q\in Q_0} EC(q), \{S\in 2^Q: S\cap F \neq \emptyset\}\rangle$, where $\kappa(T, \sigma) = \bigcup\limits_{q\in T} \bigcup\limits_{s\in \delta(q, \sigma)} EC(s)$, for every set of states $T\in 2^Q$, and letter $\sigma \in \Sigma$. That is, given a non-$\epsilon$ transition $\langle q, \sigma, s \rangle$, we move not only to $s$, but also to all the states that are reachable from $s$ via $\epsilon$-transitions.

So indeed, in the first attached image, the state $3$ has no outgoing transitions that are labeled with letters from $\Sigma$, and thus, for every set $T$ that contains state $3$, the state $3$ does not contribute any states in the set $\kappa(T, \sigma)$, for every letter $\sigma$. However, in the 2nd attached image, there is a 0-labeled transition going out from the state 3. So, in the DFA $B$, when you read the letter 0 from a set $T$ that contains the state $3$, then the state $3$ contributes the states $\kappa(\{3\}, 0)$ in $\kappa(T, 0)$. Now since $\delta(3, 0) = \{4\}$, we have that $\kappa(\{3\}, 0) = EC(4) = \{2, 4, 5, 6\}$, that is, you're right that the state $3$, upon reading the letter 0, contributes the states $2, 4, 5, $ and $6$.

Important note: she is doing $\epsilon$-transitions removal and the subset construction at the same time. I don't like that at all as it is confusing, also I don't see the formal construction itself in the video. I suggest that you try to do $\epsilon$-removal alone (without determinization). That may give you a better understanding as a start.

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