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I have looked up a question like this online and found this solution. I couldn't understand what the new NFA A' looks like.

Could anyone give me an example of NFA A'?

Sloan Kim
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  • The best way to understand what $A'$ looks like is to start with some $A$ and to construct $A'$ according to the very clear explanation in the answer. I don't see what's stopping you from doing this. – Yuval Filmus Sep 20 '20 at 09:34
  • @YuvalFilmus So far my learning is not complete to the extent where I can create A' from the very clear explanation in the answer. – Sloan Kim Sep 20 '20 at 09:59
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    Please credit the original source of all copied material. – D.W. Sep 20 '20 at 17:58
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    @SloanKim The beauty of formal notation, as used in the proof, is that you don't need to fully understand a construction to replicate it. In order to try an example, you can (a) start with a DFA, (b) write the DFA down in formal notation, using the automaton tuple and defining its contents, then (c) writing down the new NFA in the same same formal notation, using the definition of the new set Delta, and (d) optionally visualize the NFA. If step (c) is too difficult, then you'll need to exercise your use of formal notation a bit, independently of learning about DFAs and NFAs. – DCTLib Sep 21 '20 at 07:29

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Maybe the formal proof below will help you to understand the construction using a NFA.

Let $\mathcal{A} = (Q, A, \cdot, i, F)$ be a deterministic automaton recognising your language. For each pair of states $(p, q)$, let $L_{p,q}$ be the language recognised by the automaton $(Q, A, \cdot, p, \{q\})$. Thus $L_{p,q}$ is the set of labels of all paths from $p$ to $q$. By construction, each language $L_{p,q}$ is regular. Let $T$ be the set of transitions of $\mathcal{A}$, that is, $T = \{ (p,a,q) \in Q \times A \times Q \mid p \cdot a = q \}.$ Let $$ R = \bigcup_{(p,a,q) \in T} \bigcup_{f \in F} L_{i,p}L_{q,f} $$

Claim. $D(L) = R$.

Indeed, let $w \in D(L)$. By definition, there exist some words $u$ and $v$ and a letter $a$ such that $w = uv$ and $uav \in L$. Let $p = i\cdot u$, $q = p \cdot a$ and $f = q \cdot v$. By construction, $u \in L_{i,p}$, $v \in L_{q,f}$, $(p,a,q) \in T$ and $f \in F$. Thus $w \in R$.

Let now $w \in R$. Then there exist some states $p, q \in Q$ and $f \in F$ and some transition $(p,a,q) \in T$, such that $w \in L_{i,p}L_{q,f}$. It follows that $w = uv$ for some $u \in L_{i,p}$, and $v \in L_{q,f}$. Now $i \xrightarrow{u} p \xrightarrow{a} q \xrightarrow{v} f \in F$, which means that $uav \in L$. Consequently, $w \in D(L)$, which proves the claim.

Coming back to your question, the first copy of $\cal A$ in ${\cal A}'$ is used to accept $L_{i,p}$ and the second one to accept $L_{q,f}$. The $\epsilon$-transitions are used to pass from the first copy to the second one.

J.-E. Pin
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  • I don't get how the $\epsilon$-transitions correspond to skipping a symbol, since by definition, an $\epsilon$-transition does not consume/read any symbol from the input. Your elaboration on this is highly appreciated. – ultrajohn Oct 24 '23 at 13:17