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Let $L$ be a regular language over $\Sigma=\{a,b,c\}$. Build a finite automaton for $L/\{a\}$.

Because $L$ is regular then a DFA exists for it: $A=(\Sigma, Q, q_0, F, \delta)$.

Let $M$ be a finite automaton, $L(M)=L/\{a\}$.

$M=(\Sigma, Q\times\{0,1\}, (q_0,0), \delta', F\times\{1\})$ with the transition function defined below for all $q\in Q, \sigma \in \Sigma$: $$ \delta'((q,0),\sigma)=(\delta(q,\sigma),0)\\ \delta'((q,0),\epsilon)=(\delta(q,a),1) $$

The reasoning is that in state $0$ we just read letters in $M$ because those exact letters also exist in $L$. But when we reach end of input in $M$ via $\epsilon$ we still need to read $a$ in $L$. I'm not sure if it's valid to assume that $\epsilon$ means end of input?

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  • $\begingroup$ Try proving that your construction works. This is how you can verify that it works. $\endgroup$ – Yuval Filmus Jan 29 at 16:16
  • $\begingroup$ To me it seems correct, for example if $aba\in L\implies ab \in L/\{a\}$. Using the derivation rules in the automaton $a$ and $b$ are read in $0$ state. Then the word ends and final $a$ is read in $L$ and the state is changed to $1$ which means we're in accepting state. $\endgroup$ – Yos Jan 29 at 16:20
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First of all, note that you are constructing an NFA, and so $\delta'$ should output a set of possible transitions.

You can prove by induction the following identities: $$ \begin{align*} \delta'((q,0),w) &= \{ (\delta(q,w),0), (\delta(q,wa),1)\}, \\ \delta'((q,1),w) &= \begin{cases} \{(q,1)\} & w = \epsilon, \\ \emptyset & w \neq \epsilon. \end{cases} \end{align*} $$ In particular, $\delta'((q_0,0),w)$ intersects $F \times \{1\}$ iff $\delta(q_0,wa) \in F$. In other words, $w$ is accepted by your NFA iff $wa \in L$, so your automaton is computing $L/a$.

You can simplify your construction by taking the original DFA and simply modifying the set of accepting states, replacing $F$ with $$ F' = \{ q : \delta(q,a) \in F \}. $$ The new automaton accepts a word $w$ iff $\delta(q_0,w) \in F'$ iff $\delta(q_0,wa) = \delta(\delta(q_0,w),a) \in F$ iff $wa \in L$.

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  • $\begingroup$ Does $\delta$ function always produce a set when building an NFA? $\endgroup$ – Yos Jan 30 at 5:34
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    $\begingroup$ Yes, this is the way it is defined. It is a function $Q \times \Sigma \to 2^Q$. $\endgroup$ – Yuval Filmus Jan 30 at 5:47
  • $\begingroup$ and what made my finite automaton an NFA, the fact that I "guess" the end of the word in $L/\{a\}$? $\endgroup$ – Yos Jan 30 at 6:04
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    $\begingroup$ An $\epsilon$-NFA actually has a transition function of type $Q \times (\Sigma \cup \{\epsilon\}) \to 2^Q$. Your automaton has $\epsilon$-transitions, so in particular it must be an ($\epsilon$-)NFA. $\endgroup$ – Yuval Filmus Jan 30 at 6:25
  • $\begingroup$ and in your example of automaton, what made it an NFA? $\endgroup$ – Yos Jan 30 at 6:27

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