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[Solved]: DFA's with more than one initial state

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( This may be related to NFAs with more than one initial state )

Consider a dfa $M$ with alphabet $\Sigma$ and states $Q$, typically also characterized by a specific initial state $q_0\in Q$. As the above-cited question points out, nfa's that nondeterministically select among several possible initial states can be converted to equivalent (wrt to their accepted language) nfa's with a unique initial state. And I'm asking if the same is true of dfa's, as follows.

Instead of $M$'s input being only a tape/string $s\in\Sigma^\ast$, let its input be $(q_0,s)\in Q\times\Sigma^\ast$, i.e., you give it the initial state $q_0\in Q$ that you want it to start in, as well as a string.

I'd think each possible $q_0\in Q$ defines a different language, call it $L_{q_0}\subseteq\Sigma^\ast$, accepted by $M$. But $Q$ can possibly be partitioned into equivalence classes, where all the states in a class accept the same $L$. But if there's more than one such class, then no single dfa with a unique $q_0$ starting state could represent this $M$.

So, assuming that's more-or-less correct, my additional question is how to characterize the equivalence classes. That is, given $M$'s transition function, $\delta:\Sigma\times Q\to Q$, how would you determine whether $q\equiv q^\prime$ belong to the same class (besides, that is, individually working out the language each accepts)? Is there a more elegant way to do that?

Asked By : John Forkosh

Answered By : David Richerby

As Lukas points out, having multiple initial states is something of a red herring. You've defined your automaton to accept a subset of the strings that match the regular expression $Q\Sigma^*$ and that regular expression matches the language accepted by some ordinary DFA using the usual techniques. (Lukas gives an explicit construction in terms of the original automaton.)

For equivalence, there is a standard polynomial-time technique used for state minimization for DFAs. Luca Trevisan has a good set of notes (PDF) on this, which I'll briefly summarize. Let the automaton have state set $Q$, alphabet $\Sigma$ and transition function $\delta$.

For a state $q$ and string $w$, write $\delta^*(q,w)$ for the state that you reach by starting at $q$ and reading $w$. For states $p$ and $q$, write $p\equiv q$ to mean that $\delta^*(p,w)$ and $\delta^*(q,w)$ are either both accepting or both rejecting. This gives the same relation $\equiv$ that you give in the question. We'll show how to compute $\equiv$.

To do this, for any $n\geq 0$, define $p\equiv_n q$ in the same way as $p\equiv q$, except restricting $w$ to be of length at most $n$. We'll compute the relations $\equiv_n$ inductively, and show that knowing a finite number of them is the same as knowing $\equiv$.

First, note that $p\equiv_0 q$ if, and only if, $p$ and $q$ are both accepting or both rejecting so we can compute $\equiv_0$ easily. Now, suppose we've already computed $\equiv_n$. We claim that $p\equiv_{n+1}q$ if, and only if, $p\equiv_n q$ and $\delta(p,u)\equiv_n\delta(q,u)$ for all $u\in\Sigma$. Obviously, if $p$ and $q$ "agree" on all strings of length at most $n+1$, they must also agree on all strings of length at most $n$. For the second part, consider a string $uw$ where $|u|=1$ and $|w|=n$. We're saying that, for $p$ and $q$ to agree on $uw$, it must be that, after reading $u$, they move to states that agree on $w$.

This allows us to construct the relations $\equiv_n$ for all $n$. Finally, we claim that $\equiv_{|Q|}$ is actually the same relation as $\equiv$. To see this, observe that every $\equiv_n$ is an equivalence relation and, furthermore, the procedure from moving from $\equiv_n$ to $\equiv_{n+1}$ can only split equivalence classes into smaller ones: it can never merge them. Also, if $\equiv_n$ and $\equiv_{n+1}$ are the same relation, then $\equiv_{n+i}$ is the same relation for all $i\geq 0$. $\equiv_0$ has at most two equivalence classes. Every step of the process either splits at least one equivalence class or is the end of the game: it doesn't split any class and nor does any future step. We can't split classes more than $|Q|$ times because, if we did, we'd have more equivalence classes than states, which is impossible. So it must be that $\equiv_{|Q|}$ is the same thing as $\equiv$ and we're done.

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Question Source : http://cs.stackexchange.com/questions/62490

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