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Let $X = \{\langle M \rangle\ |\ M\text{ is a finite state machine and }L(M) = \emptyset\}$ where $\langle M \rangle$ is an encoding of the machine $M$. Is $X$ Turing decidable? Why or why not?

Raphael
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Joe
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4 Answers4

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Re-read your textbook chapter on DFAs and NFAs -- you'll find everything you need to know there.

See also the following questions, which both have answers to your question embedded in their answers (make sure to read carefully):

Next time, tell us what you have tried, to solve the exercise on your own.

Community
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D.W.
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Emptiness problem for DFAs: Given a DFA $D$ determine if $D$ accepts any strings at all, i.e. if $L(D) = ∅$.

The language:

$\qquad E_{DFA}$ = {$\langle D \rangle$ | $D$ is a DFA and $L(D) = ∅$}

Idea for the Turing Machine for $E_{DFA}$:

$\qquad$ Check if any of the accept states are reachable from the start state.

Algorithm for $E_{DFA}$ on input $ D=(Q,Σ,δ,q_0,q_A)$:

$\qquad$ 1) If $D$ is not proper encoding of DFA, reject.

$\qquad$ 2) Mark the start state of $D, q_0$.

$\qquad$ 3) Repeat until no new states are marked:

$\qquad$ $\qquad$ a) Mark any states that can be δ-reached from any marked state.

$\qquad$ 4) If no accept state is marked, accept. Else reject.

Hence, $E_{DFA}$ is decidable as there exists a valid algorithm for it.

Nabhan Abdulla
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So the question you are asking is basically, is there an algorithm that can decide is a DFA accepts no word. I assume you know something about minimizations of DFAs. And you probably know a method how to get from a DFA to a minimal DFA. Such a method can be carried out by a TM.

Now ask yourself, what is the minimal DFA $Y$ for $\emptyset$. Got it? You are left with finding an algorithm that compares the minimization of $ X $ with $Y$, but thats not too hard.

As a remark, the minimization trick works in general if you want to check if two DFAs accept the same language.

A.Schulz
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    That's an extremely complicated way of doing it! Reachability of an accepting state is much simpler. – David Richerby Dec 20 '13 at 07:39
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    Well it is basically applying what you know about minimization. We are not talking about the most efficient method here. My method teaches the OP a more general concept of how to handle similar problems. – A.Schulz Dec 20 '13 at 07:47
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Show that an NFA accepts at least one word if and only if there is a (directed) path from the initial to an accepting state.

Then it's easy to decide the property, e.g. by BFS from the initial state.

Raphael
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