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I try to understand something:
At Turing machine we have two stats: $q_{accept}$ and $q_{reject}$.

Now, if machine $M$ runs on word $w$ (I hope I write it right...) and the final configuration is: $C(w,q_t,\varepsilon )$,
and $q_t =Q-\{q_{accept},q_{reject}\}$.

Of course $M$ don't accept $w$ and don't reject $w$, but my question is:
We can say that $M$ Stops on $w$?

I try to look for an answer but I didn't found...

Thank you!

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No, Turing machine are defined in a different way than finite-automata: they don't "stop" at the end of the input, they stop whenever they reach a final state $q\in F$. Usually, there are two final states, $q_{acc},q_{rej}$ – if the machine transitions to one of these final states it stops; but until then, it keeps running.

Therefore, $\delta(q,\cdot)$ must be defined for any $q\in Q\setminus F$ and and letter the head points on. This will tell us how the machine behaves when in the state $q_t$ (using your symbol, assuming it is not a final state; see below if it is).


regarding halting without accepting/rejecting: the answer is yes. If, say, $F=\{q_{acc},q_{rej}, q_{inconclusive}\}$, and the machine reached the third final state, we can say it neither accepted nor rejected, but it halts indeed since it reached a final state in $F$. Of course, $F$ can be as large as $Q$. It may make sense for computing other task than yes/no decision problems.

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  • $\begingroup$ What is the "$\cdot$" symbol means? Thank you!! $\endgroup$ – Yoar Apr 14 '16 at 21:04
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    $\begingroup$ @Yoar I meant $\delta(q,b)$ for any state $q\in Q$ and any letter in the tape's alphabet $b\in \Gamma$. Just the standard definition of the transition function. $\endgroup$ – Ran G. Apr 14 '16 at 21:16

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