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The following theorem from Michael Sipser's book "Introduction to the Theory of Computation" states:

$A_{\textrm{LBA}}= \{ \langle M, w \rangle \mid \text{$M$ is an LBA that accepts string $w$} \}$.

THEOREM: $A_{\mathrm{LBA}}$ is decidable.

On the proof part, it states:

The idea for detecting when $M$ is looping is that as $M$ computes on $w$, it goes from configuration to configuration. If $M$ ever repeats a configuration, it would go on to repeat this configuration over and over again and thus be in a loop.

I do not understand this: "If $M$ ever repeats a configuration, it would go on to repeat this configuration over and over again". What if $M$ only repeat one configuration, then halts?

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The machine $M$ is deterministic. This means that, if $M$ is in a certain configuration $c$, then there is a single fixed configuration $c'$ (determined by the rules of $M$) which the execution of one step will lead it to. If $M$ ever reaches the configuration $c$ again, then the configuration $c'$ will follow no matter what. Hence, if the computation of $M$ causes it to be assume configurations $c_1, \dots, c_n$ and $c_n = c_1$, then $M$ will repeat the loop $c_1, \ldots, c_n$ indefinitely.

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  • $\begingroup$ What if we change the $M$ to a normal deterministic turing machine, which is allowed to move off the input portion. if $C_{1}$ = $C_{n}$, does $M$ will repeat the loop too? $\endgroup$ – Anonemous Aug 7 at 15:05
  • $\begingroup$ "Configuration" comprises the contents of all tapes of $M$ as well as the positions of every head and $M$'s internal state. Hence, it does not matter. $\endgroup$ – dkaeae Aug 7 at 15:56
  • $\begingroup$ If $M$ goes to an infinite loop on $w$, does it mean that : $w$ does not belong to the language which $M$ recognize? $\endgroup$ – Anonemous Aug 8 at 2:10
  • $\begingroup$ You should review the definition of Turing-recognizable in Sipser's book. (In the 3rd edition I have here at hand it's Definition 3.5, p.170; see also the discussion in p. 169.). If $M$ accepts $w$, then it must halt in the accept state; if it loops, then it doesn't even halt. $\endgroup$ – dkaeae Aug 8 at 7:18
  • $\begingroup$ So, if $M$ does not halt on $w$, then $w$ does not belong to the language of $M$, right? $\endgroup$ – Anonemous Aug 8 at 7:27

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