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So, this question is taken from here

 

https://gateoverflow.in/151057/langle-rangle-there-exist-input-whose-length-than-which-halts%24

 

$L=\{<M>| $M is a Turing machine and there exists an input whose length is less than 100 on which M halts$\}$

I got to know that for semi-decidability, since we have finite number of strings of length less than 100, for all such strings, we run TM in interleaved mode and if for any input TM halts, we say Yes.But my question is, for how long will we wait?

Suppose TM may take too long before it says yes, can we be sure whether we are really reaching an answer or TM has got into infinite loop?

in Theory of Computation by Boss (27.2k points) | 63 views
0
It should be accepting the strings of that language. If we give any string other than the language it may or may not halt i.e. recursively enumerable.

In the worst case scenario I would run one TM for each string as we have finite number of strings.
0

For running in interleaved mode they use $dovetailing$.

I don't know what the process does. You can chek it here

Similar ques : https://gateoverflow.in/72527/re-language-interleaved-mode

0

@Shobhit Joshi

yes RE, $\epsilon \subset \Sigma ^{*}$. So, 2nd Rice theorem not applicable here

https://gatecse.in/rices-theorem/

this one too RE

https://gateoverflow.in/283972/self-doubt

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