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An Exercise from a Textbook:

L3 := {ε,01,0011}

is L3 ∈ DTime(log(n))

The answer says yes it is, it is even in DTime(1).

How is this possible? I would say it takes at least n steps to check the word. How is it possible to do it in less than n steps?

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To add to old's answer, saying a TM runs in $O(1)$ means that the number of steps the TM makes cannot become infinitely large depending on the size of the input, and is always bounded by some constant. It is true that in this case, for some inputs, the number of steps will be exactly the size of the input, and technically depend on the size of the input. However, this isn't the case for any input of size $>4$, and therefore the $O(1)$ notation is correct.

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Your Turing machine just needs to read the first four symbols one the input tape that will be sufficient to to solve the problem $L3$.

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  • $\begingroup$ yes thats what I thought too and this means n-symbols not log(n) symbols? $\endgroup$ – simplesystems Jun 25 '18 at 9:14
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    $\begingroup$ This means $O(1)$ symbols $\endgroup$ – old Jun 25 '18 at 9:16
  • $\begingroup$ ok, O(1) means constant, but if the input is the empty Word or 01, than it wouldnt have to read 4 symbols? So it would be a bit faster? $\endgroup$ – simplesystems Jun 25 '18 at 9:19

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