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I have been working in the time analysis for an algorithm and finally I got a curve that fits:

$O(2^{(\log_2(N)^{2.01})})$

N is the number of elements.

I'm right to say the above time complexity is quasi-polynomial?

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  • $\begingroup$ What "elements"? That is crucial here. (Also, don't use "time complexity"; you have an asymptotic upper bound.) $\endgroup$ – Raphael Dec 3 '14 at 11:29
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    $\begingroup$ Oh, I just saw this: "I got a curve that fits" -- oh no, don't do that! What you are doing is not analysis but fortune-telling. If you want to learn about "real" algorithm analysis, you may want to head over to our reference questions on the matter. $\endgroup$ – Raphael Dec 3 '14 at 16:04
  • $\begingroup$ @Raphael log2 = log (base 2) (i.e. Log2(32) = 5 ), N = input size (number of input elements). $\endgroup$ – Jesus Salas Dec 3 '14 at 17:57
  • $\begingroup$ @Raphael I was focusing to double check the resultant time complexity. There is no curve that fits" the result is an asymptotic worst case analysis for an algorithm $\endgroup$ – Jesus Salas Dec 3 '14 at 17:59
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    $\begingroup$ "input size" and "number of input elements" is not necessarily the same. Which size does each element have? Are they numbers? $\endgroup$ – Raphael Dec 3 '14 at 18:00
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Quasi-polynomial means different things to different people, but in many contexts it means a running time of the form $O(2^{O(\log^{O(1)} N)})$, to which your example conforms.

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  • $\begingroup$ Is $N$ for you the size of the input (encoding), or what is it? $\endgroup$ – Raphael Dec 3 '14 at 11:31
  • $\begingroup$ $N$ is always the size of the input, unless stated otherwise. $\endgroup$ – Yuval Filmus Dec 3 '14 at 15:59
  • $\begingroup$ I only wish that were true. That said, note that the OP binds $N$ to be the "number of elements". $\endgroup$ – Raphael Dec 3 '14 at 16:01

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