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+16 votes
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The running time of the following algorithm

Procedure $A(n)$

If $n \leqslant 2$ return ($1$) else return $(A( \lceil  \sqrt{n}  \rceil))$;

is best described by

  1. $O(n)$
  2. $O(\log n)$
  3. $O(\log \log n)$
  4. $O(1)$
asked in Algorithms by Veteran (68.8k points)
edited by | 1.7k views
how to solve this question..?

2 Answers

+34 votes
Best answer
The complexity will be the number of times the recursion happens which is equal to the number of times we can take square root of n recursively, till n becomes $2$.

$T(n) = T\left(\lceil \sqrt n \rceil \right) + 1$

$T(2) = 1$
$T\left(2^2\right) =  T(2) + 1 = 2$
$T\left(2^{2^2}\right) =  T(4) + 1 = 3$
$T\left(2^{2^3}\right) =  T(16) + 1 = 4$

So, $T(n) = \lg\lg n + 1 = O\left(\log \log n\right)$
answered by Veteran (332k points)
edited by
why is it +1 and not some constant c in recurrence relation
why we are adding constant in recurrence relation?
because u will do some constant amount of work at every recurrence

Read this article once, then you may not need to solve this again ever, it becomes very intuitive. 

https://stackoverflow.com/questions/16472012/what-would-cause-an-algorithm-to-have-olog-log-n-complexity

+9 votes
log log n.
substitute root n = m then proceed.
answered by Veteran (49.2k points)


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