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Finding the running time of the following algorithm.

$Procedure$ $A(n)$
   $\textrm{ if (n}<=\textrm{2) then return 1 ;}$
   $Return(A(\left \lceil \sqrt{n} \right \rceil))$
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Please refer this

$T(n) = T(\sqrt{n}) +1;$

Let $n = 2^{m} $

$T(2^{m}) = T(\sqrt{2^{m}}) +1;$

$T(2^{m}) = T(2^{m/2}) +1;$

Now let $T(2^{m}) = S(m)$

So $S(m) = S(\frac{m}{2} ) +1$

It is binary search recurrence relation

so $S(m) = O(log m)= O(log log n)$   ($\because n = 2^{m}$)

since $T(n) = T(2^{m}) = S(m)$

T(n) is $O(log log n)$

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