# Time Complexity for Recurrence relation

687 views
What will be the time complexity for the following recurrence relation?

$T(n) = 8\sqrt{n} T(\sqrt{n})+(log n)^{2}$

According to me it is $\Theta (n(logn)^{3})$ . Please confirm.

Correct me if i am wrong 2

Thanks for the answer. I am having a little difficulty in understanding. Please have a look at my answer using Master Theorem. 1
we cannot apply master's theorem here because the no of subproblems should be >=1 or (a>=1).
1

when you reduce size of subproblem by log in s function why you have not taken log of m^2.

it should be ,P(m)=8P(m/2)+2logm/m

0
@arnab ...could you please elaborate last step ...
1

n(1/2+1/4+1/8+1/8....1/2^k) < n(1/2+1/4+1/8+...) = n(0.5/0.5)= n

T(n) <= 8k*n

T(n) = O(n(logn)3)
I hope that you have understood now

0

## Related questions

1
421 views
What is the time complexity of the following recurrence relation and step to derive the same $T(n) = T(\sqrt{n}) + log(logn)$
recurrence relation for the functional value of F(n) is given below : $F(n) = a_{1}F(n-1) + a_{2}F(n-2) + a_{3}F(n-3) + ....... + a_{k}F(n-k)$ where $a_{i} =$ ... $O(k_{2}r^{k_{1}n})$) B. Linear ( $O(n)$ ) C. Logarithmic ( $O(\log n)$ ) D. $O(n \log n)$