How would i go about calculating the Big Theta of the following nested loop?

Question

for i = 0 ... log(n) - 1
    for j = 2^i + 1 ... 2^(i + 1)
         [code with Big-Theta(1) running time]

Given this piece of code the outer loop runs log(n) times. But I'm confused as to how to try to calculate the running time of the second loop so as to multiply them. Thank you for reading.

David Taub · Accepted Answer · 2017-12-28T11:05:58.853

1

Each internal loop does $2^i$ iterations of $\Theta(1)$ so it takes $\Theta(2^i)$. Therefore the time complexity of the code is the $\Theta(\Sigma_{i=0}^{log(n)} 2^i)=\Theta(n)$.

You could intuitively think of summation here as writing in binary base a number where all its digits are 1, and has the same number of digits as n, it is linear to the size of n.

edited Dec 28 '17 at 11:05

answered Dec 28 '17 at 09:07

David Taub

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Hi, thanks for your answer. I'm slightly confused as to how you got that the sum from 0 to log(n) is n – Raees Rajwani Dec 28 '17 at 09:10
Added a small explanation. Hope it helps! – David Taub Dec 28 '17 at 09:15

How would i go about calculating the Big Theta of the following nested loop?

1 Answers1