GATE CSE 2002 | Question: 2.11

Question

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

• A. $O(n)$
• B. $O(\log n)$
• C. $O(\log \log n)$
• D. $O(1)$

Comments

how to solve this question..?

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https://math.stackexchange.com/questions/538268/can-you-help-me-to-solve-the-recurrence-relation-tn-t-sqrt-n-1

Answer 1

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, $T(n) = \lg\lg n + 1 = O\left(\log \log n\right)$

Answer : Option C

Comments

why is it +1 and not some constant c in recurrence relation

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why we are adding constant in recurrence relation?

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because u will do some constant amount of work at every recurrence

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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

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@Hemant Parihar why this is not following O(lognlogn) complexity though problem is diveded into √n each time?

https://gateoverflow.in/197521/recurrence-relation

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@soumayan bandhu because in the question you mentioned square root is calculated twice at each step.

Answer 2

T(n)=T(sqrt(n))+1

We substitute n=2^m. Then T(2^m)=T(2^m/2) +1

we substitute T(2^m)=S(m).

S(m)=S(m/2)+1.

Now applying masters theoram we get S(m)=O(log m). As m=logn.

T(n)=O(log log n)

Comments

@teluguenglish You wrote k = 1 which is wrong, k should be 0 as 1 = n^0.

Then you will get S(m) = O(logm) by extended master's theorem.

As n = 2 ^ m, we will get m = log n.

Then T(n) = S(m) = log(m) = log(log(n)) = loglogn.

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Why is 1 added here T(n)=T(sqrt(n))+1 , in original question we don't have it ?

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@KavitaRana2406 Every recursive call contributes one constant step, so we always add +1 (or +c) even if not printed in the code.

Answer 3

Another way to solve this is:

T(n) = T(n^1/2)+c

T(n) = T(n^1/4)+c+c

T(n) = T(n^1/8)+c+c+c

T(n) = T(n^1/16)+c+c+c+c

................

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T(n) = T(n^(1/2^k))+k.c

So, n^(1/2^k) = 2

taking log will give us:-      log(n) = 2^k--------------futher taking log --------- k =loglogn

T(n) = T(2)+loglogn . c

so , ans will be C) O(loglogn)

Comments

best explanation...@lakshaysaini2013

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This is best

Answer 4

$Ans:C$

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$T(n) = T(\sqrt n)  + c$

Assume $n=2^k$

$T(2^k)=T(2^{k/2})+c$

Assume $T(2^k)=S(k)$

$S(k)=S(k/2)+c$

Apply Master Theorem

$S(k)=\Theta(1.\log\ k)$

Now just do the reverse

$T(2^k)=\Theta(\log\ k)$

Now replace with $k=\log_{2}n$

$T(2^{\log_{2}n})=\Theta(\log(\log_{2}n))$

$T(n)=\Theta(\log(\log_{2}n))$

Comments

Guys please upvote this, best of the best answer!!!!

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how did you applied master's theorem?
can you please tell me

Answer 5

log log n.
substitute root n = m then proceed.

Answer 6

We will solve this question using recursive tree method