7.6k views

Consider the following recurrence:

$T(n)=2T\left ( \sqrt{n}\right )+1,$ $T(1)=1$

Which one of the following is true?

1. $T(n)=\Theta (\log\log n)$
2. $T(n)=\Theta (\log n)$
3. $T(n)=\Theta (\sqrt{n})$
4. $T(n)=\Theta (n)$
edited | 7.6k views
+5
Using some substitutions along with applying master theorem answer seems to be O(logn). But i have a doubt. For n=2 ceil(sqrt(2)) will be 2 always. Since this will never converge, recurrence wont terminate. Any ideas what should be done here.
+2
You are correct. It should be a mistake in question. Case 2 should have been specially mentioned.
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hello, can you give the brief idea what's going in this recurrence ?
+1
B. $\log n$
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I am not clear what you are trying to say
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If someone is getting confused in the  S(m) = 2S(m / 2) + 1  line in jatin's comment below,

then pls see below explanation,

T(2m) = 2T(2m/2) + 1

=>Let , T(2m) = S(m)

=> 2T(2m / 22) + 1

=> 2S(m / 2) + 1 (2m == m & 22 == 2)

$T(n) = 2T\left(n^{\frac{1}{2}}\right)+1$

$= 2\left(2T\left(n^{\frac{1}{2^2}}\right) + 1\right) + 1$

$= 4\times T\left(n^{\frac{1}{2^2}}\right) + 5$

$= 8\times T\left(n^{\frac{1}{2^3}}\right) + 13 \\ \cdots$

$=2^{(\lg \lg n)} + 2 \times \lg \lg n + 1\text{ (Proved below)}$

$= \Theta(\lg n)$

$n^{\frac{1}{2^k}} = 2 \\ \text{(Putting 2 so that we can take log.}\\\text{One more step of recurrence can't change the complexity.)} \\\implies \frac{1}{{2^k}} \lg n = 1 \text{(Taking log both sides)}\\\implies \lg n = 2^k \\\implies k = \lg \lg n$

So, answer is B, $T(n) = \Theta(\log n)$

edited by
+85
$T(n)=2T({\sqrt{n}})+1$

Put, $n=2^m$

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

put,$T(2^m) =s(m)$

$s(m)=2s(m/2)+1$

Using case 1 of master method ,

$=\Theta(m) = \Theta(\log n)$
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@asu what is correct answer here ?
+2
it is B, both methods are fine - it was marked wrongly as A.
+4
\begin{align*} \\ T(n) & =2T(\sqrt{n}) +1 \\ & =2T(n^{\frac{1}{2}})+1 \\ & = 4T(n^{\frac{1}{4}})+3 \\ & = 8T(n^{\frac{1}{8}})+ 7 \\ & = 2^{k} T(n^{\frac{1}{2^{k}}})+ (2^{k} -1 ) \end{align*}

When we put  $k= \log \log n$

$2^{\log \log n}T(2)+2^{\log \log n} -1$
How this is $\Theta( \log n )$  ?
+1
@Arjun sir
I think u have missed 2 in the first step pls check that
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Thanks. Corrected..
+4
@Dulqar since $\alpha^{log_{\alpha}x} = x$, we get $2^{log_{2}log_{2}n} + T(2) + 2^{log_{2}log_{2}n}-1= log_2n + T(2) + log_2n - 1$

Constants are ignored, and we are left with $log_2n$.
+1
But in Official Key of ISRO, answer is Option A
+1

I'm getting same answer- option (b), but my second last step is different. Please someone guide me where am I getting wrong?

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@jatin_saini, are you sure, according to your method  i am getting O(n), can you show your procedure once ..!
+1
here que case 2 ie,T(1)=1 is given as wrong so T(2)=1 is correctone, now you could get the ans
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How you got O(n).

You can check procedure above.
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How to get that m/2 after substituting T(2m) =  S(m)

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i get Θ(n) then how you got log n using master theoram step 1. plz explain this i am confusing only last step.
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I am not getting the second last step of the solution , how T(n ^ (1/2^k)) becomes equal to T(1) as it will be possible only when T(1)=2, but it is given in question that T(1)=1.

0

@dharmesh7

T(n)=2T(√ n )+1

Put, n=2^m

T(2^m)=2T(2^m/2)+1

put,T(2^m)=s(m)

s(m)=2s(m/2)+1

Using case 1 of master method ,

=Θ(m)

because of ,

n = 2^m (take log both of the side )

logn = m log2

m = log n
so
=Θ(logn)

0
thanks

T(n)=2T($\sqrt{n}$)+1

=2T(2m)+1     $\sqrt{n}$=2m , m=log n

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

=O(m)

=O(log n)

0
√n=2^m then m=log n

Logic is not clear.
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see jatin saini's answer. He applied correct logic.

srestha , your answer is correct but i doubt if procedure and logic correct too...

asume n=2k and k=log n

T(2k)=T(2k/2)+1

asume T(2)=S(k)

now S(k)=s(k/2)+1

use master theorm a=1 andb=2

T(2k)=logk

T(n)=log log(n)

0

It is not right solution,bcoz 2 is miss in the equation.

Right equation is , S(k)= 2s(k/2)+1

By Master theorem  ,S(k)= k

then T(n)= logn

Hence ans is B) logn

Unrolling the recursion,

T(n)  =  2T(n^(1/2)) + 1
=  2^2T(n^(1/4)) + 2
= 2^3T(n^(1/8)) + 3
.
.    k  steps
.
=  2^kT(n^(1/2k)) + k              …………. (1)

Using the Base case,

n^(1/2k) = 2
Taking log on both sides
log2n = 2k
k = log2log2n

From (1),

T(n) =  log2n  +  log2log2n
= Theta(log2n)

Here log2n : log(base 2) n

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