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Poly-Poly + a[i] * power 6. (15%) Consider the nested loops shown below, where N is assumed to be a power of 2. i.e., N-2 for
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Answer #1

`Hey,

Note: If you have any queries related the answer please do comment. I would be very happy to resolve all your queries.

a)

Outer loop iterations will get over when

N/2^k=1 because at every iteration i=i/2 where k is number of iterations-1

So,

k=log2(N)

So, iterations of outer loop will be log2(N)+1

b)

Outer loop iterations will get over when

i*2^k=N because at every iteration j=j*2 where k is number of iterations-1

So,

k=log2(N/i)

So, iterations of outer loop will be log2(N/i)+1

c)

For i=1, inner loop will run log2(N)+1 times

For i=2, inner loop will run log2(N/2)+1 times

For i=2^2, inner loop will run log2(N/2^2)+1 times

..

..

So, Total=log2(N)+log2(N/2)+......log2(N/2^(k-1))+N=log2(N^k/2^(1+2+3....k-1))+N

=log2(N^(log2(N)+1)/2^(log2(n)*(log2(n)+1)/2))+N

Kindly revert for any queries

Thanks.

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