How to solve this recurrence relation: T(n) = 4*T(

Question

I know how to solve the recurrence relations using Master Method. Also I'm aware of how to solve the recurrences below:

T(n) = sqrt(n)*T(sqrt(n)) + n

T(n) = 2*T(sqrt(n)) + lg(n)

In the above two recurrences there is same amount of work at each level of the recursion tree. And there are a total of log log n levels in the recursion tree.

I'm having trouble in solving this one: T(n) = 4*T(sqrt(n)) + n

EDIT: Here n is a power of 2

Answer 1

   T(n) = 4 T(sqrt(n)) + n
   4 [ 4 T(sqrt(sqrt(n) + n ] + n
   4^k * T(n^(1/2^k)) +kn because n is power of 2.
   4^k * T(2^(L/2^k)) +kn   [  Let n = 2^L , L= logn]
   4^k * T(2) +kn   [  Let L = 2^k,  k = logL = log log n]
   2^2k * c +kn
   L^2 * c + nloglogn 
   logn^2 * c + nloglogn
   = O(nloglogn)

Answer 2

Suppose that n = 2^k. We have T(2^k) = 4*T(2^(k/2)) + 2^k. Let S(k) = T(2^k). We have S(k) = 4S(k/2) + 2^k. By using Mater Theorem, we get S(k) = O(2^k). Since S(k) = O(2^k) and S(k) = T(2^k), T(2^k) = O(2^k) which implies T(n) = O(n).

Answer 3

I'm having trouble in solving this one: T(n) = 4*T(sqrt(n)) + n

EDIT: Here n is a power of 2

This edit is important. So lets say that the recurrence stops at 2.

So the question now is how deep the recursion tree is. Well, that is the number of times that you can take the square root of n before n gets sufficiently small (say, less than 2). If we write

n = 2^{lg n}

then on each recursive call n will have its square root taken. This is equivalent to halving the above exponent, so after k iterations we have that

n^1/(2k) = 2^{lg n/(2k)}

We want to stop when this is less than 2, giving

2^{lg n/(2k)} = 2

lg n/(2^k) = 1

lg n = 2^k

lg lg n = k

So after lg lg n iterations of square rooting the recursion stops. (source)

For each recursion we will have 4 new branches, the total of branches is 4 ^ (depth of the tree) therefore 4^(lg lg n).

EDIT:

Source