如何解决下面的递归关系？

Question

This question came in the examination and I don't know how to do it can anyone help me or give some hint. I think the master's method is not applicable here? Please help.

T(n)=T(n/2)+ θ(logn)

Answer 1

Assuming n is a power of 2 , say n = 2^k , and for simplicity, let's say T(n) = T(n/2) + lg(n) where lg is logarithm base 2 , and T(1) = lg(1) = 0 .

T(n) = lg(n) + lg(n/2) + lg(n/4) + ... + lg(1)
     = lg(2^k) + lg(2^{k-1}) + ... + lg(2^0)
     = k.lg(2) + (k-1)lg(2) + ... + 0.lg(2)
     = (k + (k-1) + ... + 0) lg(2)
     = k(k+1)/2
     = lg(n)(lg(n)+1)/2
     = Theta(lg(n)^2).

For n not a power of 2 , one can note that T is an increasing function, and so T(2^k) <= T(n) <= T(2^{k+1}) where k = floor(lg(n)) . From the exact result above, we get

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

and so

T(n) = Theta(floor(lg(n))^2) = Theta(lg(n)^2)