thanks in advance for your help in figuring this out. I'm taking an algorithms class and I'm stuck on something. According to the professor, the following holds true where C(1)=1 and n is a power of 2:
C(n) = 2 * C(n/2) + n
resolves to C(n) = n * lg(n) + n
C(n) = 2 * C(n/2) + lg(n)
resolves to C(n) = 3 * n - lg(n) - 2
The first one I completely grok. As I understand the form, what's stated is that C(n) resolves to two sub-problems, each of which requires n/2 work to solve, and an additional n amount of work to split and merge everything. As such, for every division of the problem, the constant 2 is increased by a factor of ^k (where k is the number of splits), the 2 in n/2 is also increased by a factor of ^k for much the same reason, and the last n is multiplied by a factor of k because each split creates a multiple of k extra work.
My confusion stems from the second relation. Given that the first and second relations are almost identical, why isn't the result of the second something like nlgn+(lgn^2)?
The general result is the Master Theorem
But in this specific case, you can work out the math for a power of 2:
C(2^k)
= 2 * C(2^(k-1)) + lg(2^k)
= 4 * C(2^(k-2)) + lg(2^k) + 2 * lg(2^(k-1))
= ... repeat ...
= 2^k * C(1) + sum (from i=1 to k) 2^(k-i) * lg 2^i
= 2^k + lg(2) * sum (from i=1 to k) 2^(i) * i
= 2^k - 2 + 2^k+1 - k
= 3 * 2^k - k - 2
= 3 * n - lg(n) - 2