递归关系：T（n）= 2T（n / 2）+ log（n）

Question

I have a recurrence relation which is like the following:

T(n) = 2T(n/2) + log ₂ n

I am using recursion tree method to solve this. And at the end, i came up with the following equation:

T(n)=(2log ₂ n)(n-1)-(1*2 + 2*2 ² + ... + k*2 ^k ) where k=log ₂ n.

I am trying to find a theta notation for this equation. But i cannot find a closed formula for the sum (1*2 + 2*2 ² + ... + k*2 ^k ). How can i find a big theta notation for T(n)?

Answer 1

If you know some calculus you should be able to solve that easily.

1 + x + x^2 + ... + x^(n+1) = (x^(n+2) - 1) / (x - 1)

Multiplying by x, x + x^2 + x^3 + ... + x^(n + 2) = (x^(n + 3) - x) / (x - 1)

Differentiating the LHS will give you your series for x = 2. Differentiating the RHS will give you the closed form.

Answer 2

您应该使用主定理来计算您的复杂度，这会容易得多。

Answer 3

I suppose you are doing some mistake, as per my calculations:

T(n)=(k-1)*log(n/(2^(k-1)))+2^k*T(n/2^k).

Put k=log(n)

I can post an image of my solution if you like. :)

Answer 4

These recurrence can be solved with a masters theorem . In your case a = 2 , b = 2 and therefore c = logb(a) = 1 .

Your f(n) = log n and because n^c grows faster than your f , it dominates the solution and you fall in the first case. So the complexity is O(n) .