Вопрос

I came across this weird recurrence equation:

T(n,h) = T(n/2, h1) + T(n/2, h-h1) + nh

and:

T(1,h) = O(h)

I need to find the asymptotic upper bound. I have never come across a recurrence relation with two arguments.

After much searching I found this set of slides which indicates to me that it is related to computational geometry.

It would be nice if someone could point me in the right direction.

Это было полезно?

Решение

Assuming that n is a power of 2, 0 <= h1 <= h, T(0, h) = 0, and T(1, h) = h, an upper bound is 2nh by the following inductive proof.

Basis: T(0, h) <= 0 <= 2(0)h, and T(1, h) <= h <= 2(1)h.

Inductive step: T(n, h) = T(n/2, h1) + T(n/2, h - h1) + nh <= 2(n/2)h1 + 2(n/2)(h - h1) + nh <= 2nh, regardless of h1.

Другие советы

Assuming the first recursive call includes h instead of h1, and replacing h1 by c (for the sake of readability), I decided to proceed like the following:

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