递归函数 g(N) = (1/c)g(N/2) 的 Big-Theta(Θ)

Question

I'm trying to analyse the recursive function g(N) = (1/c)g(N/2), for N >= 2, with the terminating case being g(1) = 1, and then find the big-theta bounding it. Bare in mind, this problem treats n as if it were some power of 2, so N = 2 ⁿ for example, which means n = lg(N).

What I found when doing this has me questioning whether or not I'm correct in my assumption. The i ^th case expands to g(N) = (c ^-i )g(2 ^{n - i} ). Then set n - i = 0, making i = n and g(N) = (c ^-n )g(2 ^{n - n} ). Finally, g(2 ⁰ ) = 1, so g(N) = c ^-n = c ^-lg(N) = N ^-lg(c) = (1/N) ^lg(c)

What I gather out of this is that (1/N) ^p = Θ(0), since the limit of (1/N) ^p as N -> ∞ is equal to 0.

I'm really just wondering if I'm doing this right, since in real world applications I wouldn't think it's possible to have an algorithm with a time complexity of 0.

Answer 1

You write

real world applications... algorithm with a time complexity...

so it seems that the recurrence g(n) = (1/c)g(n/2) stands for the running time of a real-world algorithm. As such, this makes sense only if 0 < c < 1 , since, otherwise, it's an algorithm that works faster on larger inputs, which doesn't make much sense in the real world.

Your analysis makes sense, up to the point where you analyze Θ( ^-lg(c) ) , and decide this tends to 0. For the reasonable values of c , it naturally enough tends to infinity.