Big O符号问题求助：如何证明教授关于时间复杂度的结论有误？

Hey there! Let's walk through this problem clearly—since you're building up your CS fundamentals while coming from a UI dev and English lit background, I'll make sure the reasoning is straightforward but rigorous.

First, let's formalize the problem based on the common scenario that leads to a Θ(n log n) result (since you mentioned your class is landing on that answer): we're assuming the pseudocode follows a recursive structure like this:

function recursiveFunc(n):
    if n <= 1:
        return base case
    f(n)  // Θ(n) execution time
    recursiveFunc(n/2)
    recursiveFunc(n/2)

结论 #

The total asymptotic running time of this code is Θ(n log n).

证明方法1：递归树分析法 #

Let's break down the time cost at each level of the recursion tree:

• Level 0 (top level): We call f(n) once, which takes Θ(n) time.
• Level 1: We have two recursive calls to recursiveFunc(n/2), each calling f(n/2). The total time here is 2 * Θ(n/2) = Θ(n) (the 2 and denominator cancel out).
• Level 2: Four calls to recursiveFunc(n/4), each with f(n/4). Total time: 4 * Θ(n/4) = Θ(n).
• ... and so on: Every level of the recursion tree will have a total time cost of Θ(n).

Now, how many levels are there? Since we keep dividing n by 2 until we hit the base case (n ≤ 1), the number of levels is log₂n (this is the number of times you can halve n to get to 1).

Multiplying the time per level by the number of levels gives us: Θ(n) * log₂n = Θ(n log n) (the constant factor from log base 2 doesn't matter for asymptotic notation).

证明方法2：主定理验证 #

We can also confirm this using the Master Theorem, which applies to recursive functions of the form:
T(n) = aT(n/b) + f(n)
For our problem:

• a = 2 (two recursive calls per level)
• b = 2 (each call processes n/2 elements)
• f(n) = Θ(n)

The Master Theorem's second case applies here: when f(n) = Θ(n^log_b a). Calculating log_b a gives log₂2 = 1, so n^1 = n, which matches f(n) = Θ(n).

By the second case of the Master Theorem, this means T(n) = Θ(n log n).

That's why your class is landing on the n log n result—both recursive tree analysis and the Master Theorem confirm it!

内容的提问来源于stack exchange，提问作者jakePhilly