求解下述嵌套循环的时间复杂度与渐近符号（Tilde）

Let's break this down step by step, starting with calculating the closed-form of the execution count, then moving to asymptotic notation.

Step 1: Closed-Form Calculation of Execution Count #

You stated the total number of iterations is the sum:

N/2 + (N/2 +1) + (N/2 +2) + ... + (N-1)

This is an arithmetic series where:

• The first term a = N/2
• The last term b = N-1
• Number of terms n = (N-1 - N/2 +1) = N/2 (counting every integer from N/2 to N-1 inclusive)

The formula for the sum of an arithmetic series is:

Sum = (number of terms) * (first term + last term) / 2

Plugging in our values:

Sum = (N/2) * (N/2 + (N-1)) / 2

Simplify the expression inside the parentheses first:
N/2 + N -1 = (3N/2) -1 = (3N -2)/2

Substitute back to get the closed-form:

Sum = (N/2) * (3N -2)/2 / 2 = N*(3N -2)/8 = (3N² - 2N)/8

Let's verify with your example N=100:
(3*(100)² -2*100)/8 = (30000 -200)/8 = 29800/8 = 3725, which matches the sum 50+51+...+99 (calculated as (50+99)*50/2=3725). Perfect!

Step 2: Time Complexity and Asymptotic Tilde Notation #

Big-O Time Complexity #

Time complexity focuses on the dominant term as N grows to infinity. In our closed-form (3N² -2N)/8, the highest-degree term is 3N²/8. Lower-order terms (like -2N/8) become negligible compared to the quadratic term for large N.

We drop the constant coefficient and lower-order terms, so the time complexity is O(N²).

Tilde Notation (Asymptotic Approximation) #

Tilde notation captures the leading term with its exact coefficient, giving a precise asymptotic scaling.

As N → ∞, the -2N term becomes insignificant relative to 3N², so we can write:

Tilde(S) ~ (3/8)N²

This means that for very large N, the number of iterations is approximately (3/8) times N squared, and the ratio of the actual count to this approximation approaches 1 as N grows.

Quick Note on Code vs Stated Execution Count #

There's a small mismatch between the code you provided and the execution count you described. The code:

for (int i = N/2; i < N; i++) { 
    for (int j = i; j < N; j++) { 
        doSomething(i, j); 
    } 
}

would actually result in a sum of 1+2+...+N/2 = N(N+2)/8 (for even N), with a tilde notation of ~N²/8. But since you explicitly specified the execution count as the sum from N/2 to N-1, we focused on that scenario. If you meant the code's actual iteration count, feel free to ask for that derivation!

内容的提问来源于stack exchange，提问作者bls