为什么排序不及时获取O(n log(n)) [英] Why is sorting not taking O(n log (n)) in time

问题描述

在以下代码段中，std::sort的时间消耗.这应该需要 O(nlog(n))时间. std::chrono仅用于测量std::sort.

In the following snippet the time consumption of std::sort. This should take O(nlog(n)) time. std::chrono is used solely measuring std::sort.

我使用优化级别为-O3的Intel编译器18.0.3编译了以下代码.我使用Redhat6.

I compiled the following code with the Intel compiler 18.0.3 with optimization level -O3. I use Redhat6.

#include <vector>
#include <random>
#include <limits>
#include <iostream>
#include <chrono>
#include <algorithm>

int main() {
    std::random_device dev;
    std::mt19937 rng(dev());
    std::uniform_int_distribution<std::mt19937::result_type> dist(std::numeric_limits<int>::min(),
                                                                  std::numeric_limits<int>::max());

    int ret = 0;

    const unsigned int max = std::numeric_limits<unsigned int>::max();
    for (auto j = 1u; j < max; j *= 10) {
        std::vector<int> vec;

        vec.reserve(j);

        for (int i = 0; i < j; ++i) {
            vec.push_back(dist(rng));
        }

        auto t_start = std::chrono::system_clock::now();
        std::sort(vec.begin(), vec.end());
        const auto t_end = std::chrono::system_clock::now();
        const auto duration = std::chrono::duration_cast<std::chrono::duration<double>>(t_end - t_start).count();
        std::cout << "Time measurement: j= " << j << " took " << duration << " seconds.\n";
        ret + vec[0];
    }
    return ret;
}

该程序的输出为

Time measurement: j= 1 took 1.236e-06 seconds.
Time measurement: j= 10 took 5.583e-06 seconds.
Time measurement: j= 100 took 1.0145e-05 seconds.
Time measurement: j= 1000 took 0.000110649 seconds.
Time measurement: j= 10000 took 0.00142651 seconds.
Time measurement: j= 100000 took 0.00834339 seconds.
Time measurement: j= 1000000 took 0.098939 seconds.
Time measurement: j= 10000000 took 0.938253 seconds.
Time measurement: j= 100000000 took 10.2398 seconds.
Time measurement: j= 1000000000 took 114.214 seconds.
Time measurement: j= 1410065408 took 163.824 seconds.

这似乎非常接近线性行为.

This seems to be very close to linear behaviour.

为什么std::sort需要O(n)而不是O(nlog(n))?

推荐答案

您呈现的图形非常适合y = x log (x).与x相比，log(x)的影响很小.我认为您的结果将通过具有很好意义的x log (x)的卡方.

The graph you present is a good fit for y = x log (x). Compared to x, log(x) has a small effect. I conject your results would pass a chi square for x log (x) with good significance.

这里没有惊喜.

这是您欣赏O(n log n)并不比O(n)差很多的试金石.

This is a touchstone for your appreciation that O(n log n) isn't much worse than O(n).

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