更少的恒定时间迭代需要更多时间 - c++ 编译器依赖性？

Question

我想在我的课堂上演示使用预先排序的概率进行抽样可以缩短执行时间。在下面的代码中，sample() 函数是工作的马。相同的随机变量分布以两种形式存储：未排序的概率（数组p 和x）和排序的概率（数组p1 和x1） - 请参阅main() 函数。循环迭代的计数器变量计数。

结果：使用(p,x) 输入，sample() 花费的时间是(p1, x1) 的两倍，但经过的执行时间相同甚至更长。我在家用笔记本电脑 Kubuntu 18.04 上尝试了 g++ 7.4.0 编译器，并在 (wandbox dot org) 尝试了不同的 g++ 版本，结果基本相同。

我不明白这怎么可能：更少的恒定时间迭代需要更长的时间。

代码：

#include <iostream>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <chrono>

using namespace std;

inline double runif(){return rand()/double(RAND_MAX);}

double sample(double* p, double* x, int N, double u, unsigned long* count)
{
    int k;
    for(k=0; (k<N) && (u>p[k]); k++, (*count)++)
        u -= p[k];
    return x[k];
}

double sample_alias(double* p, double* x, int N, double u)
{
    double u1 = u * N;
    int K = floor(u1);
    double u2 = u1 - K;
    return (u2<p[K]) ? *(x+2*K) : *(x+2*K+1);
}


int main()
{
    double p[] = {0.2, 0.05, 0.125, 0.5, 0.125};
    double x[] = {0,   -3,   1,     -2,  3};

    double p1[] = {0.5, 0.2, 0.125, 0.125, 0.05};
    double x1[] = {-2,   0,    3,    1,    -3};

    double sum;
    unsigned long counter;

#define NN 4000000
    double *u;
    u = (double*)calloc(NN, sizeof(double));
    if(u==NULL) perror("Not enough mem!");
    srand(5647892);
    for (int i=0; i<NN; i++) u[i]=runif();

    cout << "Test 1 (unsorted)" << endl;
    sum=0.0; counter = 0;
    auto begt = std::chrono::steady_clock::now();
    for(int i=0; i<NN; i++) sum+=sample(p,x,5,u[i], &counter);
    auto endt = std::chrono::steady_clock::now();
    auto elapsed = endt - begt;
    cout<<sum/double(NN)<<endl<<"Run took "<<elapsed.count()<<", total loop: "<< counter<<endl;

    cout << "Test 1 (sorted)" << endl;
    sum=0.0; counter = 0;
    begt = std::chrono::steady_clock::now();
    for(int i=0; i<NN; i++) sum+=sample(p1,x1,5,u[i],&counter);
    endt = std::chrono::steady_clock::now();
    elapsed = endt - begt;
    cout<<sum/double(NN)<<endl<<"Run took "<<elapsed.count()<<", total loop: "<< counter<<endl;

    free(u);
    return 0;
}

我的测试输出：

Test 1 (unsorted)
-0.650426
Run took 32114525, total loop: 9205058
Test 1 (sorted)
-0.649237
Run took 40915156, total loop: 4101917

Answer 1

事实证明，这是编译器功能和算法复杂性之间的权衡。 5 元素数组太小，无法展示排序的好处，而 CPU 机制正在超越这一优势。只有在使用更多数据（大约 30 个元素）初始化所有数组（p、x、p1 和 x1）之后，排序数组生成的输出时间比未排序数组快。

证明（一个新的main()函数）：

int main()
{
  // R: p1 <- dhyper( 0:30, 100, 200, 30)
  double p[]  = {2.365460e-06, 4.149930e-05, 3.463503e-04, 1.831185e-03, 6.890624e-03,
         1.965600e-02, 4.420738e-02, 8.049383e-02, 1.209103e-01, 1.519072e-01,
         1.612748e-01, 1.458034e-01, 1.128909e-01, 7.516566e-02, 4.315606e-02,
         2.139919e-02, 9.167998e-03, 3.391496e-03, 1.081390e-03, 2.963208e-04,
         6.947942e-05, 1.385778e-05, 2.332594e-06, 3.278979e-07, 3.795897e-08,
         3.550624e-09, 2.612802e-10, 1.454013e-11, 5.743665e-13, 1.433179e-14,
         1.695929e-16};
  double x[]  = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
         20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30};
  // R: p1.sort( p1, decreasing=TRUE, index=TRUE)
  double p1[] = {1.612748e-01, 1.519072e-01, 1.458034e-01, 1.209103e-01, 1.128909e-01,
         8.049383e-02, 7.516566e-02, 4.420738e-02, 4.315606e-02, 2.139919e-02,
         1.965600e-02, 9.167998e-03, 6.890624e-03, 3.391496e-03, 1.831185e-03,
         1.081390e-03, 3.463503e-04, 2.963208e-04, 6.947942e-05, 4.149930e-05,
         1.385778e-05, 2.365460e-06, 2.332594e-06, 3.278979e-07, 3.795897e-08,
         3.550624e-09, 2.612802e-10, 1.454013e-11, 5.743665e-13, 1.433179e-14,
         1.695929e-16};
  double x1[] = {10,  9, 11,  8, 12,  7, 13,  6, 14, 15,  5, 16,  4, 17,  3, 18,  2, 19,
         20,  1, 21,  0, 22, 23, 24,  25, 26, 27, 28, 29, 30};

  double sum;
  unsigned long counter;

#define NN 1000000
  double *u;
  u = (double*)calloc(NN, sizeof(double));
  if(u==NULL) perror("Not enough mem!");
  srand(5647892);
  for (int i=0; i<NN; i++) u[i]=runif();

  int sz = sizeof(p)/sizeof(p[0]);

  cout << "Test 1 (unsorted)" << endl;
  sum=0.0; counter = 0;
  srand(5647892);
  auto begt = std::chrono::steady_clock::now();
  for(int i=0; i<NN; i++) sum+=sample(p,x,sz,u[i], &counter);
  auto endt = std::chrono::steady_clock::now();
  auto elapsed = endt - begt;
  cout<<sum/double(NN)<<endl<<"Run took "<<elapsed.count()<<", total loop: "<< counter<<endl;

  cout << "Test 1 (sorted)" << endl;
  sum=0.0; counter = 0;
  srand(5647892);
  begt = std::chrono::steady_clock::now();
  for(int i=0; i<NN; i++) sum+=sample(p1,x1,sz,u[i],&counter);
  endt = std::chrono::steady_clock::now();
  elapsed = endt - begt;
  cout<<sum/double(NN)<<endl<<"Run took "<<elapsed.count()<<", total loop: "<< counter<<endl;

  free(u);
  return 0;
}

示例运行时间：

Test 1 (unsorted)
10.0038
Run took 23076167, total loop: 10003850
Test 1 (sorted)
10.0047
Run took 14010650, total loop: 3442722