更少的恒定时间迭代花费更多时间-c ++编译器依赖性？

Question

我想在我的课堂上证明，按预定概率进行抽样可以缩短执行时间。在下面的代码中，sample()功能是工作马。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）-请参见main()函数。该计数器变量用于循环迭代计数。

结果：在输入(p,x)的情况下，sample()花费的时间是(p1, x1)的两倍，但是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。

我不知道这是怎么可能的：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，这是编译器功能与算法复杂度之间的权衡。 5个元素的数组太小，无法通过顺序展示来获得好处，而CPU机制正赶超这一优势。仅在初始化具有更多数据（大约30个元素）的所有数组（p，x，p1和x1）之后，排序后的数组才能比未排序的数组更快地产生输出时间。

证明（新的main()功能）：

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;
}

样品运行时间：

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