更少的恆定時間迭代需要更多的時間 - C++ 編譯器依賴？

Question

我想在我的課堂上證明使用預先排序的概率進行采樣可以縮短執行時間。 在下面的代碼中， sample()函數是工作的馬。 相同的隨機變量分布以兩種形式存儲：未排序的概率（數組p和x ）和排序的概率（數組p1和x1 ） - 請參閱main()函數。 計數器變量計數循環迭代。

結果：對於(p,x)輸入， sample()花費的時間是(p1, x1)兩倍(p1, x1)但執行時間相同甚至更長。 我在家用筆記本電腦 Kubuntu 18.04 上嘗試了 g++ 7.4.0 編譯器，並在（ wandbox dot org ）嘗試了不同的 g++ 版本，結果基本相同。

我不明白這怎么可能：更少的恆定時間迭代需要更長的時間。

編碼：

#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;
}

我的測試輸出：

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

Answer 1

事實證明，這是編譯器的能力和算法的復雜性之間的權衡。 5 元素陣列太小，無法展示訂購的好處，而 CPU 機制正在超越這一收益。 只有在使用更多數據（大約 30 個元素）初始化所有數組（ p 、 x 、 p1和x1 ）之后，排序數組生成的輸出時間比未排序數組快。

證明（一個新的main()函數）：

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;
}

樣品運行時間：

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