使用TBB的并行性-我们的检查列表中应该包含哪些内容

直到最近，并行编程的前景才引起我的注意。从那以后，我使用了各种并行编程库。也许我第一站是英特尔线程构建块(TBB)。但是，经常成为瓶颈的是由于舍入等因素造成的错误，以及这些程序在不同处理器架构中的不可预测行为。下面是一段代码，用于计算两组值的Pearsons相关系数。它采用了TBB的非常基本的并行模式——*parallel_for*和*parallet_reduce*：

// A programme to calculate Pearsons Correlation coefficient 
#include <math.h>
#include <stdlib.h>
#include <iostream>
#include <tbb/task_scheduler_init.h>
#include <tbb/parallel_for.h>
#include <tbb/parallel_reduce.h>
#include <tbb/blocked_range.h>
#include <tbb/tick_count.h>


using namespace std;
using namespace tbb;
const size_t n=100000;
double global=0;
namespace s //Namesapce for serial part
{
double *a,*b;
int j;
double mean_a,mean_b,sd_a=0,sd_b=0,pcc=0;
double sum_a,sum_b,i;
}
namespace p //Namespace for parallel part
{
double *a,*b;
double mean_a,mean_b,pcc;
double sum_a,sum_b,i;
double sd_a,sd_b;
}

class serials
{
public:
void computemean_serial()
{
using namespace s;
sum_a=0,sum_b=0,i=0;
a=(double*) malloc(n*sizeof(double));
b=(double*) malloc(n*sizeof(double));
for(j=0;j<n;j++,i++)
{ 
a[j]=sin(i);
b[j]=cos(i);
sum_a=sum_a+a[j];
sum_b=sum_b+b[j];
}
mean_a=sum_a/n;
mean_b=sum_b/n;
cout<<"nMean of a :"<<mean_a;
cout<<"nMean of b :"<<mean_b;
}
void computesd_serial()
{
using namespace s;
for(j=0;j<n;j++)
{sd_a=sd_a+pow((a[j]-mean_a),2);
sd_b=sd_b+pow((b[j]-mean_b),2);
}
sd_a=sd_a/n;
sd_a=sqrt(sd_a);
sd_b=sd_b/n;
sd_b=sqrt(sd_b);
cout<<"nStandard deviation of a :"<<sd_a;
cout<<"nStandard deviation of b :"<<sd_b;
}
void pearson_correlation_coefficient_serial()
{
using namespace s;
pcc=0;
for(j=0;j<n;j++)
{
pcc+=(a[j]-mean_a)*(b[j]-mean_b);
}
pcc=pcc/(n*sd_a*sd_b);
cout<<"nPearson Correlation Coefficient: "<<pcc;
}
};

class parallel
{
public:
class compute_mean 
{
double *store1,*store2;
public: 
double mean_a,mean_b;
void operator()( const blocked_range<size_t>& r)
{
double *a= store1;
double *b= store2;
for(size_t i =r.begin();i!=r.end(); ++i)
{    
mean_a+=a[i];
mean_b+=b[i];
}
}
compute_mean( compute_mean& x, split) : store1(x.store1),store2(x.store2),mean_a(0),mean_b(0){}
void join(const compute_mean& y) {mean_a+=y.mean_a;mean_b+=y.mean_b;}
compute_mean(double* a,double* b): store1(a),store2(b),mean_a(0),mean_b(0){}
};
class read_array
{
double *const a,*const b;
public:
read_array(double* vec1, double* vec2) : a(vec1),b(vec2){}  // constructor copies the arguments into local store 
void operator() (const blocked_range<size_t> &r) const {              // opration to be used in parallel_for 
for(size_t k = r.begin(); k!=r.end(); k++,global++)
{   
a[k]=sin(global);
b[k]=cos(global);
}
}};
void computemean_parallel()
{
using namespace p;
i=0;
a=(double*) malloc(n*sizeof(double));
b=(double*) malloc(n*sizeof(double));
parallel_for(blocked_range<size_t>(0,n,5000),read_array(a,b));
compute_mean sf(a,b);
parallel_reduce(blocked_range<size_t>(0,n,5000),sf);
mean_a=sf.mean_a/n;
mean_b=sf.mean_b/n;
cout<<"nMean of a :"<<mean_a;
cout<<"nMean of b :"<<mean_b;
}
class compute_sd 
{
double *store1,*store2;
double store3,store4;
public: 
double sd_a,sd_b,dif_a,dif_b,temp_pcc;
void operator()( const blocked_range<size_t>& r)
{
double *a= store1;
double *b= store2;
double mean_a=store3;
double mean_b=store4;
for(size_t i =r.begin();i!=r.end(); ++i)
{ 
dif_a=a[i]-mean_a;
dif_b=b[i]-mean_b;
temp_pcc+=dif_a*dif_b;
sd_a+=pow(dif_a,2);
sd_b+=pow(dif_b,2);
}}
compute_sd( compute_sd& x, split) : store1(x.store1),store2(x.store2),store3(p::mean_a),store4(p::mean_b),sd_a(0),sd_b(0),temp_pcc(0){}
void join(const compute_sd& y) {sd_a+=y.sd_a;sd_b+=y.sd_b;}
compute_sd(double* a,double* b,double mean_a,double mean_b): store1(a),store2(b),store3(mean_a),store4(mean_b),sd_a(0),sd_b(0),temp_pcc(0){}
};

void computesd_and_pearson_correlation_coefficient_parallel()
{
using namespace p;
compute_sd obj2(a,b,mean_a,mean_b);
parallel_reduce(blocked_range<size_t>(0,n,5000),obj2);
sd_a=obj2.sd_a;
sd_b=obj2.sd_b;
sd_a=sd_a/n;
sd_a=sqrt(sd_a);
sd_b=sd_b/n;
sd_b=sqrt(sd_b);
cout<<"nStandard deviation of a :"<<sd_a;
cout<<"nStandard deviation of b :"<<sd_b;
pcc=obj2.temp_pcc;
pcc=pcc/(n*sd_a*sd_b);
cout<<"nPearson Correlation Coefficient: "<<pcc;
}
};
main()
{       
serials obj_s;
parallel obj_p;
cout<<"nSerial Part";
cout<<"n-----------";
tick_count start_s=tick_count::now();
obj_s.computemean_serial();
obj_s.computesd_serial();
obj_s.pearson_correlation_coefficient_serial();
tick_count end_s=tick_count::now();
cout<<"n";
task_scheduler_init init;
cout<<"nParallel Part";
cout<<"n-------------";
tick_count start_p=tick_count::now();
obj_p.computemean_parallel();
obj_p.computesd_and_pearson_correlation_coefficient_parallel();
tick_count end_p=tick_count::now();
cout<<"n";
cout<<"nTime Estimates";
cout<<"n--------------";
cout<<"nSerial Time :"<<(end_s-start_s).seconds()<<" Seconds";
cout<<"nParallel time :"<<(end_p-start_p).seconds()<<" Secondsn";
}

好吧！它在装有酷睿i5的Windows机器上运行得很好。它为输出中的每个参数提供了绝对相同的值，并行代码比串行代码快。这是我的输出：

操作系统:Windows 7 Ultimate 64位处理器：核心i5

Serial Part
-----------
Mean of a :1.81203e-05
Mean of b :1.0324e-05
Standard deviation of a :0.707107
Standard deviation of b :0.707107
Pearson Correlation Coefficient: 3.65091e-07
Parallel Part
-------------
Mean of a :1.81203e-05
Mean of b :1.0324e-05
Standard deviation of a :0.707107
Standard deviation of b :0.707107
Pearson Correlation Coefficient: 3.65091e-07
Time Estimates
--------------
Serial Time : 0.0204829 Seconds
Parallel Time : 0.00939971 Seconds

那么其他机器呢？如果我说它会很好用，那么至少我的一些朋友会说："伙计，等等！有点可疑。"在不同的机器中，答案(由并行代码和串行代码产生的答案之间)有微小的差异，尽管并行代码总是比串行代码快。那么，是什么造成了这些差异呢？我们得出的结论是，这种异常行为是以过度并行和处理器架构差异为代价的舍入误差。

这引出了我的问题：

• 当我们使用并行时，我们需要采取哪些预防措施处理我们代码中的库以利用多核处理器
• 在什么情况下，我们甚至不应该使用平行方法尽管有多个处理器可用
• 为了避免舍入误差，我们能做的最好的事情是什么？(让我来指定我不是在谈论强制执行互斥和屏障有时可能会限制平行度的延伸，但大约简单的编程技巧有时很方便)

我很高兴看到你对这些问题的建议。请随意回答如果你有时间限制，最适合你的部分。

编辑-我在此处包含了更多结果

操作系统:Linux Ubuntu 64位处理器：核心i5

Serial Part
-----------
Mean of a :1.81203e-05
Mean of b :1.0324e-05
Standard deviation of a :0.707107
Standard deviation of b :0.707107
Pearson Correlation Coefficient: 3.65091e-07
Parallel Part
-------------
Mean of a :-0.000233041
Mean of b :0.00414375
Standard deviation of a :2.58428
Standard deviation of b :54.6333
Pearson Correlation Coefficient: -0.000538456
Time Estimates
--------------
Serial Time :0.0161237 Seconds
Parallel Time :0.0103125 Seconds

操作系统:Linux Fedora 64位处理器：核心i3

Serial Part
-----------
Mean of a :1.81203e-05
Mean of b :1.0324e-05
Standard deviation of a :0.707107
Standard deviation of b :0.707107
Pearson Correlation Coefficient: 3.65091e-07
Parallel Part
-------------
Mean of a :-0.00197118
Mean of b :0.00124329
Standard deviation of a :0.707783
Standard deviation of b :0.703951
Pearson Correlation Coefficient: -0.129055
Time Estimates
--------------
Serial Time :0.02257 Seconds
Parallel Time :0.0107966 Seconds

编辑：更改后哪一天建议

操作系统:Linux Ubuntu 64位处理器:corei5

Serial Part
-----------
Mean of a :1.81203e-05
Mean of b :1.0324e-05
Standard deviation of a :0.707107
Standard deviation of b :0.707107
Pearson Correlation Coefficient: 3.65091e-07
Parallel Part
-------------
Mean of a :-0.000304446
Mean of b :0.00172593
Standard deviation of a :0.708465
Standard deviation of b :0.7039
Pearson Correlation Coefficient: -0.140716
Time Estimates
--------------
Serial Time :0.0235391 Seconds
Parallel time :0.00810775 Seconds

谨致问候。

注意1：我不能保证上面的代码是正确的。我相信是的。

注意2：这段代码也在Linux盒子上进行了测试

注意3：尝试了不同的粒度组合和自动分区选项