矩阵反转使用线程较慢

Question

I made a function that makes the inverse and then another multithreaded, as long I have to make inverse of arrays >2000 x 2000. A 1000x1000 array unthreated takes 2.5 seconds (on a i5-4460 4 cores 2.9ghz) and multithreaded takes 7.25 seconds

I placed the multithreads in the part that most time consumption is taken. Whai is wrong? Is due vectors are used instead of 2 dimensions arrays?

This is the minimum code to test both versions:

#include<iostream>
#include <vector>
#include <stdlib.h>
#include <time.h>
#include <chrono>
#include <thread>
const int NUCLEOS = 8;

#ifdef __linux__ 
#include <unistd.h>    //usleep()
typedef std::chrono::system_clock t_clock;    //try to use high_resolution_clock on  new linux x64 computer!
#else
typedef std::chrono::high_resolution_clock t_clock;
#pragma warning(disable:4996)
#endif
using namespace std;


std::chrono::time_point<t_clock> start_time, stop_time = start_time; char null_char = '\0';
void timer(char *title = 0, int data_size = 1) { stop_time = t_clock::now(); double us = (double)chrono::duration_cast<chrono::microseconds>(stop_time - start_time).count();   if (title) printf("%s time = %7lgms = %7lg MOPs\n", title, (double)us*1e-3, (double)data_size / us); start_time = t_clock::now(); }



//makes columns 0
void colum_zero(vector< vector<double> > &x, vector< vector<double> > &y, int pos0, int pos1,int dim, int ord);

//returns inverse of x, x is not modified, not threaded
vector< vector<double> > inverse(vector< vector<double> > x)
{
    if (x.size() != x[0].size())
    {
        cout << "ERROR on inverse() not square array" << endl; getchar(); return{};//returns a null
    }

    size_t dim = x.size();
    int i, j, ord;
    vector< vector<double> > y(dim,vector<double>(dim,0));//initializes output = 0
    //init_2Dvector(y, dim, dim);
    //1. Unity array y: 
    for (i = 0; i < dim; i++)
    {
        y[i][i] = 1.0;
    }

    double diagon, coef;
    double *ptrx, *ptry, *ptrx2, *ptry2;
    for (ord = 0; ord<dim; ord++)
    {
        //2 Hacemos diagonal de x =1
        int i2;
        if (fabs(x[ord][ord])<1e-15) //If that element is 0, a line that contains a non zero is added
        {
            for (i2 = ord + 1; i2<dim; i2++)
            {
                if (fabs(x[i2][ord])>1e-15) break;
            }
            if (i2 >= dim)
                return{};//error, returns null
            for (i = 0; i<dim; i++)//added a line without 0
            {
                x[ord][i] += x[i2][i];
                y[ord][i] += y[i2][i];
            }
        }
        diagon = 1.0/x[ord][ord];
        ptry = &y[ord][0];
        ptrx = &x[ord][0];
        for (i = 0; i < dim; i++)
        {
            *ptry++ *= diagon;
            *ptrx++ *= diagon;
        }
        //uses the same function but not threaded:
        colum_zero(x,y,0,dim,dim,ord);
    }//end ord
    return y;
}

//threaded version
vector< vector<double> > inverse_th(vector< vector<double> > x)
{
    if (x.size() != x[0].size())
    {
        cout << "ERROR on inverse() not square array" << endl; getchar(); return{};//returns a null
    }

    int dim = (int) x.size();
    int i, ord;
    vector< vector<double> > y(dim, vector<double>(dim, 0));//initializes output = 0
                                                            //init_2Dvector(y, dim, dim);
                                                            //1. Unity array y: 
    for (i = 0; i < dim; i++)
    {
        y[i][i] = 1.0;
    }

    std::thread tarea[NUCLEOS];
    double diagon;
    double *ptrx, *ptry;// , *ptrx2, *ptry2;
    for (ord = 0; ord<dim; ord++)
    {
        //2 Hacemos diagonal de x =1
        int i2;
        if (fabs(x[ord][ord])<1e-15) //If a diagonal element=0 it is added a column that is not 0 the diagonal element
        {
            for (i2 = ord + 1; i2<dim; i2++)
            {
                if (fabs(x[i2][ord])>1e-15) break;
            }
            if (i2 >= dim)
                return{};//error, returns null
            for (i = 0; i<dim; i++)//It is looked for a line without zero to be added to make the number a non zero one to avoid later divide by 0
            {
                x[ord][i] += x[i2][i];
                y[ord][i] += y[i2][i];
            }
        }
        diagon = 1.0 / x[ord][ord];

        ptry = &y[ord][0];
        ptrx = &x[ord][0];
        for (i = 0; i < dim; i++)
        {
            *ptry++ *= diagon;
            *ptrx++ *= diagon;
        }

        int pos0 = 0, N1 = dim;//initial array position
        if ((N1<1) || (N1>5000))
        {
            cout << "It is detected out than 1-5000 simulations points=" << N1 << " ABORT or press enter to continue" << endl; getchar();
        }
        //cout << "Initiation of " << NUCLEOS << " threads" << endl;
        for (int thread = 0; thread<NUCLEOS; thread++)
        {
            int pos1 = (int)((thread + 1)*N1 / NUCLEOS);//next position
            tarea[thread] = std::thread(colum_zero, std::ref(x), std::ref(y), pos0, pos1, dim, ord);//ojo, coil current=1!!!!!!!!!!!!!!!!!!
            pos0 = pos1;//next thread will work at next point
        }
        for (int thread = 0; thread<NUCLEOS; thread++)
        {
            tarea[thread].join();
            //cout << "Thread num: " << thread << " end\n";
        }
    }//end ord
    return y;
}

//makes columns 0
void colum_zero(vector< vector<double> > &x, vector< vector<double> > &y, int pos0, int pos1,int dim, int ord)
{
    double coef;
    double *ptrx, *ptry, *ptrx2, *ptry2;
    //Hacemos '0' la columna ord salvo elemento diagonal:
    for (int i = pos0; i<pos1; i++)//Begin to end for every thread
    {
        if (i == ord) continue;
        coef = x[i][ord];//element to make 0 
        if (fabs(coef)<1e-15) continue; //If already zero, it is avoided
        ptry = &y[i][0];
        ptry2 = &y[ord][0];
        ptrx = &x[i][0];
        ptrx2 = &x[ord][0];
        for (int j = 0; j < dim; j++)
        {
            *ptry++ = *ptry - coef * (*ptry2++);//1ª matriz
            *ptrx++ = *ptrx - coef * (*ptrx2++);//2ª matriz
        }
    }
}


void test_6_inverse(int dim)
{
    vector< vector<double> > vec1(dim, vector<double>(dim));
    for (int i=0;i<dim;i++)
        for (int j = 0; j < dim; j++)
        {
            vec1[i][j] = (-1.0 + 2.0*rand() / RAND_MAX) * 10000;
        }
    vector< vector<double> > vec2,vec3;
    double ini, end;
    ini = (double)clock();
    vec2 = inverse(vec1);
    end = (double)clock();
    cout << "=== Time inverse unthreaded=" << (end - ini) / CLOCKS_PER_SEC << endl;
    ini=end;
    vec3 = inverse_th(vec1);
    end = (double)clock();
    cout << "=== Time inverse   threaded=" << (end - ini) / CLOCKS_PER_SEC << endl;
    cout<<vec2[2][2]<<" "<<vec3[2][2]<<endl;//to make the sw to do de inverse
    cout << endl;
}


int main()
{
    test_6_inverse(1000);
    cout << endl << "=== END ===" << endl; getchar(); 
    return 1;
}

Answer 1

After looking deeper in the code of the colum_zero() function I have seen that one thread rewrites in the data to be used by another threads, so the threads are not INDEPENDENT from each other. Fortunately the compiler detect it and avoid it.

Conclusions:

1. It is not recommended to try Gauss-Jordan method alone to make multithreads
2. If somebody detects that in multithread is slower and the initial function is spreaded correctly for every thread, perhaps is due one thread results are used by another
3. The main function inverse() works and can be used by other programmers, so this question should not be deleted

Non answered question: What is a matrix inverse method that could be spreaded in a lot of independent threads to be used in a gpu?