Question

我写了一个平行的Jacobi算法，它解决了稀疏线性系统。矩阵的所有非零系数已经以向量内的修改的CRS形式存储，每行的第一个元素是对角元素。这是主要的例程：

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <cmath>
#include <chrono>
#include <mpi.h>
using namespace std;
using namespace chrono;
using double_vector = vector<double>;
using int_vector = vector<int>;

int main() {
    MPI_Init(NULL, NULL);
    int WorldRank;
    MPI_Comm_rank(MPI_COMM_WORLD, &WorldRank);
    int WorldSize;
    MPI_Comm_size(MPI_COMM_WORLD, &WorldSize);

    int Order = 8;
    int Length = 22;
    double_vector Coefficients(Length);
    int_vector RowIndex(Order+1);
    int_vector ColumnIndex(Length);
    double_vector InitialAnswer(Order);
    double_vector IndependentTerm(Order);   

    if (WorldRank == 0) {
        Coefficients = {10, 1, 10, 1, 1, 10, 1, 1, 10, 1, 1, 10, 1, 1, 10, 1, 1, 10, 1, 1, 10, 1};
        RowIndex = {0, 2, 5, 8, 11, 14, 17, 20, 22};
        ColumnIndex = {0, 1, 1, 0, 2, 2, 1, 3, 3, 2, 4, 4, 3, 5, 5, 4, 6, 6, 5, 7, 7, 6};
        IndependentTerm = {11, 12, 12, 12, 12, 12, 12, 11};
        InitialAnswer = {5, 5, 5, 5, 5, 5, 5, 5};
    }

    auto Start = steady_clock::now();
    double_vector Answer = Jacobi(Coefficients, RowIndex, ColumnIndex, IndependentTerm, InitialAnswer);

    if (WorldRank == 0) {
        auto End = steady_clock::now();
        duration<double> ElapsedSeconds = End - Start;
        print_vector(Answer, "Answer");
        cout << "\n\tElapsed time: " << setprecision(6) << ElapsedSeconds.count() << "s" << endl << endl;       
    }

    MPI_Barrier(MPI_COMM_WORLD);
    MPI_Finalize();
    return 0;
}

以下是Jacobi例程：

double_vector Jacobi(double_vector& Coefficients, int_vector& RowIndex, int_vector& ColumnIndex, double_vector& IndependentTerm, double_vector& InitialAnswer) {
    int WorldRank;
    MPI_Comm_rank(MPI_COMM_WORLD, &WorldRank);
    int WorldSize;
    MPI_Comm_size(MPI_COMM_WORLD, &WorldSize);

    int Length = Coefficients.size();
    int Order = IndependentTerm.size();
    int Remainder = Order % WorldSize;
    int Counter = 0;
    int* Displacements = new int[WorldSize];
    int* ElementsPerProcess = new int[WorldSize];
    for (int i=0; i<WorldSize; i++) {
        ElementsPerProcess[i] = Order/WorldSize;
        if (Remainder > 0) {
            ElementsPerProcess[i]++;
            Remainder--;
        }
        Displacements[i] = Counter;
        Counter += ElementsPerProcess[i];
    }
    int iStart = Displacements[WorldRank];
    int iEnd = iStart + ElementsPerProcess[WorldRank];

    MPI_Bcast(&Coefficients[0], Length, MPI_DOUBLE, 0, MPI_COMM_WORLD);
    MPI_Bcast(&RowIndex[0], Order+1, MPI_INT, 0, MPI_COMM_WORLD);
    MPI_Bcast(&ColumnIndex[0], Length, MPI_INT, 0, MPI_COMM_WORLD);
    MPI_Bcast(&IndependentTerm[0], Order, MPI_DOUBLE, 0, MPI_COMM_WORLD);
    MPI_Bcast(&InitialAnswer[0], Order, MPI_DOUBLE, 0, MPI_COMM_WORLD);

    int MaximumIterations = 1000;
    double Summatory;
    double_vector Answer(Order);
    for (int k=0; k<MaximumIterations; k++) {
        for (int i=iStart; i<iEnd; i++) {
            Summatory = 0.0;
            for (int j=RowIndex[i]+1; j<RowIndex[i+1]; j++) {
                Summatory -= Coefficients[j] * InitialAnswer[ColumnIndex[j]];
            }
            Summatory += IndependentTerm[i];
            Answer[i] = Summatory / Coefficients[RowIndex[i]];
        }
        MPI_Allgatherv(&Answer[iStart], ElementsPerProcess[WorldRank], MPI_DOUBLE, &InitialAnswer[0], ElementsPerProcess, Displacements, MPI_DOUBLE,    MPI_COMM_WORLD);
    }

    delete Displacements;
    delete ElementsPerProcess;

    return Answer;
}

我的测试用例是一个8x8三对角线性系统，具有严格的主导对角线。未知值均为1.0。运行 mpirun -n 1 ./main ，它会输出正确的答案：

1.000   1.000   1.000   1.000   1.000   1.000   1.000   1.000

但是，当我使用多个进程执行此例程时，答案值不会收集到所有进程。也就是说，如果进程0 打印答案，则会显示以下内容：

1.000   1.000   1.000   1.000   0.000   0.000   0.000   0.000

如果我得到进程1 来打印答案：

0.000   0.000   0.000   0.000   1.000   1.000   1.000   1.000

已经确认所有进程都可以正确访问我播放的数据。我也相信我正确使用 MPI_Allgatherv 。

MPICH 3.2 - 编译器：mpicxx - 当前标志：-g -std = c ++ 14

关于我做错什么的任何想法？

谢谢。

为什么MPI_Allgatherv没有聚集到所有人？

0 个答案: