1小时前,我提出了一个问题,但问题没有得到很好的回答,所以我重新创建了一个问题。
我得到了C中的Jacobi松弛代码:
while ( error > tol && iter < iter_max ) {
error = 0.0;
for( int j = 1; j < n-1; j++)
{
for( int i = 1; i < m-1; i++ )
{
Anew[j][i] = 0.25 * ( A[j][i+1] + A[j][i-1]
+ A[j-1][i] + A[j+1][i]);
error = fmax( error, fabs(Anew[j][i] - A[j][i]));
}
}
for( int j = 1; j < n-1; j++)
{
for( int i = 1; i < m-1; i++ )
{
A[j][i] = Anew[j][i];
}
}
if(iter % 100 == 0) printf("%5d, %0.6f\n", iter, error);
iter++;
}我做这个的时候:
我已经用OpenACC并行化了这段代码。现在,我想使用MPI来尝试理解它是如何工作的。但是,对于我所做的第一次实现,我没有很好的结果(新数组的构造并不好)。如何将此代码部分与MPI并行?
发布于 2016-07-01 02:07:20
下面是我为类似情况编写的代码,您可以使用它作为指南。
do {
iter++;
MPI_Irecv(&old[1][0], 1, myHelloVector, nbrs[LEFT], 2, MPI_COMM_WORLD, \
&requestFourR[LEFT]);
MPI_Irecv(&old[1][chunkSize[1]+1], 1, myHelloVector, nbrs[RIGHT], 1, \
MPI_COMM_WORLD, &requestFourR[RIGHT]);
MPI_Irecv(&old[0][1], chunkSize[1], MPI_FLOAT, nbrs[UP], 4, \
MPI_COMM_WORLD, &requestFourR[UP]);
MPI_Irecv(&old[chunkSize[0]+1][1], chunkSize[1], MPI_FLOAT, \
nbrs[DOWN], 3, MPI_COMM_WORLD, &requestFourR[DOWN]);
MPI_Issend(&old[1][1], 1, myHelloVector, nbrs[LEFT], 1, \
MPI_COMM_WORLD, &requestFourS[LEFT]);
MPI_Issend(&old[1][chunkSize[1]], 1, myHelloVector, nbrs[RIGHT], 2, \
MPI_COMM_WORLD, &requestFourS[RIGHT]);
MPI_Issend(&old[1][1], chunkSize[1], MPI_FLOAT, nbrs[UP], 3, \
MPI_COMM_WORLD, &requestFourS[UP]);
MPI_Issend(&old[chunkSize[0]][1], chunkSize[1], MPI_FLOAT, nbrs[DOWN], 4, \
MPI_COMM_WORLD, &requestFourS[DOWN]);
calImage(old, new, edge, chunkSize[ROWS], chunkSize[COLS]);
for (itr = 0; itr < 4; itr++) {
MPI_Waitany(4, &requestFourR[0], &index, &status);
switch ( index ) { /* status.MPI_TAG) */
case 0: /* RIGHT */
j = 1;
for (i = 2; i < chunkSize[0]; i++) {
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+ \
old[i][j+1] - edge[i][j]);
}
break;
case 1: /* LEFT */
j = chunkSize[1];
for (i = 2; i < chunkSize[0]; i++) {
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+ \
old[i][j+1] - edge[i][j]);
}
break;
case 2: /* DOWN */
i = 1;
for (j = 2; j < chunkSize[1]; j++) {
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+ \
old[i][j+1] - edge[i][j]);
}
break;
case 3: /* UP */
i = chunkSize[0];
for (j = 2; j < chunkSize[1]; j++) {
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+ \
old[i][j+1] - edge[i][j]);
}
break;
}
}
i = 1; j = 1;
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+old[i][j+1] - \
edge[i][j]);
i = 1; j = chunkSize[1];
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+old[i][j+1] - \
edge[i][j]);
i = chunkSize[0]; j = 1;
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+old[i][j+1] - \
edge[i][j]);
i = chunkSize[0]; j = chunkSize[1];
new[i][j] = 0.25f*(old[i-1][j]+old[i+1][j]+old[i][j-1]+old[i][j+1] - \
edge[i][j]);
MPI_Waitall(4, requestFourS, statusS);
temp = old;
old = new;
new = temp;
} while(your_stopping_condition);calImage()函数正在执行不依赖于晕交换操作的计算。
void calImage(float **image, float **newImage, float **edge, \
int rows, int cols) {
int i, j;
for (i = 2; i < rows; i++) {
for (j = 2; j < cols; j++) {
newImage[i][j] = 0.25f * (image[i-1][j] \
+ image[i+1][j] \
+ image[i][j-1] \
+ image[i][j+1] \
- edge[i][j]);
}
}
}https://stackoverflow.com/questions/37970024
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