C语言 我的 MPI 雅可比迭代程序给出了错误的结果

Question

我在 C 语言中运行以下程序时遇到上述错误。

该程序首先生成 3x3 数组并执行雅可比迭代。它使用 MPI 库。我不知道代码的哪些部分是错误的。

#include <stdio.h>
#include <string.h>
#include <mpi.h>
#include <math.h> // l2-norm //
#include <time.h>
int main(int argc, char **argv)
{
int numprocs, myid;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
MPI_Comm_rank(MPI_COMM_WORLD, &myid);
double a[3][3];
double b[3];
double x[3]={0};
double xa[3]={0};
double xnew[3]={0};
double y[3]={0};
float sigancha;
time_t startTime=0, endTime=0;
int n=3;
int i, j =0;
int k=0;
int o;
int hoessu=300;
int minhoessu=300;
double sum=1;
int numsent =0;
int ans;
int row;
MPI_Status status;
int sender;
int po;
double *buffer;
/* synchronization */
MPI_Barrier(MPI_COMM_WORLD);
for (i=0; i<n; i++){
b[i]=i*100;
for (j=0; j<n; j++) {
a[i][j]=((i+j)%10);
if (i==j) {a[i][j]+=5000;}
}
x[i]=b[i]/a[i][i];
}
/* run if sum is greater than 0.0002 */
for (k=0; k<hoessu&&sum>0.0002||k<minhoessu; k++) {
numsent = 0;
for (o=myid+1; o<n+1; o+=numprocs) {
i=o-1;
xa[i]=b[i]+a[i][i]*x[i];
for (j=0; j<n; j++) {
xa[i]-=a[i][j]*x[j];
}
xnew[i]=xa[i]/a[i][i];
/*send xnew[i] to master*/
MPI_Send(&xnew[i],1,MPI_DOUBLE,0,i,MPI_COMM_WORLD);
}
if (myid == 0){
/*get xnew[i]*/
for (i=0; i<n; i++) {
MPI_Recv(&ans, 1, MPI_DOUBLE, MPI_ANY_SOURCE, MPI_ANY_TAG, MPI_COMM_WORLD, &status);
sender = status.MPI_SOURCE;
row = status.MPI_TAG;
xnew[row] = ans;
}
/*calculates sum at master*/
for (j=0; j<n; j++){
sum=0.0;
sum+=(xnew[j]-x[j])*(xnew[j]-x[j]);
x[j]=xnew[j];
}
sum=pow(sum,0.5);
MPI_Bcast(&x[0], n, MPI_DOUBLE, 0, MPI_COMM_WORLD);
}
}
if (myid == 0){
endTime=clock();
sigancha=(float)(endTime-startTime)/(CLOCKS_PER_SEC);
printf("finishedn");
for (j=0; j<n; j++) {
printf("x[%d]=%fln",j+1,xnew[j]);
}
printf("iteration; %d times itereation are done. n l2-norm, error is %fl .n %f seceonds are used. n",k,sum,sigancha);
}
MPI_Finalize();
}

使用 mpicc 进行编译。

mpicc mpijacobi2.c -o taskingyeje  
./taskingyeje

结果。

finished
x[1]=-1736884775.000000l
x[2]=-370936800.000000l
x[3]=2118301216.000000l
iteration; 300 times itereation are done. 
l2-norm, error is 34332272.000000l .
0.020000 seceonds are used.

但是，此结果不是预期的结果。如果这个程序运行良好，它应该给出与串行雅可比迭代相同的结果。 这将是

x[1]=-0.000020l
x[2]=-0.019968l
x[3]=0.399956l

我不知道为什么这个程序会产生错误的结果。

Answer 1

#include <stdio.h>
#include <string.h>
#include <mpi.h>
#include <math.h> // l2-norm //
#include <time.h>
int main(int argc, char **argv)
{
int numprocs, myid;
MPI_Init(&argc,&argv);
MPI_Comm_size(MPI_COMM_WORLD, &numprocs);
MPI_Comm_rank(MPI_COMM_WORLD, &myid);
double a[700][700];
double b[700];
double x[700]={0};
double xa[700]={0};
double xnew[700]={0};
double y[700]={0};
float sigancha;
time_t startTime=0, endTime=0;
int n=700;
int i, j =0;
int k=0;
int o;
int hoessu=300;
int minhoessu=300;
double sum=1;
int numsent =0;
int ans;
int row;
MPI_Status status;
int sender;
int po;
double *buffer;
/* synchronization */
MPI_Barrier(MPI_COMM_WORLD);
for (i=0; i<n; i++){
b[i]=i*100;
for (j=0; j<n; j++) {
a[i][j]=((i+j)%10);
if (i==j) {a[i][j]+=10000;}
}
x[i]=b[i]/a[i][i];
}
/* run if sum is greater than 0.0002 */
for (k=0; k<hoessu&&sum>0.0002||k<minhoessu; k++) {
numsent = 0;
for (o=myid+1; o<n+1; o+=numprocs) {
i=o-1;
xa[i]=b[i]+a[i][i]*x[i];
for (j=0; j<n; j++) {
xa[i]-=a[i][j]*x[j];
}
xnew[i]=xa[i]/a[i][i];
/*send xnew[i] to master*/
ans=xnew[i];
MPI_Allgather(&xnew[i],1,MPI_DOUBLE,&xnew[i],1,MPI_DOUBLE,MPI_COMM_WORLD);
}
if (myid == 0){
/*calculates sum at master*/
for (j=0; j<n; j++){
sum=0.0;
sum+=(xnew[j]-x[j])*(xnew[j]-x[j]);
x[j]=xnew[j];
}
sum=pow(sum,0.5);
MPI_Bcast(&x[0], n, MPI_DOUBLE, 0, MPI_COMM_WORLD);
}
}
if (myid == 0){
endTime=clock();
sigancha=(float)(endTime-startTime)/(CLOCKS_PER_SEC);
printf("finishedn");
for (j=0; j<n; j++) {
printf("x[%d]=%fln",j+1,xnew[j]);
}
printf("iteration; %d times itereation are done. n l2-norm, error is %fl .n %f seceonds are used. n",k,sum,sigancha);
}
MPI_Finalize();
}