如何使用 MPI 在 C 中对二维数组求和
How to sum a 2D array in C using MPI
这是我用来对一维数组中的所有值求和的程序,它工作正常。但是我如何修改它以在二维数组上工作?假设变量 a
类似于 a = { {1,2}, {3,4}, {5,6} };
.
我尝试了一些解决方案,但它们都不起作用,所以有人可以解释一些重要的更改以使其也与 2D 数组兼容。
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
// size of array
#define n 10
int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// Temporary array for slave process
int a2[1000];
int main(int argc, char* argv[])
{
int pid, np,
elements_per_process,
n_elements_recieved;
// np -> no. of processes
// pid -> process id
MPI_Status status;
// Creation of parallel processes
MPI_Init(&argc, &argv);
// find out process ID,
// and how many processes were started
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
// master process
if (pid == 0) {
int index, i;
elements_per_process = n / np;
// check if more than 1 processes are run
if (np > 1) {
// distributes the portion of array
// to child processes to calculate
// their partial sums
for (i = 1; i < np - 1; i++) {
index = i * elements_per_process;
MPI_Send(&elements_per_process,
1, MPI_INT, i, 0,
MPI_COMM_WORLD);
MPI_Send(&a[index],
elements_per_process,
MPI_INT, i, 0,
MPI_COMM_WORLD);
}
// last process adds remaining elements
index = i * elements_per_process;
int elements_left = n - index;
MPI_Send(&elements_left,
1, MPI_INT,
i, 0,
MPI_COMM_WORLD);
MPI_Send(&a[index],
elements_left,
MPI_INT, i, 0,
MPI_COMM_WORLD);
}
// master process add its own sub array
int sum = 0;
for (i = 0; i < elements_per_process; i++)
sum += a[i];
// collects partial sums from other processes
int tmp;
for (i = 1; i < np; i++) {
MPI_Recv(&tmp, 1, MPI_INT,
MPI_ANY_SOURCE, 0,
MPI_COMM_WORLD,
&status);
int sender = status.MPI_SOURCE;
sum += tmp;
}
// prints the final sum of array
printf("Sum of array is : %d\n", sum);
}
// slave processes
else {
MPI_Recv(&n_elements_recieved,
1, MPI_INT, 0, 0,
MPI_COMM_WORLD,
&status);
// stores the received array segment
// in local array a2
MPI_Recv(&a2, n_elements_recieved,
MPI_INT, 0, 0,
MPI_COMM_WORLD,
&status);
// calculates its partial sum
int partial_sum = 0;
for (int i = 0; i < n_elements_recieved; i++)
partial_sum += a2[i];
// sends the partial sum to the root process
MPI_Send(&partial_sum, 1, MPI_INT,
0, 0, MPI_COMM_WORLD);
}
// cleans up all MPI state before exit of process
MPI_Finalize();
return 0;
}
您可以使用 MPI_Reduce 代替 MPI_Send
/MPI_Recv
来简化很多事情:
Reduces values on all processes to a single value
可以找到关于该例程的很好的教程 here。
因此每个进程都包含一个数组(例如, 进程 0 { 1, 2, 3, 4, 5}
和进程 1 {6, 7, 8, 9, 10 }
)并执行该数组的部分求和。最后,每个进程使用 MPI_Reduce
将所有部分和求和为主进程可用的单个值(它也可能是另一个进程)。看看这个例子:
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char* argv[]){
int np, pid;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
int partial_sum = 0;
if (pid == 0) {
int a[] = { 1, 2, 3, 4, 5};
for(int i = 0; i < 5; i++)
partial_sum += a[i];
}
else if (pid == 1){
int a[] = {6, 7, 8, 9, 10};
for(int i = 0; i < 5; i++)
partial_sum += a[i];
}
int sum;
MPI_Reduce(&partial_sum, &sum, 1, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
if (pid == 0){
printf("Sum of array is : %d\n", sum);
}
MPI_Finalize();
return 0;
}
此代码仅适用于 2 个进程(有点傻(但我用它来展示 MPI_Reduce
.
的使用
I tried few solutions but they are not working, so can someone explain
few important changes to make to make it compatible with 2D array
also.
如果您调整代码以使用我所展示的 MPI_Reduce
,那么它是一维数组还是二维数组并不重要,因为您将首先将部分求和转换为单个值,然后性能降低。
或者,您也可以将每一行分配给一个进程,然后对整个数组进行归约,然后 master 进程对结果数组进行求和。
一个例子:
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char* argv[]){
int np, pid;
MPI_Status status;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
int partial_sum = 0;
int size = 5;
int a[5] = {1, 2, 3 , 4, 5};
int sum[5] = {0};
MPI_Reduce(&a, &sum, size, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
if (pid == 0){
int total_sum = 0;
for(int i = 0; i < size; i++)
total_sum += sum[i];
printf("Sum of array is : %d\n", total_sum);
}
MPI_Finalize();
return 0;
}
输出(两个进程):
Sum of array is : 30
这是我用来对一维数组中的所有值求和的程序,它工作正常。但是我如何修改它以在二维数组上工作?假设变量 a
类似于 a = { {1,2}, {3,4}, {5,6} };
.
我尝试了一些解决方案,但它们都不起作用,所以有人可以解释一些重要的更改以使其也与 2D 数组兼容。
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
// size of array
#define n 10
int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// Temporary array for slave process
int a2[1000];
int main(int argc, char* argv[])
{
int pid, np,
elements_per_process,
n_elements_recieved;
// np -> no. of processes
// pid -> process id
MPI_Status status;
// Creation of parallel processes
MPI_Init(&argc, &argv);
// find out process ID,
// and how many processes were started
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
// master process
if (pid == 0) {
int index, i;
elements_per_process = n / np;
// check if more than 1 processes are run
if (np > 1) {
// distributes the portion of array
// to child processes to calculate
// their partial sums
for (i = 1; i < np - 1; i++) {
index = i * elements_per_process;
MPI_Send(&elements_per_process,
1, MPI_INT, i, 0,
MPI_COMM_WORLD);
MPI_Send(&a[index],
elements_per_process,
MPI_INT, i, 0,
MPI_COMM_WORLD);
}
// last process adds remaining elements
index = i * elements_per_process;
int elements_left = n - index;
MPI_Send(&elements_left,
1, MPI_INT,
i, 0,
MPI_COMM_WORLD);
MPI_Send(&a[index],
elements_left,
MPI_INT, i, 0,
MPI_COMM_WORLD);
}
// master process add its own sub array
int sum = 0;
for (i = 0; i < elements_per_process; i++)
sum += a[i];
// collects partial sums from other processes
int tmp;
for (i = 1; i < np; i++) {
MPI_Recv(&tmp, 1, MPI_INT,
MPI_ANY_SOURCE, 0,
MPI_COMM_WORLD,
&status);
int sender = status.MPI_SOURCE;
sum += tmp;
}
// prints the final sum of array
printf("Sum of array is : %d\n", sum);
}
// slave processes
else {
MPI_Recv(&n_elements_recieved,
1, MPI_INT, 0, 0,
MPI_COMM_WORLD,
&status);
// stores the received array segment
// in local array a2
MPI_Recv(&a2, n_elements_recieved,
MPI_INT, 0, 0,
MPI_COMM_WORLD,
&status);
// calculates its partial sum
int partial_sum = 0;
for (int i = 0; i < n_elements_recieved; i++)
partial_sum += a2[i];
// sends the partial sum to the root process
MPI_Send(&partial_sum, 1, MPI_INT,
0, 0, MPI_COMM_WORLD);
}
// cleans up all MPI state before exit of process
MPI_Finalize();
return 0;
}
您可以使用 MPI_Reduce 代替 MPI_Send
/MPI_Recv
来简化很多事情:
Reduces values on all processes to a single value
可以找到关于该例程的很好的教程 here。
因此每个进程都包含一个数组(例如, 进程 0 { 1, 2, 3, 4, 5}
和进程 1 {6, 7, 8, 9, 10 }
)并执行该数组的部分求和。最后,每个进程使用 MPI_Reduce
将所有部分和求和为主进程可用的单个值(它也可能是另一个进程)。看看这个例子:
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char* argv[]){
int np, pid;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
int partial_sum = 0;
if (pid == 0) {
int a[] = { 1, 2, 3, 4, 5};
for(int i = 0; i < 5; i++)
partial_sum += a[i];
}
else if (pid == 1){
int a[] = {6, 7, 8, 9, 10};
for(int i = 0; i < 5; i++)
partial_sum += a[i];
}
int sum;
MPI_Reduce(&partial_sum, &sum, 1, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
if (pid == 0){
printf("Sum of array is : %d\n", sum);
}
MPI_Finalize();
return 0;
}
此代码仅适用于 2 个进程(有点傻(但我用它来展示 MPI_Reduce
.
I tried few solutions but they are not working, so can someone explain few important changes to make to make it compatible with 2D array also.
如果您调整代码以使用我所展示的 MPI_Reduce
,那么它是一维数组还是二维数组并不重要,因为您将首先将部分求和转换为单个值,然后性能降低。
或者,您也可以将每一行分配给一个进程,然后对整个数组进行归约,然后 master 进程对结果数组进行求和。
一个例子:
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char* argv[]){
int np, pid;
MPI_Status status;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &pid);
MPI_Comm_size(MPI_COMM_WORLD, &np);
int partial_sum = 0;
int size = 5;
int a[5] = {1, 2, 3 , 4, 5};
int sum[5] = {0};
MPI_Reduce(&a, &sum, size, MPI_INT, MPI_SUM, 0, MPI_COMM_WORLD);
if (pid == 0){
int total_sum = 0;
for(int i = 0; i < size; i++)
total_sum += sum[i];
printf("Sum of array is : %d\n", total_sum);
}
MPI_Finalize();
return 0;
}
输出(两个进程):
Sum of array is : 30