Reputation: 33679
Assume I have a function f(i)
which depends on an index i
(among other values which cannot be precomputed).
I want to fill an array a
so that a[n] = sum(f(i)) from i=0 to n-1
.
Edit: After a comment by Hristo Iliev I realized what I am doing is a cumulative/prefix sum.
This can be written in code as
float sum = 0;
for(int i=0; i<N; i++) {
sum += f(i);
a[i] = sum;
}
Now I want to use OpenMP to do this in parallel. One way I could do this with OpenMP is to write out the values for f(i)
in parallel and then take care of the dependency in serial. If f(i)
is a slow function then this could work well since the non-paralleled loop is simple.
#pragma omp parallel for
for(int i=0; i<N; i++) {
a[i] = f(i);
}
for(int i=1; i<N; i++) {
a[i] += a[i-1];
}
But it's possible to do this without the non-parallel loop with OpenMP. The solution, however, that I have come up with is complicated and perhaps hackish. So my question is if there is a simpler less convoluted way to do this with OpenMP?
The code below basically runs the first code I listed for each thread. The result is that values of a
in a given thread are correct up to a constant. I save the sum for each thread to an array suma
with nthreads+1
elements. This allows me to communicate between threads and determine the constant offset for each thread. Then I correct the values of a[i]
with the offset.
float *suma;
#pragma omp parallel
{
const int ithread = omp_get_thread_num();
const int nthreads = omp_get_num_threads();
const int start = ithread*N/nthreads;
const int finish = (ithread+1)*N/nthreads;
#pragma omp single
{
suma = new float[nthreads+1];
suma[0] = 0;
}
float sum = 0;
for (int i=start; i<finish; i++) {
sum += f(i);
a[i] = sum;
}
suma[ithread+1] = sum;
#pragma omp barrier
float offset = 0;
for(int i=0; i<(ithread+1); i++) {
offset += suma[i];
}
for(int i=start; i<finish; i++) {
a[i] += offset;
}
}
delete[] suma;
A simple test is just to set f(i) = i
. Then the solution is a[i] = i*(i+1)/2
(and at infinity it's -1/12).
Upvotes: 4
Views: 7269
Reputation: 440
For completeness sake, I add the code of OP's MWE when Hristo's remark is taken into account:
#include <iostream>
#include <omp.h>
using std::cout;
using std::endl;
const int N = 10;
const int Nthr = 4;
float f(int i) {return (float)i;}
int main(void) {
omp_set_num_threads(Nthr);
float* a = new float[N];
float *suma = new float[Nthr+1];
suma[0] = 0.0;
float sum = 0.0;
#pragma omp parallel for schedule(static) firstprivate(sum)
for (int i=0; i<N; i++) {
sum += f(i);
a[i] = sum;
suma[omp_get_thread_num()+1] = sum;
}
// this for-loop is also a commulative sum, but it has only Nthr iterations
for (int i=1; i<Nthr;i++)
suma[i] += suma[i-1];
#pragma omp parallel for schedule(static)
for(int i=0; i< N; i++) {
a[i] += suma[omp_get_thread_num()];
}
for (int i=0; i<N; i++) {
cout << a[i] << endl;
}
delete[] suma;
int n = 95;
cout << a[n] << endl << n*(n+1)/2 << endl;
delete[] a;
return 0;
}
Upvotes: 0
Reputation: 8042
You can extend your strategy to an arbitrary number of sub-regions, and reduce them recursively, using tasks:
#include<vector>
#include<iostream>
using namespace std;
const int n = 10000;
const int baseLength = 100;
int f(int ii) {
return ii;
}
int recursiveSumBody(int * begin, int * end){
size_t length = end - begin;
size_t mid = length/2;
int sum = 0;
if ( length < baseLength ) {
for(size_t ii = 1; ii < length; ii++ ){
begin[ii] += begin[ii-1];
}
} else {
#pragma omp task shared(sum)
{
sum = recursiveSumBody(begin ,begin+mid);
}
#pragma omp task
{
recursiveSumBody(begin+mid,end );
}
#pragma omp taskwait
#pragma omp parallel for
for(size_t ii = mid; ii < length; ii++) {
begin[ii] += sum;
}
}
return begin[length-1];
}
void recursiveSum(int * begin, int * end){
#pragma omp single
{
recursiveSumBody(begin,end);
}
}
int main() {
vector<int> a(n,0);
#pragma omp parallel
{
#pragma omp for
for(int ii=0; ii < n; ii++) {
a[ii] = f(ii);
}
recursiveSum(&a[0],&a[n]);
}
cout << n*(n-1)/2 << endl;
cout << a[n-1] << endl;
return 0;
}
Upvotes: 5