Harmonic progression sum c++ MPI - c++

Question

I'm trying to make a parallel version of "Harmonic Progression Sum" problem using MPI.
But I'm new with MPI and I don't know how to run this method with MPI, because it isn't work.
Parallel Program:
//#include "stdafx.h"
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <mpi.h>
#define d 10 //Numbers of Digits (Example: 5 => 0,xxxxx)
#define n 1000 //Value of N (Example: 5 => 1/1 + 1/2 + 1/3 + 1/4 + 1/5)
using namespace std;
int numProcess, rank, msg, source, dest, tag, qtd_elemento;
int escravo(long unsigned int *digits, int ValueEnd)
{
MPI_Status status;
MPI_Recv(digits, (d + 11), MPI_INT, MPI_ANY_SOURCE, MPI_ANY_TAG, MPI_COMM_WORLD, &status);
for (int i = 1; i <= ValueEnd; ++i) {
long unsigned int remainder = 1;
for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit) {
long unsigned int div = remainder / i;
long unsigned int mod = remainder % i;
digits[digit] += div;
remainder = mod * 10;
}
}
MPI_Send(&digits, 1, MPI_INT, 0, 1, MPI_COMM_WORLD);
}
void HPSSeguencial(char* output) {
long unsigned int digits[d + 11];
int DivN = n / 4; //Limiting slave.
for (int digit = 0; digit < d + 11; ++digit)
digits[digit] = 0;
if (rank != 0){
escravo(digits, (DivN * 1 ) );
escravo(digits, (DivN * 2 ) );
escravo(digits, (DivN * 3 ) );
escravo(digits, (DivN * 4 ) );
}
for (int i = d + 11 - 1; i > 0; --i) {
digits[i - 1] += digits[i] / 10;
digits[i] %= 10;
}
if (digits[d + 1] >= 5) {
++digits[d];
}
for (int i = d; i > 0; --i) {
digits[i - 1] += digits[i] / 10;
digits[i] %= 10;
}
stringstream stringstreamA;
stringstreamA << digits[0] << ",";
for (int i = 1; i <= d; ++i) {
stringstreamA << digits[i];
}
string stringA = stringstreamA.str();
stringA.copy(output, stringA.size());
}
int main() {
MPI_Init(&argc,&argv);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &numProcess);
char output[d + 10];
HPSSeguencial(output);
cout << output << endl;
MPI_Finalize();
system("PAUSE");
return 0;
}
Original Code
#include "stdafx.h"
#include <iostream>
#include <sstream>
#include <time.h>
#define d 10 //Numbers of Digits (Example: 5 => 0,xxxxx)
#define n 1000 //Value of N (Example: 5 => 1/1 + 1/2 + 1/3 + 1/4 + 1/5)
using namespace std;
void HPS(char* output) {
long unsigned int digits[d + 11];
for (int digit = 0; digit < d + 11; ++digit)
digits[digit] = 0;
for (int i = 1; i <= n; ++i) {
long unsigned int remainder = 1;
for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit) {
long unsigned int div = remainder / i;
long unsigned int mod = remainder % i;
digits[digit] += div;
remainder = mod * 10;
}
}
for (int i = d + 11 - 1; i > 0; --i) {
digits[i - 1] += digits[i] / 10;
digits[i] %= 10;
}
if (digits[d + 1] >= 5) {
++digits[d];
}
for (int i = d; i > 0; --i) {
digits[i - 1] += digits[i] / 10;
digits[i] %= 10;
}
stringstream stringstreamA;
stringstreamA << digits[0] << ",";
for (int i = 1; i <= d; ++i) {
stringstreamA << digits[i];
}
string stringA = stringstreamA.str();
stringA.copy(output, stringA.size());
}
int main() {
char output[d + 10];
HPS(output);
cout << output<< endl;
system("PAUSE");
return 0;
}
Examples:
Input:
#define d 10
#define n 1000
Output:
7,4854708606╠╠╠╠╠╠╠╠╠╠╠╠
Input:
#define d 12
#define n 7
Output:
2,592857142857╠╠╠╠╠╠╠╠╠╠╠╠╠╠ÀÂ♂ü─¨#
Regards
Original Code
http://regulus.pcs.usp.br/marathon/current/warmup.pdf

Answer 1

I am assuming that you want to parallelize this part:
for (int i = 1; i <= ValueEnd; ++i)
{
long unsigned int remainder = 1;
for (long unsigned int digit = 0; digit < d + 11 && remainder; ++digit)
{
long unsigned int div = remainder / i;
long unsigned int mod = remainder % i;
digits[digit] += div;
remainder = mod * 10;
}
}
You can divide each for iteration by each MPI process:
int idP = getProcessId(), numP = numberProcess();
for (int i = idP; i <= ValueEnd; i+=numP)
{
...
}
The getProcessId() gives you the process ID and numberProcess() gives you the number of process:
int getProcessId(){
int rank;
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
return rank;
}
// Get number of process
int numberProcess(){
int numProc;
MPI_Comm_size(MPI_COMM_WORLD, &numProc);
return numProc;
}
Each process will have a copy of the array digits; After the parallel for, the master process collects the results from all slaves process using a MPI_reduce. Or if you want to combine the values from all processes and distribute the result back to all processes you can use MPI_Allreduce.
long unsigned int digits[d + 11];
int DivN = n / 4; //Limiting slave.
for (int digit = 0; digit < d + 11; ++digit)
digits[digit] = 0;
if (rank != 0){
escravo(digits, (DivN * 1 ) );
escravo(digits, (DivN * 2 ) );
escravo(digits, (DivN * 3 ) );
escravo(digits, (DivN * 4 ) );
}
According to the code above process 0 will not execute the method escravo. Furthermore, you are not distributing correctly the work among processes. Process 1 will execute the out for loop inside the method escravo from 1 until n/4, but then process 2 will execute from 1 until 2n/4... Thus, you have different processes executing the same iterations, when what you really want is to divide these iterations among process.