#include <iostream>
#include <iomanip>
#include <mpi.h>

/*

# Goal

Estimate pi using Leibniz formula:
 pi/4 = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{sn + 1}.

Example from:
 Pitt-Francis and Whiteley, Guide to Scientific Computing in C++, Springer 2012, ch.11


# Compilation

$ mpic++ -o leibniz.out leibniz.cpp


# Execution, and expected (approximate) result.

Result of my laptop below.
Since this machine has four processors, there is no gain increasing
the number of proccesses to five or more.

Speedup    = T_1 / T_p   = [1, 1.846, 2.599, 3.313]
Efficiency = Speedup / p = [1, 0.923, 0.866, 0.828]


$ time mpirun -np 1 ./leibniz.out
pi is approximated as 3.141592653!

real    0m14.397s
user    0m14.222s
sys     0m0.048s


$ time mpirun -np 2 ./leibniz.out
pi is approximated as 3.141592653!

real    0m7.801s
user    0m15.170s
sys     0m0.073s


$ time mpirun -np 3 ./leibniz.out
pi is approximated as 3.141592653!

real    0m5.539s
user    0m15.671s
sys     0m0.105s


$ time mpirun -np 4 ./leibniz.out
pi is approximated as 3.141592653!

real    0m4.346s
user    0m15.925s
sys     0m0.174s


$ time mpirun -np 8 ./leibniz.out
pi is approximated as 3.141592653!

real    0m4.411s
user    0m16.172s
sys     0m0.280s

*/


int main(int argc, char* argv[]) {

  long long max_n = 1000000000;
  MPI::Init(argc, argv);

  int num_procs = MPI::COMM_WORLD.Get_size();
  int rank = MPI::COMM_WORLD.Get_rank();

  double sum = 0;
  for (long long n=rank; n<max_n; n+=num_procs) {
    double tmp = 1.0 / ( 2.0 * (double)n + 1);
    if (n % 2 == 0) {
      sum += tmp;
    } else {
      sum -= tmp;
    }
  }

  double global_sum;
  MPI::COMM_WORLD.Reduce(&sum, &global_sum, 1, MPI::DOUBLE, MPI::SUM, 0);
  // 1 : count, 0: destination

  if (rank == 0) {
    std::cout << std::setprecision(10) <<
      "pi is approximated as " << 4.0*global_sum << "!\n";
  }

  MPI::Finalize();
  return 0;
}