/* use csbackup.joensuu.fi instead of cs.joensuu.fi
   
   compile: /usr/local/mpich-1.2.0/bin/mpicc ds.ex2.6-7.example.c
   run : /usr/local/mpich-1.2.0/bin/mpirun -np 5 a.out
   for nice printing:
   /usr/local/mpich-1.2.0/bin/mpirun -np 5 a.out | sort

   To understand the program, draw the processors and slowly draw each
   message.

6) Write an algorithm (a working C-program) for counting and
   distributing a sum. At the beginning, there is an integer on each
   process. At the end, each process will have a copy of the sum of
   all integers. What is the time complexity of your answer?
*/
   
#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>

/* a variable needeed for call, result not used */
static MPI_Status tmpstat;

/* logical directions */
#define Left (PID-1)
#define Right (PID+1)
#define Any MPI_ANY_SOURCE

/* Send and Receive operations
   DEST & SOURCE can be Left/Right or a PID (int 0..P-1)
   SOURCE can be Any
*/

/* Send the value of an int _variable_ VAR to destination process DEST */
#define Send(DEST, VAR) MPI_Send(&VAR, 1, MPI_INT, DEST, 0, MPI_COMM_WORLD)

/* Receive an integer to an int variable VAR from destination process SOURCE */
#define Receive(SOURCE, VAR) MPI_Recv(&VAR, 1, MPI_INT, SOURCE, 0, MPI_COMM_WORLD, &tmpstat)



int main(int argc, char **argv)
{
  int PID,	/* process-ID */
	  P;	/* number of processes */ 

  /* own variables */
  int sum, mess, rnd;

  /* keep these 3 calls */
  MPI_Init(&argc, &argv);
  MPI_Comm_rank(MPI_COMM_WORLD, &PID);
  MPI_Comm_size(MPI_COMM_WORLD, &P);

  /* actual code */

  srand(PID);
  rnd = rand() % 10;

  /* All print just for debugging */
  printf("0Proc: %d, rnd: %d \n", PID, rnd);

  /* Receive from left, except 0 just assign sum */
  if (PID != 0) {
	  Receive(Left, mess);
	  sum = rnd + mess;
  } else {
	  sum = rnd;
  }


  /* send sum right, except P-1 */
  /* receive final sum from right */
  if (PID != P-1) {
	  /* |-> */
	  Send(Right, sum);
	  /* |<- */
	  Receive(Right, sum);
  }

  /* All print just for debugging */
  printf("1Proc: %d, sum: %d \n", PID, sum);

  /* send final sum to left */
  if (PID != 0) {
	  /* <-| */
	  Send(Left, sum);
  }

  /* keep this call */
  MPI_Finalize();

  exit(0);

}