An approximation to the value of π can be calculated from the following expression
where the answer becomes more accurate with increasing N. As each term is independent, the summation over i can be parallelized nearly trivially.
Starting from the serial code pi.cpp (or [pi.F90}(pi.F90) for Fortran), make a version that performs the calculation in parallel.
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Divide the range over N in
ntasks, so that rank 0 does i=1, 2, ..., N / ntasks, rank 1 does i=N / ntasks + 1, N / ntasks + 2, ... , etc.. You may assume that N is evenly divisible by the number of processes. -
All tasks calculate their own partial sums
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Once finished with the calculation, all ranks expect rank 0 send their partial sum to rank 0, which then calculates the final result and prints it out.
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Run the code with different number of processes, do you get exactly the same result? If not, can you explain why?
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Make a version where rank 0 receives the partial sums with
MPI_ANY_SOURCE. When running multiple times with the same number of processes, do you get always exactly the same result? If not, can you explain why? -
(Bonus) Make a version that works with arbitrary number of processes. Now, if N cannot be divided evenly by
ntasks, some processes calculate more terms than others.