Calculation of Pi with multiple threads( no acceleration - what to do )

Question

I'm trying to calculate pi with arbitrary precision on Python using one of Ramanujan's formulas: http://en.wikipedia.org/wiki/Approximations_of_%CF%80#20th_century. It basically requires lots of factorials and high-precision floating numbers division.

I'm using multiple threads to divide infinite series calculation by giving each thread all the members that have a certain modulus when divided by the number of threads. So if you have 3 threads, the sum should be divided like this Thread 1 ---> 1, 4, 7... members Thread 2 ---->2, 5, 8... Thread 3 ---->3, 6, 9...

Here's my code so far:

from decimal   import *
from math      import sqrt, ceil
from time      import clock
from threading import *
import argparse

memoizedFactorials = []
memoizedFactorials.append( 1 )
memoizedFactorials.append( 1 )

class Accumulator:
    def __init__( self ):
        self._sum = Decimal( 0 )

    def accumulate( self, decimal ):
        self._sum += decimal

    def sum( self ):
        return self._sum

def factorial( k ):
    if k < 2: return 1
    elif len(memoizedFactorials) <= k:
        product = memoizedFactorials[ - 1 ] #last element 
        for i in range ( len(memoizedFactorials), k+1 ):
            product *= i;
            memoizedFactorials.append(product)

    return memoizedFactorials[ k ]

class Worker(Thread):
    def __init__( self, startIndex, step, precision, accumulator ):
        Thread.__init__( self, name = ("Thread - {0}".format( startIndex ) ) )
        self._startIndex = startIndex
        self._step = step
        self._precision = precision
        self._accumulator = accumulator

    def run( self ):
        sum = Decimal( 0 )
        result = Decimal( 1 )
        zero = Decimal( 0 )

        delta = Decimal(1)/( Decimal(10)**self._precision + 1 )
        #print "Delta - {0}".format( delta ) 
        i = self._startIndex
        while( result - zero > delta ):
            numerator = Decimal(factorial(4 * i)*(1103 + 26390 * i))
            denominator = Decimal((factorial(i)**4)*(396**(4*i)))
            result =  numerator / denominator
            print "Thread - {2} --- Iteration - {0:3} --->{1:3}".format( i, result, self._startIndex )
            sum += result
            i += self._step

        self._accumulator.accumulate( sum ) 
        print 

def main( args ):
    numberOfDigits = args.numberOfDigits;
    getcontext().prec = numberOfDigits + 8
    zero = Decimal(1) / Decimal( 10**( numberOfDigits + 1 ) )

    start = clock()
    accumulator = Accumulator()

    threadsCount = args.numberOfThreads;
    threadPool = []
    for i in range(0, threadsCount ):
        worker = Worker( i, threadsCount, numberOfDigits, accumulator )
        worker.start()
        threadPool.append( worker )

    for worker in threadPool:
        worker.join()

    sum = accumulator.sum();

    rootOfTwo = Decimal(2).sqrt()

    result = Decimal( 9801 ) / ( Decimal( 2 ) * rootOfTwo * sum ) 
    end = clock();

    delta = end - start;

    print result;
    print ("Took it {0} second to finish".format( delta ) )

    #testing the results
    #realPiFile = open("pi.txt");
    #myPi = str(result)
    #realPi = realPiFile.read( len(myPi) - 1 )

    #if ( myPi[:-1] != realPi ):
    #    print "Answer not correct!"
    #    print "My pi   - {0}".format(myPi)
    #    print "Real pi - {0}".format(realPi)

if __name__ == '__main__':
    parser = argparse.ArgumentParser(description = 'Calculate Pi at with arbitrary precision')
    parser.add_argument('-p',            dest = 'numberOfDigits',  default=20, type = int, help ='Number of digits in pi ')
    parser.add_argument('-t', '--tasks', dest = 'numberOfThreads', default=1,  type = int, help ='Number of tasks( threads )')
    parser.add_argument('-o',            dest = 'outputFileName',  type = str,             help ='Connect to VCS testing servers')
    parser.add_argument('-q', '--quet',  dest = 'quetMode'      ,  action='store_true',    help ='Run in quet mode')

    args = parser.parse_args()

    print args
    main(args)
    a = raw_input("Press any key to continue...")

What concerns me thati have very small or no acceleration in time when using multiple threads. For example 1000 digits of pi: 1 Thread --> 0.68 seconds 2 Threads --> 0.74 seconds 4 Threads --> 0.75 seconds 10 threads --> 0.96 seconds

Do you have any ideas on how to decrease the time. I see on task manager that when using four threads both of my cores get involved on 100%. However time seems to be the same.

PS: It's a homework assignment so i can't use another formula. PSS: I'm using python 2.7

Thanks:)

Answer 1

Threads do not speed things up because of the GIL (Global Interpret Lock). Use multiprocessing for this kind of task. Its usage is very similar to threads.

Answer 2

Python has a GIL(Global Interpreter Lock) that prevents more than one thread to execute python code at the same time, i.e. you cannot obtain a speed up on CPU-bound tasks using multiple threads. You must use multiple processes.

Answer 3

Instead to brute-force your way through the series and all those nasty factorials you shall definitely learn about the Binary Splitting algorithm.

http://numbers.computation.free.fr/Constants/Algorithms/splitting.html

This guy already did the work for you. It has python implementations of the binary splitting structure applied to the Chudnovsky formula:
http://www.craig-wood.com/nick/articles/pi-chudnovsky/

The main advantage of such a structure is that it eliminates the need for divisions, factorials and any floating point calculations while calculating the series. Then you perform a single, final, supermassive division between a numerator and a denominator. Actually I have no idea how to multithread it but that's a start.