用多线程计算Pi(无加速度 - 做什么)

Question

我正在尝试使用Ramanujan的公式之一在Python上以任意精度计算pi:http://en.wikipedia.org/wiki/Approximations_of_%CF%80#20th_century.它基本上需要大量因子和高精度浮点数除法.

我使用多个线程来划分无限级数计算,方法是给每个线程赋予所有具有一定模数的成员除以线程数.因此,如果你有3个线程,总和应该像这样划分线程1 ---> 1,4,7 ...成员线程2 ----> 2,5,8 ...线程3 ---- > 3,6,9 ......

到目前为止,这是我的代码:

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...")

我担心的是,当使用多个线程时,它的加速度非常小或没有.例如pi的1000位数:1线程 - > 0.68秒2线程 - > 0.74秒4线程 - > 0.75秒10线程 - > 0.96秒

您对如何减少时间有任何想法吗？我在任务管理器上看到,当使用四个线程时,我的两个内核都会100%参与其中.但是时间似乎是一样的.

PS:这是一个家庭作业,所以我不能使用另一个公式.PSS:我正在使用python 2.7

谢谢:)

Answer 1

Python有一个GIL(全局解释器锁),它可以防止多个线程同时执行python代码,即你无法使用多个线程获得CPU绑定任务的加速.您必须使用多个进程.