Python pi calculation?

Question

I am a python beginner and I want to calculate pi. I tried using the Chudnovsky algorithm because I heard that it is faster than other algorithms.

This is my code:

Answer 1

from decimal import *

#Sets decimal to 25 digits of precision
getcontext().prec = 25

def factorial(n):
    if n<1:
        return 1
    else:
        return n * factorial(n-1)

def plouffBig(n): #http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(1)/(16**k))*((Decimal(4)/(8*k+1))-(Decimal(2)/(8*k+4))-(Decimal(1)/(8*k+5))-(Decimal(1)/(8*k+6)))
        k += 1
    return pi

def bellardBig(n): #http://en.wikipedia.org/wiki/Bellard%27s_formula
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(-1)**k/(1024**k))*( Decimal(256)/(10*k+1) + Decimal(1)/(10*k+9) - Decimal(64)/(10*k+3) - Decimal(32)/(4*k+1) - Decimal(4)/(10*k+5) - Decimal(4)/(10*k+7) -Decimal(1)/(4*k+3))
        k += 1
    pi = pi * 1/(2**6)
    return pi

def chudnovskyBig(n): #http://en.wikipedia.org/wiki/Chudnovsky_algorithm
    pi = Decimal(0)
    k = 0
    while k < n:
        pi += (Decimal(-1)**k)*(Decimal(factorial(6*k))/((factorial(k)**3)*(factorial(3*k)))* (13591409+545140134*k)/(640320**(3*k)))
        k += 1
    pi = pi * Decimal(10005).sqrt()/4270934400
    pi = pi**(-1)
    return pi
print "\t\t\t Plouff \t\t Bellard \t\t\t Chudnovsky"
for i in xrange(1,20):
    print "Iteration number ",i, " ", plouffBig(i), " " , bellardBig(i)," ", chudnovskyBig(i)

Answer 2

It seems you are losing precision in this line:

pi = pi * Decimal(12)/Decimal(640320**(1.5))

Try using:

pi = pi * Decimal(12)/Decimal(640320**Decimal(1.5))

This happens because even though Python can handle arbitrary scale integers, it doesn't do so well with floats.

Bonus

A single line implementation using another algorithm (the BBP formula):

from decimal import Decimal, getcontext
getcontext().prec=100
print sum(1/Decimal(16)**k * 
          (Decimal(4)/(8*k+1) - 
           Decimal(2)/(8*k+4) - 
           Decimal(1)/(8*k+5) -
           Decimal(1)/(8*k+6)) for k in range(100))

Answer 3

For people who come here just to get a ready solution to get arbitrary precision of pi with Python (source with a couple of edits):

import decimal

def pi():
    """
    Compute Pi to the current precision.

    Examples
    --------
    >>> print(pi())
    3.141592653589793238462643383

    Notes
    -----
    Taken from https://docs.python.org/3/library/decimal.html#recipes
    """
    decimal.getcontext().prec += 2  # extra digits for intermediate steps
    three = decimal.Decimal(3)      # substitute "three=3.0" for regular floats
    lasts, t, s, n, na, d, da = 0, three, 3, 1, 0, 0, 24
    while s != lasts:
        lasts = s
        n, na = n + na, na + 8
        d, da = d + da, da + 32
        t = (t * n) / d
        s += t
    decimal.getcontext().prec -= 2
    return +s               # unary plus applies the new precision

decimal.getcontext().prec = 1000
pi = pi()