OverflowError: (34, 'Result too large')

Question

I'am getting an overflow error(OverflowError: (34, 'Result too large')
I want to calculate pi to 100 decimals here's my code:

def pi(): 
    pi = 0 
    for k in range(350): 
        pi += (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k 
    return pi 
print(pi())

Answer 1

Python floats are neither arbitary precision nor of unlimited size. When k = 349, 16.**k is much too large - that's almost 2^1400. Fortunately, the decimal library allows arbitrary precision and can handle the size:

import decimal
decimal.getcontext().prec = 100
def pi():
    pi = decimal.Decimal(0)
    for k in range(350):
        pi += (decimal.Decimal(4)/(decimal.Decimal(8)*decimal.Decimal(k+1))...)

Answer 2

You reached the limits of your platform's float support, probably after k = 256:

>>> k = 256
>>> (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: (34, 'Result too large')
>>> k = 255
>>> (4./(8.*k+1.) - 2./(8.*k+4.) - 1./(8.*k+5.) - 1./(8.*k+6.)) / 16.**k
3.19870064997e-313

See sys.float_info for the exact limitations, but you are unlikely to run into a current CPU and OS combination that'll give you 100 significant digits in any case; my MacBook Pro with 64-bit OS X will only support 15.

Use the decimal module to go beyond your hardware limitations.

from decimal import Decimal, localcontext

def pi(): 
    with localcontext() as ctx:
        ctx.prec = 100  # 100 digits precision
        pi = Decimal(0) 
        for k in range(350): 
            pi += (Decimal(4)/(Decimal(8)*k+1) - Decimal(2)/(Decimal(8)*k+4) - Decimal(1)/(Decimal(8)*k+5) - Decimal(1)/(Decimal(8)*k+6)) / Decimal(16)**k 
    return pi 

Answer 3

16.**256 is too large to be stored in double precision float. I suggest that you run your cycle for less, like range(250), because larger k values will not contribute to the first hundred digits anyway.

Another thing you might try is to multiply by 16.*(-k) instead of dividing by 16.*k. This number will be rounded to zero for large k, therefore will not give you runtime errors.

I suggest that you use numpy.power instead of **, it handles overflows better. For example, in your code numpy.power(16.,256) would evaluate to inf, and dividing a finite number by inf gives zero, which avoids runtime errors just like the method suggested in the previous paragraph.

Answer 4

I use python3.6 AMD64,I also meet this problem,this is because python built-in float is double-precision-float,it's 64 bit,in most progamming task,64 bit is enough,but in some extra task,it's not enough(like scitific computing,big data compute)

Answer 5

This is a python solution to this problem using the decimal library. This code counts one thousand digits of pi.

import decimal
def pi( prec = 10 ** 3 ):
    decimal.getcontext().prec = prec
    b =  decimal.Decimal(1)
    pi = 0
    for k in range(prec):
        pi += ( b*4/(8*k+1) - b*2/(8*k+4) - b*1/(8*k+5) - b*1/(8*k+6)) / 16**k
    return pi
print(pi())

This is a solution using only the built-in any size integers. It works much more efficiently and allows you to count ten thousand digits of pi.

def pi( prec = 10 ** 4 ):
    b = 10 ** prec
    pi = 0
    for k in range(prec):
        pi += ( b*4//(8*k+1) - b*2//(8*k+4) - b*1//(8*k+5) - b*1//(8*k+6)) // 16**k
    return pi
print(pi())

By starting this code, you can brag to your friends that you have counted ten thousand as pi :).