Using the sqrt function of math module for long numbers in python

Question

I was working with numbers of 200 digits in python. When finding the square root of a number using math.sqrt(n) I am getting a wrong answer.

In[1]: n=9999999999999999999999999999999999999999999999999999999999999999999999
       999999999999999999999999998292000000000000000000000000000000000000000000
       0000000000000000000000000000000000000000000000000000726067

In[2]: x=int(math.sqrt(n))

In[3]: x
Out[1]: 10000000000000000159028911097599180468360808563945281389781327
        557747838772170381060813469985856815104L

In[4]: x*x
Out[2]: 1000000000000000031805782219519836346574107361670094060730052612580
        0264077231077619856175974095677538298443892851483731336069235827852
        3336313169161345893842466001164011496325176947445331439002442530816L

In[5]: math.sqrt(n)
Out[3]: 1e+100

The value of x is coming larger than expected since x*x (201 digits) is larger than n (200 digits). What is happening here? Is there some concept I am getting wrong here? How else can I find the root of very large numbers?

Asked By: Juan John Mathews

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Source

Answer 1

math.sqrt returns an IEEE-754 64-bit result, which is roughly 17 digits. There are other libraries that will work with high-precision values. In addition to the decimal and mpmath libraries mentioned above, I maintain the gmpy2 library (https://code.google.com/p/gmpy/).

>>> import gmpy2
>>> n=gmpy2.mpz(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
>>> gmpy2.get_context().precision=2048
>>> x=gmpy2.sqrt(n)
>>> x*x
mpfr('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0',2048)
>>>

The gmpy2 library can also return integer square roots (isqrt) or quickly check if an integer is an exact square (is_square).

Answered By: casevh

Answer 2

Here’s an integer square root program using Hero’s method that I wrote a while back. For the initial approximation it uses a number of half the bit length of the input value, so it starts converging pretty quickly. However, I haven’t timed it to see if it’s faster in Python than just using a simpler initial approximation. 🙂

#! /usr/bin/env python

''' Long integer square roots. Newton's method.

    Written by PM 2Ring. Adapted from C to Python 2008.10.19
'''

import sys

def root(m):
    # Get initial approximation
    n, a, k = m, 1, 0
    while n > a:
        n >>= 1
        a <<= 1
        k += 1
        #print k, ':', n, a

    # Go back one step & average
    a = n + (a>>2)
    #print a

    # Apply Newton's method
    while k:
        a = (a + m // a) >> 1
        k >>= 1
        #print k, ':', a
    return a

def main():
    m = len(sys.argv) > 1 and int(sys.argv[1]) or 2*10L**100
    print "The Square Root of", m
    print root(m)

if __name__ == '__main__':
    main()

Answered By: PM 2Ring

Answer 3

Using the decimal module:

import decimal
D = decimal.Decimal
n = D(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)

with decimal.localcontext() as ctx:
    ctx.prec = 300
    x = n.sqrt()
    print(x)
    print(x*x)
    print(n-x*x)

yields

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999983754999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998612677

99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

0E-100

Answered By: unutbu

Answer 4

You can find the root of very large numbers with the python3 num7 package as following:

#square root of very large numbers
from num7 import Num

n = Num('99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067.0')
r1 = n.sqrt()
print('n=>', n)
print('DIGITS', n.len()) #DIGITS (200, 0)
print('r1=>', r1)#9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999145.99999999999999999999999999999999999999999999999999999999999999999999999999999999
print('DIGITS', r1.len())#DIGITS (100, 80)
print('--'*20) 

x2 = r1*r1     
print('x2=>', x2)#99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982919999999999999999999999999999999999999999999999999999999999999999999999999999999800000000000000729316.0000000000000000000000000000000000000000000000000000000000000000000000000000170800000000000000000000000000000000000000000000000000000000000000000000000000000001
print('DIGITS', x2.len())#DIGITS (200, 160)
print('--'*20)
print('difference n-x2=>', n-x2) #199999999999999996750.9999999999999999999999999999999999999999999999999999999999999999999999999999829199999999999999999999999999999999999999999999999999999999999999999999999999999999

Answered By: giocip

Using the sqrt function of math module for long numbers in python

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