I'd like to calculate the square root of a number bigger than 10^2000 in Python. If I treat this number like a normal integer, I will always get this result back:
Traceback (most recent call last):
File "...", line 3, in <module>
print( q*(0.5) )
OverflowError: int too large to convert to float
How do I fix this? Or does a possibilty other than using Python exist to calculate this square root?
Just use the decimal module:
>>> from decimal import *
>>> Decimal(10**2000).sqrt()
Decimal('1.000000000000000000000000000E+1000')
>>> Decimal(10**200000).sqrt()
Decimal('1.000000000000000000000000000E+100000')
>>> Decimal(15**35315).sqrt()
Decimal('6.782765081358674922386659760E+20766')
You can also use the gmpy2 library.
>>> import gmpy2
>>> n = gmpy2.mpz(99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999982920000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000726067)
>>> gmpy2.get_context().precision=2048
>>> x = gmpy2.sqrt(n)
Useful links:
The usual square root methods convert the parameter to a float value before doing the calculation. As you saw, this does not work well with very large integers.
So use a function that is designed to work on arbitrarily large integers. Here is one, guaranteed to return correct integer part of the square root of any positive integer. This function drops the fractional part of the result, which may or may not be what you want. Since this function uses iteration it is also slower than the built-in square root routines. The Decimal module works on larger integers than the built-in routines but the precision of the values must be defined in advance so it does not work on arbitrarily large values.
import math
_1_50 = 1 << 50 # 2**50 == 1,125,899,906,842,624
def isqrt(x):
"""Return the integer part of the square root of x, even for very
large integer values."""
if x < 0:
raise ValueError('square root not defined for negative numbers')
if x < _1_50:
return int(math.sqrt(x)) # use math's sqrt() for small parameters
n = int(x)
if n <= 1:
return n # handle sqrt(0)==0, sqrt(1)==1
# Make a high initial estimate of the result (a little lower is slower!!!)
r = 1 << ((n.bit_length() + 1) >> 1)
while True:
newr = (r + n // r) >> 1 # next estimate by Newton-Raphson
if newr >= r:
return r
r = newr
When using sqrt from the library math, before it proceeds to square root it, it will convert the value to a float.
If we manually try to convert the 10**2000 to a float, it also triggers an error
>>> float(10**2000)
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
<ipython-input-14-6ac81f63106d> in <module>
----> 1 math.sqrt(10**2000)
OverflowError: int too large to convert to float
If we were speaking of a big number, but with the square equals or less than 308, the Decimal module would do the work as follows
>>> from decimal import Decimal
>>> Decimal(math.sqrt(10**308))
Decimal('10000000000000000369475456880582265409809179829842688451922778552150543659347219597216513109705408327446511753687232667314337003349573404171046192448274432')
However, as the number is way square is way bigger than 308, in this case, 2000, one would have to do as follows
>>> from decimal import Decimal
>>> Decimal(10**2000).sqrt()
Decimal('1.000000000000000000000000000E+1000')
Let's see the output if one tries to convert the Decimal(10**2000) to float
>>> float(Decimal(10**2000))
inf
One might also use the decimal module when working with factorials, as they tend to get large really fast.