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pythonmath
Rave
Rave

Reputation: 843

counting the number of digits in a larger number

I am currently trying out Project Euler and one of the question is calculate 2^1000, and count the number of digits. I can easily do it for 2^15 but the issue with 2^1000 is that when I calculate it, it's given in scientific notation, so it's hard to sum the digits.

import math

def power(x):
    y_p=1000*math.log(x,10.0)
    y=math.pow(10,y_p)
    return y

if __name__=="__main__":

    ans=power(2)
    a=str(ans)
    print a
    sum=0

    for i in a:
       if i == ".":
          print "encountered ."
       elif i == "e":
          break
       else:
          sum=sum+int(i)

    print sum

Upvotes: 2

Views: 3963

Answers (8)

squikel
squikel

Reputation: 1

Another option to count the number of digits of a big integer is

int(math.ceil(math.log10(big_integer)))

For example:

>>> big_integer=2**100
>>> print int(math.ceil(math.log10(big_integer)))
31
>>> print big_integer
1267650600228229401496703205376

Upvotes: 0

Sudhir
Sudhir

Reputation: 31

Actually the easier method is to use log (or log10) in python. log10 has enough precision for this and can even solve 2**1000000 instantly without taking any space.

from math import log10
ans=int(1000*log10(2))+1

Upvotes: 3

Johan Lundberg
Johan Lundberg

Reputation: 27038

The other answers address how to sum the digits in a large number, but if your question is indeed on how to get the number of digits then just do

largenumber=2*1000
int(math.log(largenumber,10)+1)

or 

len(str(largenumber))

Upvotes: 2

Raymond Hettinger
Raymond Hettinger

Reputation: 226256

To work this problem, use integer math and convert the result to a string:

>>> digits = str(2 ** 1000)
>>> len(digits)              # count the digits
302
>>> sum(map(int, digits))    # sum the digits
1366

Upvotes: 0

thesamet
thesamet

Reputation: 6582

To calculate 2^1000 in Python use 2**1000. Using floating point functions like math.log and math.pow you are likely to get inaccurate results.

Now, here is how to do it:

l = str(2**1000)
digits = [int(digit) for digit in l]
print sum(digits)

The first line converts the number to a string in base 10 representation.The second line iterates on the characters and transforms the string to a list of digits. And the third prints their sum.

Upvotes: 6

danr
danr

Reputation: 2415

Avoid using double precision and use arbitrary length integers instead:

  sum([ int(i) for i in str(2 ** 1000) ])

Upvotes: 0

Sufian Latif
Sufian Latif

Reputation: 13356

Use integer arithmetic, ints in python does not overflow, so no need to do the floating point calculations. Calculate the power yourself:

def pow(a, b):
    n = 1
    for i in range(b):
        n *= a
    return a

which is O(n). You can also try the O(lg n) method:

def pow(a, b):
    if b == 0:
        return 1
    temp = pow(a, b/2)
    if b % 2 == 0:
        return temp * temp
    return temp * temp * a

Calculate the sum of digits as you are doing now.

Upvotes: 0

Nathan Binkert
Nathan Binkert

Reputation: 9124

What about using long()?

import math

def power(x):
    y_p = 1000 * math.log(x, 10.0)
    y = math.pow(10, y_p)
    return long(y)  # convert to long since we know it is an integral value

if __name__ == "__main__":
    ans = power(2)
    a = str(ans)
    print a
    print len(a)

Upvotes: 0

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