euler problem 57

57.py

"""
It is possible to show that the square root of two can be expressed as an infinite continued fraction.

 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?
"""

import sys
from fractions import Fraction
"""
 1+1/2 = 3/2
 1+1/(2+1/2) = 7/5
 1+1/(2+1/(2+1/2)) = 17/12
 so i tried to represent  1/(2+1/2) recurrence as a function!!! 
"""
def Irrational(counter):
    if counter==1:
        return Fraction(1,2)
    return Fraction(1,1)/(Fraction(2,1) + Irrational(counter=counter-1))

if __name__ =="__main__":
   sys.setrecursionlimit(1500)
   number=0
   for counter  in range(1,1001):
       ##we need to add 1 to irrational part.
       fraction = Fraction(1,1)+Irrational(counter)
       if(len(str(fraction.numerator))>len(str(fraction.denominator))):
           number=number+1
   print number