problem057.py

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

from common_funcs import answer
import fractions

def solve():
    a = fractions.Fraction(1,1)

    count = 0
    for _ in range(1000):
        a = 1 + fractions.Fraction(1,1+a)
        if len(str(a.numerator)) > len(str(a.denominator)):
            count += 1
    return count
    
answer(solve)