Euler 57: Continued Fractions

gistfile1.txt

# http://projecteuler.net/problem=58

# i found this confusing, so i worked through dreamshire's analysis
# numerator is always = numerator + denominator * 2
# denominator is = numerator + denominator 
# starting with numerator 3 and denominator 2
# which is round 1, so start at round 2

numerator = 3
denominator = 2
count = 0

for x in range (2,1000):
	# this is a smart way to set them so you don't have to use an intermediate var
	numerator, denominator = numerator + denominator * 2, numerator + denominator
	if len(str(numerator)) > len(str(denominator)):
		count = count + 1
		
print count
		
	

interestingly, before i understood dreamshire's version, i ended up computing a recursive version of the sqrt of 2-

def a(n):
    if n == 1:
        return 1.5
    return (a(n-1)+2/a(n-1))/2

print a(10)```