Project Euler Problem 71 Solution

Question

Consider the fraction, nd\frac{n}{d}, where nn and dd are positive integers. If n<dn \lt d and HCF(n,d)=1\mathrm{HCF}(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d8d \leq 8 in ascending order of size, we get:

18,17,16,15,14,27,13,38,25,37,12,47,35,58,23,57,34,45,56,67,78\displaystyle \frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}

It can be seen that 25\frac{2}{5} is the fraction immediately to the left of 37\frac{3}{7}.

By listing the set of reduced proper fractions for d1,000,000d \leq 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 37\frac{3}{7}.

Haskell

prevFrac :: Int -> Int -> Int -> (Int, Int)
prevFrac num den limit = inner 1 limit 1 where
    inner n d i | i == limit = (n, d)
                | den*m < num*i && m*d > n*i = inner m i (i+1)
                | otherwise = inner n d (i+1)
                where m = num*i `quot` den

main :: IO ()
main = print $ fst $ prevFrac 3 7 1000000
$ ghc -O2 -o ordered-fractions ordered-fractions.hs
$ time ./ordered-fractions
real   0m0.012s
user   0m0.008s
sys    0m0.000s

Python

#!/usr/bin/env python
from fractions import Fraction

def main():
    three_sevenths = 3.0 / 7
    closest = Fraction(three_sevenths).limit_denominator(1000000)
    while closest == Fraction(3, 7):
        three_sevenths -= 1e-6
        closest = Fraction(three_sevenths).limit_denominator(1000000)
    print(closest.numerator)

if __name__ == "__main__": main()
$ time python3 prop-frac.py
real   0m0.030s
user   0m0.024s
sys    0m0.004s