Project Euler Problem 73 Solution
Consider the fraction, , where and are positive integers. If and , it is called a reduced proper fraction.
If we list the set of reduced proper fractions for in ascending order of size, we get:
It can be seen that there are 3 fractions between and .
How many fractions lie between and in the sorted set of reduced proper fractions for ?
import qualified Data.Set as Set primes :: [Int] primes = 2 : sieve primes [3,5..] where sieve (p:ps) xs = h ++ sieve ps [x | x <- t, rem x p /= 0] where (h, t) = span (< p*p) xs factorize :: Int -> [Int] factorize n = primeFactors n primes where primeFactors 1 _ =  primeFactors _  =  primeFactors m (p:ps) | m < p * p = [m] | r == 0 = p : primeFactors q (p:ps) | otherwise = primeFactors m ps where (q, r) = quotRem m p uniq :: Ord a => [a] -> [a] uniq xs = uniq' Set.empty xs where uniq' _  =  uniq' set (y:ys) | Set.member y set = uniq' set ys | otherwise = y : uniq' (Set.insert y set) xs mobius :: Int -> Int mobius n | not squarefree = 0 | otherwise = (-1)^r where factors = factorize n uniqFactors = uniq factors r = length uniqFactors squarefree = r == length factors f :: Int -> Int f m = sum [((n - 1) `quot` 2) - (n `quot` 3) | n <- [1..m]] r :: Int -> Int r limit = sum [(mobius m) * f (limit `quot` m) | m <- [1..limit]] main :: IO () main = print $ r 12000
$ ghc -O2 -o fraction-range fraction-range.hs $ time ./fraction-range real 0m0.009s user 0m0.008s sys 0m0.000s
#!/usr/bin/env python from fractions import gcd print(sum(1 for d in range(2, 12001) for n in range(1, d) if (n*3 > d) and (n*2 < d) and gcd(n, d) == 1))
$ time python3 fraction-range.py real 0m47.171s user 0m46.848s sys 0m0.012s
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