## Project Euler Problem 243 Solution

## Question

A positive fraction whose numerator is less than its denominator is called a proper fraction.

For any denominator, $d$, there will be $d - 1$ proper fractions; for example, with $d = 12$:

$\displaystyle \frac{1}{12}, \frac{2}{12}, \frac{3}{12}, \frac{4}{12}, \frac{5}{12}, \frac{6}{12}, \frac{7}{12}, \frac{8}{12}, \frac{9}{12}, \frac{10}{12}, \frac{11}{12}.$

We shall call a fraction that cannot be cancelled down a *resilient fraction*.

Furthermore we shall define the *resilience* of a denominator, $R(d)$, to be the ratio of its proper fractions that are resilient; for example, $R(12) = \frac{4}{11}$.

In fact, $d = 12$ is the smallest denominator having a resilience $R(d) < \frac{4}{10}$.

Find the smallest denominator $d$, having a resilience $R(d) < \frac{15499}{94744}$.

## Haskell

```
import Data.List (union)
import qualified Data.Set as Set
pairwise :: (a -> a -> a) -> [a] -> [a]
pairwise f (xs:ys:t) = f xs ys : pairwise f t
pairwise _ t = t
primes :: [Int]
primes = 2 : _Y ((3 :) . gaps 5 . _U . map (\p-> [p*p, p*p+2*p..]))
where
_Y g = g (_Y g) -- recursion, Y combinator
_U ((x:xs):t) = x : (union xs . _U . pairwise union) t -- ~= nub.sort.concat
gaps k s@(x:xs)
| k < x = k : gaps (k+2) s -- ~= [k,k+2..]\\s, when
| otherwise = gaps (k+2) xs -- k <= head s && null(s\\[k,k+2..])
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
totient :: Int -> Double
totient 1 = 1.0
totient n = (fromIntegral n) * product [1.0 - (1.0 / (fromIntegral p)) | p <- uniq $ factorize n]
resilience :: Int -> Double
resilience d = (totient d) / (fromIntegral (d - 1))
primorials :: [Int]
primorials = scanl1 (*) primes
candidates :: [Int]
candidates = expand 1 primorials where
expand m ps@(x:y:_) | m * x < y = m * x : expand (m+1) ps
| otherwise = expand 1 (tail ps)
main :: IO ()
main = print $ head [d | d <- candidates, resilience d < 15499 / 94744]
```

```
$ ghc -O2 -o resilience resilience.hs
$ time ./resilience
real 0m0.002s
user 0m0.000s
sys 0m0.000s
```

Questions? Comments? Send me an email: z@chdenton.com