So today, I was pondering How difficult it’d be to implement the Sieve of Eratosthenes in Haskell.

Turns out, very easy after all.. and more elegant.

sieve [] = []
sieve (x:xs) = x : sieve [p | p <- xs, p `mod` x /= 0]

So With that figured out, A new challenge was to solve this for any n. O’course, my earlier much simpler solution would have sufficed –

foldl (lcm) 1 [1..20]

But that doesn’t quite use my own implementation of the Sieve of Eratosthenes algorithm, does it?

So that needed to be corrected 😉

So here’s my solution –

(The biggest Haskell program I’ve ever written :D)

import IO

{- Implements the Sieve of Eratosthenes -}

sieve :: [Integer] -> [Integer]
sieve [] = []
sieve (x : xs) = x : sieve [p | p <- xs, p `mod` x /= 0]

{- Lists all primes <= n -}
primes :: Integer -> [Integer]
primes n = (2 : sieve [3, 5 .. n])

{- Returns maximum exponent m in the prime factorization of n
– with the prime a as base,
– i.e. a^m * d = n, where a is prime, m is max
– for e.g. for 2, 24 it returns 2.. because 12 = 2^2 * 3
-}
nExponent :: Integer -> Integer -> Integer
nExponent a n =
if n `mod` a /= 0
then 0
else (1 + nExponent a (n `div` a))

{- Returns Maximum p’s exponent of a given set of numbers
– for e.g. for [2,4,6,8,10] and p=2 the exponents list’d be [1,2,1,3,1]
– the exponents of prime factorization where 2 is the prime number.
– And max’d be 3 for (8 = 2^3)
-}
maxExponent :: [Integer] -> Integer -> Integer
maxExponent numList p = maximum (map (nExponent p) numList)

{-
– Get list of Primes <= n
– Get the maximum exponent for each prime p
– in the prime factorizations of [1 .. n], (well [2 .. n])
– foreach (m, n) m <- primes, n <- exponents, calculate m^n and
– multiply the results
-}
main = do