Testing primality

Testing Primality

(for a context to this see Prime numbers).

Primality Test and Integer Factorization

Simplest primality test and integer factorization is by trial division:

import Data.List (unfoldr)
import Data.Maybe (listToMaybe)

factors :: Integer -> [Integer]
factors n 
  = unfoldr (\n -> listToMaybe 
       [(x,     div n x ) | x <- [2..n], mod n x==0])    n
  = unfoldr (\(d,n) -> listToMaybe 
       [(x, (x, div n x)) | x <- [d..n], mod n x==0]) (2,n)
  = unfoldr (\(d,n) -> listToMaybe 
       [(x, (x, div n x)) | x <- takeWhile ((<=n).(^2)) 
                       [d..] ++ [n|n>1], mod n x==0]) (2,n)
isPrime n = n > 1 && head (factors n) == n

The factors produced by this code are all prime by construction, because we enumerate possible divisors in ascending order while dividing each found factor out of the number being tested.

Of course there's no need to try any even numbers above 2. Given an infinite list of primes we can avoid any composites, not just evens:

pfactors primes n = unfoldr (\(ds, n) -> listToMaybe  
    [(x, (dropWhile (< x) ds, div n x)) 
       | x <- takeWhile ((<=n).(^2)) ds ++ [n|n>1], mod n x==0]) 
    (primes,n)
primes :: [Integer]
primes = 2 : 3 : [x | x <- [5,7..]
                    , head (pfactors (tail primes) x) == x]

Re-writing the above as a recursive code, we get:

isPrime n = n > 1 &&
    foldr (\p r -> p*p > n || ((n `rem` p) /= 0 && r))
          True primes

primeFactors n | n > 1 = go n primes   -- or go n (2:[3,5..])
   where                               -- for one-off invocation
     go n ps@(p:t)
        | p*p > n    = [n]
        | r == 0     =  p : go q ps
        | otherwise  =      go n t
                where
                  (q,r) = quotRem n p

When trying to factorize only one number or two, it might be faster to just use (2:[3,5..]) as a source of possible divisors instead of calculating the prime numbers first, depending on the speed of your primes generator. For more than a few factorizations, when no other primes source is available, just use

primes = 2 : filter isPrime [3,5..]

More at Prime numbers#Optimal trial division.

Miller-Rabin Primality Test

-- (eq. to) find2km (2^k * n) = (k,n)
find2km :: Integral a => a -> (a,a)
find2km n = f 0 n
    where 
        f k m
            | r == 1 = (k,m)
            | otherwise = f (k+1) q
            where (q,r) = quotRem m 2        

-- n is the number to test; a is the (presumably randomly chosen) witness
millerRabinPrimality :: Integer -> Integer -> Bool
millerRabinPrimality n a
    | a <= 1 || a >= n-1 = 
        error $ "millerRabinPrimality: a out of range (" 
              ++ show a ++ " for "++ show n ++ ")" 
    | n < 2 = False
    | even n = False
    | b0 == 1 || b0 == n' = True
    | otherwise = iter (tail b)
    where
        n' = n-1
        (k,m) = find2km n'
        b0 = powMod n a m
        b = take (fromIntegral k) $ iterate (squareMod n) b0
        iter [] = False
        iter (x:xs)
            | x == 1 = False
            | x == n' = True
            | otherwise = iter xs

-- (eq. to) pow' (*) (^2) n k = n^k
pow' :: (Num a, Integral b) => (a->a->a) -> (a->a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
    where 
        f x n y
            | n == 1 = x `mul` y
            | r == 0 = f x2 q y
            | otherwise = f x2 q (x `mul` y)
            where
                (q,r) = quotRem n 2
                x2 = sq x
 
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c = (b * c) `mod` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a

-- (eq. to) powMod m n k = n^k `mod` m
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)

Example:

-- check if '1212121' is prime with several witnesses
> map (millerRabinPrimality 1212121) [5432,1265,87532,8765,26]
[True,True,True,True,True]