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99 questions/Solutions/31

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(**) Determine whether a given integer number is prime.

Well, a natural number k is a prime number if it is larger than 1 and no natural number n >= 2 with n^2 <= k is a divisor of k. However, we don't actually need to check all natural numbers n <= sqrt k. We need only check the primes p <= sqrt k:

isPrime :: Integral a => a -> Bool
isPrime k = k > 1 &&
   foldr (\p r -> p*p > k || k `rem` p /= 0 && r)
      True primesTME

This uses

{-# OPTIONS_GHC -O2 -fno-cse #-}
-- tree-merging Eratosthenes sieve
--  producing infinite list of all prime numbers
primesTME = 2 : gaps 3 (join [[p*p,p*p+2*p..] | p <- primes'])
    primes' = 3 : gaps 5 (join [[p*p,p*p+2*p..] | p <- primes'])
    join  ((x:xs):t)        = x : union xs (join (pairs t))
    pairs ((x:xs):ys:t)     = (x : union xs ys) : pairs t
    gaps k xs@(x:t) | k==x  = gaps (k+2) t 
                    | True  = k : gaps (k+2) xs

The tree-merging Eratosthenes sieve here seems to strike a good balance between efficiency and brevity. More at Prime numbers haskellwiki page. The semi-standard union function is readily available from
 package, put here just for reference:

-- duplicates-removing union of two ordered increasing lists
union (x:xs) (y:ys) = case (compare x y) of 
           LT -> x : union  xs  (y:ys)
           EQ -> x : union  xs     ys 
           GT -> y : union (x:xs)  ys

Here is another solution, intended to be extremely short while still being reasonably fast.

isPrime :: (Integral a) => a -> Bool
isPrime n | n < 4 = n > 1
isPrime n = all ((/=0).mod n) $ 2:3:[x + i | x <- [6,12..s], i <- [-1,1]]
            where s = floor $ sqrt $ fromIntegral n

This one does not go as far as the previous, but it does observe the fact that you only need to check numbers of the form 6k +/- 1 up to the square root. And according to some quick tests (nothing extensive) this version can run a bit faster in some cases, but slower in others; depending on optimization settings and the size of the input.