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99 questions/Solutions/31

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

Solution 1

isPrime :: Integral a => a -> Bool
isPrime p = p > 1 && 
            (all ((/= 0).(p `rem`)) $ candidateFactors p)
candidateFactors p = takeWhile ((<= p).(^2)) [2..]

Well, a natural number p is a prime number if it is larger than 1 and no natural number n >= 2 with n^2 <= p is a divisor of p. That's exactly what is implemented: we take the list of all integral numbers starting with 2 as long as their square is at most p and check that for all these n there is a non-zero remainder concerning the division of p by n.

However, we don't actually need to check all natural numbers <= sqrt P. We need only check the primes <= sqrt P:

candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in
                     takeWhile (<= z) 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, and semi-standard union function is readily available from
 package, put here just for reference. More at Prime numbers haskellwiki page.
-- 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

Solution 2

The following is faster (in Hugs, Nov 2002 version):

isPrime n = n > 1 &&
   foldr (\p r -> p*p > n || n `rem` p /= 0 && r)
      True primesTME
This reuses same primesTME list on subsequent invocations, but the
(takeWhile ...)
 list in the first solution seems to be recreated anew for each call to isPrime (i.e. it may or may not be eliminated by a compiler, while the second solution explicitly uses the same primesTME list so there's no problem to begin with).