# 99 questions/Solutions/31

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(putting in a better alternative for allPrimes) |
m |

## Revision as of 11:08, 31 May 2011

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

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*:

-- Infinite list of all prime numbers {-# OPTIONS_GHC -O2 -fno-cse #-} candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in takeWhile (<= z) primesTME -- tree-merging Eratosthenes sieve primesTME = 2 : gaps 3 (join [[p*p,p*p+2*p..] | p <- primes']) where 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 -- 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

The tree-merging Eratosthenes sieve here seems to strike a good balance between efficiency and brevity. More at Prime numbers haskellwiki page.