# 99 questions/Solutions/31

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(additional solution) |
(change the 2nd version code) |
||

Line 43: | Line 43: | ||

'''Solution 2''' |
'''Solution 2''' |
||

− | The following seems faster (in Hugs, Nov 2002 version), with the use of Q.35 solution for <code>primeFactors</code> (which itself uses <code>primesTME</code> from here, above): |
+ | The following is faster (in Hugs, Nov 2002 version): |

<haskell> |
<haskell> |
||

− | isPrime n = n > 1 && n == head (primeFactors n) |
+ | isPrime n = n > 1 && |

+ | foldr (\p r -> p*p > n || n `rem` p /= 0 && r) |
||

+ | True primesTME |
||

</haskell> |
</haskell> |
||

− | The <code>primeFactors</code> function reuses same <code>primesTME</code> list on subsequent invocations, but the <hask>(takeWhile ...)</hask> list in the first solution seems to be recreated anew for each call to <code>isPrime</code> (i.e. it may or may not be eliminated by a compiler, while the second solution explicitly uses the same <code>primesTME</code> list so there's no problem to begin with). |
+ | This reuses same <code>primesTME</code> list on subsequent invocations, but the <hask>(takeWhile ...)</hask> list in the first solution seems to be recreated anew for each call to <code>isPrime</code> (i.e. it may or may not be eliminated by a compiler, while the second solution explicitly uses the same <code>primesTME</code> list so there's no problem to begin with). |

## Revision as of 09:04, 1 June 2011

(**) 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']) 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

`union`

function is readily available from Data.List.Ordered

-- 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

`primesTME`

list on subsequent invocations, but the (takeWhile ...)

`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).