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(putting in a better alternative for allPrimes)
 
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(**) Determine whether a given integer number is prime.
 
(**) Determine whether a given integer number is prime.
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  +
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'':
   
 
<haskell>
 
<haskell>
 
isPrime :: Integral a => a -> Bool
 
isPrime :: Integral a => a -> Bool
isPrime p = p > 1 &&
+
isPrime k = k > 1 &&
(all ((/= 0).(p `rem`)) $ candidateFactors p)
+
foldr (\p r -> p*p > k || k `rem` p /= 0 && r)
+
True primesTME
candidateFactors p = takeWhile ((<= p).(^2)) [2..]
 
 
</haskell>
 
</haskell>
   
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''.
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This uses
 
However, we don't actually need to check all natural numbers ''<= sqrt P''. We need only check the ''primes <= sqrt P'':
 
   
 
<haskell>
 
<haskell>
-- Infinite list of all prime numbers
 
 
{-# OPTIONS_GHC -O2 -fno-cse #-}
 
{-# OPTIONS_GHC -O2 -fno-cse #-}
candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in
 
takeWhile (<= z) primesTME
 
 
 
-- tree-merging Eratosthenes sieve
 
-- tree-merging Eratosthenes sieve
  +
-- producing infinite list of all prime numbers
 
primesTME = 2 : gaps 3 (join [[p*p,p*p+2*p..] | p <- primes'])
 
primesTME = 2 : gaps 3 (join [[p*p,p*p+2*p..] | p <- primes'])
 
where
 
where
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gaps k xs@(x:t) | k==x = gaps (k+2) t
 
gaps k xs@(x:t) | k==x = gaps (k+2) t
 
| True = k : gaps (k+2) xs
 
| True = k : gaps (k+2) xs
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</haskell>
   
  +
  +
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 <code>union</code> function is readily available from <hask>Data.List.Ordered</hask>&nbsp;package, put here just for reference:
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  +
  +
<haskell>
 
-- duplicates-removing union of two ordered increasing lists
 
-- duplicates-removing union of two ordered increasing lists
 
union (x:xs) (y:ys) = case (compare x y) of
 
union (x:xs) (y:ys) = case (compare x y) of
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</haskell>
 
</haskell>
   
The tree-merging Eratosthenes sieve here seems to strike a good balance between efficiency and brevity. More at [[Prime numbers]] haskellwiki page.
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Here is another solution, intended to be extremely short while still being reasonably fast.
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  +
<haskell>
  +
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
  +
</haskell>
  +
  +
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.
  +
  +
''There is a subtle bug in the version above. I'm new here (the wiki and the language) and don't know how corrections are best made (here, or on discussion?). Anyway, the above version will fail on 25, because the bound of s is incorrect. It is x+i that is bounded by the sqrt of the argument, not x. This version will work correctly:''
  +
  +
<haskell>
  +
isPrime n | n < 4 = n /= 1
  +
isPrime n = all ((/=0) . mod n) $ takeWhile (<= m) candidates
  +
where candidates = (2:3:[x + i | x <- [6,12..], i <- [-1,1]])
  +
m = floor . sqrt $ fromIntegral n
  +
</haskell>
  +
  +
  +
[[Category:Programming exercise spoilers]]

Latest revision as of 19:42, 18 January 2014

(**) 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'])
  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


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
Data.List.Ordered
 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.

There is a subtle bug in the version above. I'm new here (the wiki and the language) and don't know how corrections are best made (here, or on discussion?). Anyway, the above version will fail on 25, because the bound of s is incorrect. It is x+i that is bounded by the sqrt of the argument, not x. This version will work correctly:

isPrime n | n < 4 = n /= 1 
isPrime n = all ((/=0) . mod n) $ takeWhile (<= m) candidates 
        where candidates = (2:3:[x + i | x <- [6,12..], i <- [-1,1]])
              m = floor . sqrt $ fromIntegral n