Difference between revisions of "99 questions/Solutions/31"

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(**) Determine whether a given integer number is prime.
  
(**) 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'':
   
 
<haskell>
  
<haskell>
 
isPrime :: Integral a => a -> Bool
  
isPrime :: Integral a => a -> Bool
−
isPrime p = p > 1 && (all (\n -> p `mod` n /= 0 ) $ takeWhile (\n -> n*n <= p) [2..])
 +
isPrime k = k > 1 &&
  +
foldr (\p r -> p*p > k || k `rem` p /= 0 && r)
  +
True primesTME
 
</haskell>
  
</haskell>
   
  +
This uses
⚫
Well, a natural number p is a prime number iff it is larger than 1 and no natural number n with n >= 2 and 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 remainder concerning the division of p by n.
  
  +
  +
<haskell>
  +
{-# 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
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gaps k xs@(x:t) | k==x = gaps (k+2) t
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| True = k : gaps (k+2) xs
  +
</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:
   
−
However, we don't actually need to check all natural numbers <= sqrt P. We need only check the all natural primes <= sqrt P.
  
   
 
<haskell>
  
<haskell>
  +
-- duplicates-removing union of two ordered increasing lists
⚫
-- Infinite list of all prime numbers
  
  +
union (x:xs) (y:ys) = case (compare x y) of
−
allPrimes :: [Int]
  
  +
LT -> x : union xs (y:ys)
−
allPrimes = filter (isPrime) [2..]
  
  +
EQ -> x : union xs ys
  +
GT -> y : union (x:xs) ys
  +
</haskell>
   
  +
Here is another solution, intended to be extremely short while still being reasonably fast.
−
isPrime :: Int -> Bool
  
  +
−
isPrime p
  
  +
<haskell>
−
| p < 2 = error "Number too small"
  
−
| p == 2 = True
 +
isPrime :: (Integral a) => a -> Bool
  +
isPrime n | n < 4 = n > 1
−
| p > 2 = all (\n -> p `mod` n /= 0) (getPrimes sqrtp)
  
  +
isPrime n = all ((/=0).mod n) $ 2:3:[x + i | x <- [6,12..s], i <- [-1,1]]
−
where getPrimes z = takeWhile (<= z) allPrimes
  
−
sqrtp = floor . sqrt $ fromIntegral p
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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>
  
</haskell>
   
−
Note that the mutual dependency of allPrimes and isPrime would result in an infinite loop if we weren't careful. But since we limit our observation of allPrimes to <= sqrt x, we avoid infinite recursion.
  
   
  +
[[Category:Programming exercise spoilers]]
−
While the mutual dependency is interesting, this second version is not necessarily more efficient than the first. Though we avoid checking all natural numbers <= sqrt P in the isPrime method, we instead check the primality of all natural numbers <= sqrt P in the allPrimes definition.
  

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
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