Difference between revisions of "99 questions/Solutions/31"
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<haskell> |
<haskell> |
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{-# OPTIONS_GHC -O2 -fno-cse #-} |
{-# OPTIONS_GHC -O2 -fno-cse #-} |
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candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in |
candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in |
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-- tree-merging Eratosthenes sieve |
-- tree-merging Eratosthenes sieve |
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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']) |
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where |
where |
Revision as of 12:49, 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:
{-# OPTIONS_GHC -O2 -fno-cse #-}
candidateFactors p = let z = floor $ sqrt $ fromIntegral p + 1 in
takeWhile (<= z) primesTME
-- 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
-- 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.