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99 questions/Solutions/39

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Another way to compute the claimed list is done by using the ''Sieve of Eratosthenes''. The <hask>primes</hask> function generates a list of all (!) prime numbers using this algorithm and <hask>primesR</hask> filter the relevant range out. [But this way is very slow and I only presented it because I wanted to show how nice the ''Sieve of Eratosthenes'' can be implemented in Haskell :)]
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Another way to compute the claimed list is done by using the ''Sieve of Eratosthenes''. The <hask>primes</hask> function generates a list of all (!) prime numbers using this algorithm and <hask>primesR</hask> filter the relevant range out. [But this way is very slow and I only presented it because I wanted to show how nicely the ''Sieve of Eratosthenes'' can be implemented in Haskell :)]

Revision as of 22:38, 19 January 2011

(*) A list of prime numbers.

Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range.

Solution 1: 

primesR :: Integral a => a -> a -> [a]
primesR a b = filter isPrime [a..b]

If we are challenged to give all primes in the range between a and b we simply take all number from a up to b and filter the primes out.

Solution 2: 

primes :: Integral a => [a]
primes = let sieve (n:ns) = n:sieve [ m | m <- ns, m `mod` n /= 0 ] in sieve [2..]
 
primesR :: Integral a => a -> a -> [a]
primesR a b = takeWhile (<= b) $ dropWhile (< a) primes
Another way to compute the claimed list is done by using the Sieve of Eratosthenes. The
primes
function generates a list of all (!) prime numbers using this algorithm and
primesR
filter the relevant range out. [But this way is very slow and I only presented it because I wanted to show how nicely the Sieve of Eratosthenes can be implemented in Haskell :)]
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