Difference between revisions of "99 questions/Solutions/39"
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Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range. |
Given a range of integers by its lower and upper limit, construct a list of all prime numbers in that range. |
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− | Solution 1: |
+ | '''Solution 1:''' |
<haskell> |
<haskell> |
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primesR :: Integral a => a -> a -> [a] |
primesR :: Integral a => a -> a -> [a] |
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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. |
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. |
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− | Solution 2: |
+ | '''Solution 2:''' |
<haskell> |
<haskell> |
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primes :: Integral a => [a] |
primes :: Integral a => [a] |
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Another way to compute the claimed list is done by using the ''Sieve of Eratosthenes''. The <code>primes</code> function generates a list of all (!) prime numbers using this algorithm and <code>primesR</code> 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 :)] |
Another way to compute the claimed list is done by using the ''Sieve of Eratosthenes''. The <code>primes</code> function generates a list of all (!) prime numbers using this algorithm and <code>primesR</code> 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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− | Solution 3: |
+ | '''Solution 3:''' |
− | Use the proper Sieve of Eratosthenes from e.g. [http://www.haskell.org/haskellwiki/99_questions/Solutions/31 31st question's solution] (instead of the above sieve of Turner), adjusted to start its multiples production from the given start point: |
+ | Use the ''proper'' Sieve of Eratosthenes from e.g. [http://www.haskell.org/haskellwiki/99_questions/Solutions/31 31st question's solution] (instead of the above sieve of Turner), adjusted to start its multiples production from the given start point: |
<haskell> |
<haskell> |
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{-# OPTIONS_GHC -O2 -fno-cse #-} |
{-# OPTIONS_GHC -O2 -fno-cse #-} |
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(54,893,566 reductions, 79,935,263 cells, 6 garbage collections) |
(54,893,566 reductions, 79,935,263 cells, 6 garbage collections) |
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− | > filter isPrime [10100..10200] |
+ | > filter isPrime [10100..10200] -- Sol.1 |
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] |
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] |
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− | (34,694 reductions, 55,146 cells) |
+ | (34,694 reductions, 55,146 cells) -- isPrime: Q.31-Sol.1 using primesTME |
+ | (18,880 reductions, 32,596 cells) -- isPrime: Q.31-Sol.2 using Q.35's |
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+ | -- primeFactors using primesTME |
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</haskell> |
</haskell> |
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Revision as of 07:11, 1 June 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 :)]
Solution 3:
Use the proper Sieve of Eratosthenes from e.g. 31st question's solution (instead of the above sieve of Turner), adjusted to start its multiples production from the given start point:
{-# OPTIONS_GHC -O2 -fno-cse #-}
-- tree-merging Eratosthenes sieve, primesTME of haskellwiki/prime_numbers,
-- adjusted to produce primes in a given range
primesR a b
| b<a || b<2 = []
| otherwise =
(if a <= 2 then [2] else []) ++
gaps a' (join [[x,x+step..b] | p <- takeWhile (<= z) (tail primesTME)
, let q = p*p ; step = 2*p
x = if a' <= q then q else
let r = rem (a'-q) step
in if r==0 then a' else a'-r+step ])
where
a' = if a<=3 then 3 else (if even a then a+1 else a)
z = floor $ sqrt $ fromIntegral b + 1
join (xs:t) = union xs (join (pairs t))
join [] = []
pairs (xs:ys:t) = (union xs ys) : pairs t
pairs t = t
gaps k xs@(x:t) | k==x = gaps (k+2) t
| True = k : gaps (k+2) xs
gaps k [] = [k,k+2..b]
-- 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
union a b = a ++ b
(This turned out to be quite a project, with a few quite subtle points.) It should be much better then taking a slice of a full sequential list of primes, as it won't try to generate any primes between the square root of b and a. To wit,
> primesR 10100 10200 -- Sol.3
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193]
(6,038 reductions, 11,986 cells)
> takeWhile (<= 10200) $ dropWhile (< 10100) $ primesTME -- TME: of Q.31
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193]
(140,313 reductions, 381,058 cells)
> takeWhile (<= 10200) $ dropWhile (< 10100) $ sieve [2..] -- Sol.2
where sieve (n:ns) = n:sieve [ m | m <- ns, m `mod` n /= 0 ]
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193]
(54,893,566 reductions, 79,935,263 cells, 6 garbage collections)
> filter isPrime [10100..10200] -- Sol.1
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193]
(34,694 reductions, 55,146 cells) -- isPrime: Q.31-Sol.1 using primesTME
(18,880 reductions, 32,596 cells) -- isPrime: Q.31-Sol.2 using Q.35's
-- primeFactors using primesTME
(testing with Hugs of Nov 2002).