99 questions/Solutions/39
From HaskellWiki
| Line 40: | Line 40: | ||
gaps a' (join [[x,x+step..b] | p <- takeWhile (<= z) primes' | gaps a' (join [[x,x+step..b] | p <- takeWhile (<= z) primes' | ||
, let q = p*p ; step = 2*p | , let q = p*p ; step = 2*p | ||
| - | x = | + | x = snapUp (max a' q) q step ]) |
where | where | ||
primes' = tail primesTME -- external unbounded list of primes | primes' = tail primesTME -- external unbounded list of primes | ||
| - | a' = | + | a' = snapUp (max 3 a) 1 2 |
z = floor $ sqrt $ fromIntegral b + 1 | z = floor $ sqrt $ fromIntegral b + 1 | ||
join (xs:t) = union xs (join (pairs t)) | join (xs:t) = union xs (join (pairs t)) | ||
| Line 52: | Line 52: | ||
| True = k : gaps (k+2) xs | | True = k : gaps (k+2) xs | ||
gaps k [] = [k,k+2..b] | gaps k [] = [k,k+2..b] | ||
| - | snapUp v origin step = let r = rem (v-origin) step | + | snapUp v origin step = let r = rem (v-origin) step -- rem OK if v>=origin |
| - | in if r==0 then v else v | + | in if r==0 then v else v+(step-r) |
-- 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 | ||
| Line 65: | Line 65: | ||
> primesR 10100 10200 -- Sol.3 | > primesR 10100 10200 -- Sol.3 | ||
[10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] | [10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] | ||
| - | (5, | + | (5,497 reductions, 11,382 cells) |
> takeWhile (<= 10200) $ dropWhile (< 10100) $ primesTME -- TME of Q.31 | > takeWhile (<= 10200) $ dropWhile (< 10100) $ primesTME -- TME of Q.31 | ||
Revision as of 10:51, 3 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 numbers from a up to b and filter all the primes through.
This is good for very narrow ranges as Q.31's isPrime tests by trial division using (up to
) a memoized primes list produced by sieve of Eratosthenes to which it refers internally. So it'll be slower, but immediate.
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 :)]
this is of course a famous case of executable specification, with all the implied pitfalls of inefficiency when (ab)used as if it were an actual code.
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 (inclusive) primesR a b | b<a || b<2 = [] | otherwise = (if a <= 2 then [2] else []) ++ gaps a' (join [[x,x+step..b] | p <- takeWhile (<= z) primes' , let q = p*p ; step = 2*p x = snapUp (max a' q) q step ]) where primes' = tail primesTME -- external unbounded list of primes a' = snapUp (max 3 a) 1 2 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] snapUp v origin step = let r = rem (v-origin) step -- rem OK if v>=origin in if r==0 then v else v+(step-r) -- 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 some 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] (5,497 reductions, 11,382 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] (15,750 reductions, 29,292 cells) -- isPrime: Q.31
(testing with Hugs of Nov 2002).
This solution is faster but not immediate. It has a certain preprocessing stage but then goes on fast to produce the whole range. To illustrate, to produce the 49 primes in 1000-wide range above 120200300100 it takes about 18 seconds on my oldish notebook for the 1st version, with the first number produced almost immediately (~ 0.4 sec); but this version spews up all 49 primes in one go after just under 1 sec.
Solution 4.
For very wide ranges, specifically when 
, we're better off just using the primes sequence itself, without any post-processing:
primes :: Integral a => [a] primes = primesTME -- of Q.31 primesR :: Integral a => a -> a -> [a] primesR a b = takeWhile (<= b) . dropWhile (< a) $ primes
