# 99 questions/Solutions/39

### From HaskellWiki

("union" removed, appears on Q.31) |
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snap v origin step = let r = rem (v-origin) step -- v >= origin |
snap v origin step = let r = rem (v-origin) step -- v >= origin |
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in if r==0 then v else v+(step-r) |
in if r==0 then v else v+(step-r) |
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− | -- duplicates-removing union of two ordered increasing lists |
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− | union (x:xs) (y:ys) = case (compare x y) of |
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− | LT -> x : union xs (y:ys) |
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− | EQ -> x : union xs ys |
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− | GT -> y : union (x:xs) ys |
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</haskell> |
</haskell> |
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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, |
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, |

## Revision as of 11:32, 5 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 | even a = filter isPrime [a+1,a+3..b] | True = filter isPrime [a,a+2..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 numbers 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, testing the numbers one by one.

**Solution 2.**

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

**Solution 3.**

Another way to compute the claimed list is done by using the *Sieve of Eratosthenes*.

primesR :: Integral a => a -> a -> [a] primesR a b = takeWhile (<= b) $ dropWhile (< a) $ sieve [2..] where sieve (n:ns) = n:sieve [ m | m <- ns, m `mod` n /= 0 ]

The `sieve [2..]`

function call generates a list of all (!) prime numbers using this algorithm and `primesR`

filters 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 (mislabeled) executable specification, with all the implied pitfalls of inefficiency when (ab)used as if it were an actual code*.

**Solution 4.**

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:

-- tree-merging Eratosthenes sieve, primesTME of Q.31, -- adjusted to produce primes in a given range (inclusive) primesR a b | b<a || b<2 = [] | otherwise = (if a <= 2 then [2] else []) ++ (takeWhile (<= b) $ gaps a' $ join [[x,x+step..] | p <- takeWhile (<= z) primes' -- p^2 <= b , let q = p*p ; step = 2*p x = snap (max a' q) q step ]) where primes' = tail primesTME -- external unbounded list of primes a' = snap (max 3 a) 3 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 snap v origin step = let r = rem (v-origin) step -- v >= origin in if r==0 then v else v+(step-r)

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.4 [10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] (4,656 reductions, 10,859 cells) > takeWhile (<= 10200) $ dropWhile (< 10100) $ primesTME -- Sol.2 [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.3 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 [10101,10103..10200] -- Sol.1 [10103,10111,10133,10139,10141,10151,10159,10163,10169,10177,10181,10193] (12,927 reductions, 24,703 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 out all 49 primes in one go after just under 1 sec.