Lazy List of Prime Numbers

Question

How would one implement a list of prime numbers in Haskell so that they could be retrieved lazily?

I am new to Haskell, and would like to learn about practical uses of the lazy evaluation functionality.

Answer 1

Here's a short Haskell function that enumerates primes from Literate Programs:

primes :: [Integer]
primes = sieve [2..]
  where
    sieve (p:xs) = p : sieve [x|x <- xs, x `mod` p > 0]

Apparently, this is not the Sieve of Eratosthenes (thanks, Landei). I think it's still an instructive example that shows you can write very elegant, short code in Haskell and that shows how the choice of the wrong data structure can badly hurt efficiency.

Answer 2

我建议采用本文中的一种实现： http : //www.cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf

Answer 3

There are a number of solutions for lazy generation of prime sequences right in the haskell wiki. The first and simplest is the Postponed Turner sieve : ^{(old revision ... NB)}

primes :: [Integer]
primes = 2: 3: sieve (tail primes) [5,7..]
 where 
  sieve (p:ps) xs = h ++ sieve ps [x | x <- t, x `rem` p /= 0]  
                                -- or:  filter ((/=0).(`rem`p)) t
                  where (h,~(_:t)) = span (< p*p) xs

Answer 4

The accepted answer from @nikie is not very efficient, is gets relatively slow after some thousands, but the answer of @sleepynate is much better. It took me some time to understand it, therefore here is the same code, but just with variables named more clearly:

lazyPrimes :: [Integer]
lazyPrimes = 2: 3: calcNextPrimes (tail lazyPrimes) [5, 7 .. ]
  where
    calcNextPrimes (p:ps) candidates =
      let (smallerSquareP, (_:biggerSquareP)) = span (< p * p) candidates in
      smallerSquareP ++ calcNextPrimes ps [c | c <- biggerSquareP, rem c p /= 0]

The main idea is that the candidates for the next primes already contain no numbers that are divisible by any prime less than the first prime given to the function. So that if you call

calcNextPrimes (5:ps) [11,13,17..]

the candidate list contains no number, that is divisible by 2 or 3 , that means that the first non-prime candidate will be 5 * 5 , cause 5* 2 and 5 * 3 and 5 * 4 are already eliminated. That allows you to take all candidates, that are smaller than the square of 5 and add them straight away to the primes and sieve the rest to eliminate all numbers divisible by 5.

Answer 5

primes = 2 : [x | x <- [3..], all (\y -> x `mod` y /= 0) 
                   (takeWhile (<= (floor . sqrt $ fromIntegral x)) primes)]

With 2 in the list initially, for each integer x greater than 2 , check if for all y in primes such that y <= sqrt(x) , x mod y != 0 holds, which means x has no other factors except 1 and itself.