Haskell代码可排除Eratosthenes筛网的数字无法正常工作

Question

I came with the following Sieve of Eratosthenes implementation:

sieve :: (Integral a) => [a] -> [a]
sieve [] = []
sieve (p:ps) = p:[x | x <- sieve ps, (rem x p) /= 0] 

primes :: (Integral a) => [a]
primes = sieve [2..100]

Calling primes from gchi prints:

[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]

Adding the optimization step of starting marking numbers from p ² , I ended up with the following code:

sieve :: (Integral a) => [a] -> [a]
sieve [] = []
sieve (p:ps) = p:[x | x <- sieve ps, x > p ^ 2, (rem x p) /= 0] 

primes :: (Integral a) => [a]
primes = sieve [2..100]

But it produces the following output:

[2]

I'm new to Haskell so I'm having problem to understand why adding the x > p ^ 2 is producing this result.

Can you please spot my mistake by explaining how is Haskell doing the evaluation of that expression?

Answer 1

You have

sieve [2..10]

Which expands to

2 : [x1 | x1 <- sieve [3..10], x1 > 4, rem x1 2 /= 0]
    = 2 : [x1 | x1 <- 3 : [x2 | x2 <- sieve [4..10],
                                x2 > 9,
                                x2 `rem` 3 /= 0],
                x1 > 4,
                x1 `rem` 2 /= 0]

So first x1 is 3 , but 3 > 4 is False so we move to the next one:

    = 2 : [x1 | x1 <- [x2 | x2 <- sieve [4..10],
                            x2 > 9,
                            x2 `rem` 3 /= 0],
                x1 > 4,
                x1 `rem` 2 /= 0]

    = 2 : [x1 | x1 <- [x2 | x2 <- 4 : [x3 | x3 <- sieve [5..10],
                                            x3 > 16,
                                            x3 `rem` 4 /= 0],
                            x2 > 9,
                            x2 `rem` 3 /= 0],
                x1 > 4,
                x1 `rem` 2 /= 0]

So if x2 is 4 , x2 > 9 is false, so we move to the next element:

    = 2 : [x1 | x1 <- [x2 | x2 <- [x3 | x3 <- sieve [5..10],
                                        x3 > 16,
                                        x3 `rem` 4 /= 0],
                            x2 > 9,
                            x2 `rem` 3 /= 0],
                x1 > 4,
                x1 `rem` 2 /= 0]

So already we can see that the only actual value we know gets returned is 2 , 3 is skipped because 3 > 4 is False . 4 gets skipped, but for the wrong reason, and 5 will get skipped because 5 > 16 is False , and so on. The problem here is that your condition x > p ^ 2 filters the entire list, but you really want to just jump ahead in your list. This means that values you're actually interested in are getting filtered out from the output.

Answer 2

This line of code:

p:[x | x <- sieve ps, x > p ^ 2, (rem x p) /= 0] 

says "take all x from sieve ps such that x > p^2 and x is not divisible by p ". Which means that all numbers less than or equal to p^2 are thrown away. It is obviously not correct(the smallest counter example: [2..3] ).