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I'm learning Haskell, and I've tried to generate an infinite list of primes, but I can't understand what my function is doing wrong.
The function:
prime = 2:3:filter (\x -> all (\y -> (mod x y) > 0) (init prime)) [5..]
I think it's the init prime, but the strange thing is that even if I set an upper bound to the range (5..10 for example), the function loops forever and never gets any result for prime !! 2
Can you please tell me what I'm doing wrong?
Well, for one let's look at what init does for a finite list:
init [1] == []
init [1,2] == [1]
init [1,2,3] == [1,2]
ok, so it gives us all but the last element of the list.
So what's init primes? Well, prime without the last element. Hopefully if we implemented prime correctly it shouldn't have a last element (because there are infinitely many primes!), but more importantly we don't quite need to care yet because we don't have the full list for now anyway - we only care about the first couple of elements after all, so for us it's pretty much the same as just prime itself.
Now, looking at all: What does this do? Well, it takes a list and a predicate and tells us if all the elements of the list satisfy the predicate:
all (<5) [1..4] == True
all even [1..4] == False
it even works with infinite lists!
all (<5) [1..] == False
so what's going on here? Well, here's the thing: It does work with infinite lists... but only if we can actually evaluate the list up to the first element of the list that violates the predicate! Let's see if this holds true here:
all (\y -> (mod 5 y) > 0) (init prime)
so to find out if 5 is a prime number, we'd have to check if there's a number in prime minus the last element of prime that divides it. Let's see if we can do that.
Now let's look at the definition of prime, we get
all (\y -> (mod 5 y) > 0) (2:3:filter (\x -> all (\y -> (mod x y) > 0) (init prime)) [5..])
So to determine whether 5 is a prime number, we only have to check if it's:
divisible by 2 - it's not, let's continue
divisible by 3 - still no
divisible by ...? Well, we're in the process of checking what the 3rd prime is so we don't know yet...
and there's the crux of the problem. With this logic, to determine the third prime number you need to know the third prime number! Of course logically, we actually don't want to check this at all, rather we only need to check if any of the smaller prime numbers are divisors of the current candidate.
So how do we go about doing that? Well, we'll have to change our logic unfortunately. One thing we can do is try to remember how many primes we already have, and only take as many as we need for our comparison:
prime = 2 : 3 : morePrimes 2 [5..]
morePrimes n (x:xs)
| all (\y -> mod x y > 0) (take n prime) = x : morePrimes (n+1) xs
| otherwise = morePrimes n xs
so how does this work? Well, it basically does what we were just talking about: We remember how many primes we already have (starting at 2 because we know we have at least [2,3] in n. We then check if our next prime is divisible by any of the of n primes we already know by using take n, and if it is we know it's our next prime and we need to increment n - otherwise we just carry on.
There's also the more well known form inspired by (although not quite the same as) the Sieve of Eratosthenes:
prime = sieve [2..] where
sieve (p:xs) = p : sieve (filter (\x -> mod x p > 0) xs)
so how does this work? Well, again with a similar idea: We know that the next prime number needs to be non-divisible by any previous prime number. So what do we do? Well, starting at 2 we know that the first element in the list is a prime number. We then throw away every number divisible by that prime number using filter. And afterwards, the next item in the list is going to be a prime number again (because we didn't throw it away), so we can repeat the process.
Neither of these are one liners like the one you were hoping for though.
If the code in the other answer is restructured under the identity
[take n primes | n <- [0..]] == inits primes
eventually we get
import Data.List
-- [ ([], 2), ([2], 3), ([2,3], 5), ... ]
primes = 2 : [ c | (ps, p) <- zip (inits primes) primes,
c <- take 1 [c | c <- [p+1..],
and [mod c p > 0 | p <- ps]]]
Further improving it algorithmically, it becomes
primes = 2 : [ c | (ps, r:q:_) <- zip (inits primes) -- [] [3,4,...]
(tails $ 3 : map (^2) primes), -- [2] [4,9,...]
c <- [r..q-1], and [mod c p > 0 | p <- ps]] -- [2,3] [9,25,...]
isqrt :: Integer -> Integer
isqrt = floor . sqrt . fromIntegral
primes :: [Integer]
primes = sieve [2..] where
sieve (p:ps) = p : sieve [x | x <- ps, x `mod` p > 0]
primeFactors :: Integer -> [Integer]
primeFactors n = takeWhile (< n) [x | x <- primes, n `mod` x == 0]
Here is my code. I think you guessed what I am trying to do: A list of prime factors of a given number using infinite list of prime numbers. But this code does not evaluate lazily.
When I use ghci and :l mycode.hs and enter primeFactors 24, the result is [2, 3 ( and the cursor constantly flashing there) there isn't a further Prelude> prompt. I think there is a problem there. What am I doing wrong?
Thanks.
takeWhile never terminates for composite arguments. If n is composite, it has no prime factors >= n, so takeWhile will just sit there.
Apply takeWhile to the primes list and then filter the result with n mod x, like this:
primeFactors n = [x | x <- takeWhile (<= n) primes, n `mod` x == 0]
(<= is used instead of < for maximum correctness, so that prime factors of a prime number would consist of that number).
Have an illustration of what happens:
http://sketchtoy.com/67338195
Your problem isn't directly takeWhile, but rather the list comprehension.
[x | x <- primes, n `mod` x == 0]
For n = 24, we get 24 `mod` 2 == 0 and 24 `mod` 3 == 0, so the value of this list comprehension starts with 2 : 3 : .... But consider the ... part.
The list comprehension has to keep pulling values from primes and checking 24 `mod` x == 0. Since there are no more prime factors of 24 nothing will ever pass that test and get emitted as the third value of the list comprehension. But since there's always another prime to test, it will never stop and conclude that the remaining tail of the list is empty.
Because this is lazily evaluated, if you only ever ask for the first two elements of this list then you're fine. But if your program ever needs the third one (or even just to know whether or not there is a third element), then the list comprehension will just spin forever trying to come up with one.
takeWhile (< 24) keeps pulling elements from its argument until it finds one that is not < 24. 2 and 3 both pass that test, so takeWhile (< 24) does need to know what the third element of the list comprehension is.
But it's not really a problem with takeWhile; the problem is that you've written a list comprehension to find all of the prime factors (and nothing else), and then trying to use a filter on the results of that to cut off the infinite exploration of all the higher primes that can't possibly be factors. That doesn't really make sense if you stop to think about it; by definition anything that isn't a prime factor can't be an element of that list, so you can't filter out the non-factors larger than n from that list. Instead you need to filter the input to that list comprehension so that it doesn't try to explore an infinite space, as #n.m's answer shows.
I need to express the sequence of prime numbers. (struggling with ex 3 in project Euler).
I have happened to this recursive definition:
is_not_dividable_by :: (Integral a) => a -> a -> Bool
is_not_dividable_by x y = x `rem` y /= 0
accumulate_and :: (Integral a) => [a] -> (a -> Bool) -> Bool
accumulate_and (x:xs) (f) = (accumulate_and xs (f)) && f(x)
accumulate_and [] f = True
integers = [2,3..]
prime_sequence = [n | n <- integers, is_prime n]
where is_prime n = accumulate_and
(takeWhile (<n) (prime_sequence))
( n `is_not_dividable_by`)
result = take 20 prime_sequence
str_result = show result
main = putStrLn str_result
Though it compiles well, but when executed, it falls into a loop, and just returns <<loop>>
My problem is that I think that I can freely express recursive definitions in Haskell.
But obviously this definition does not fit with the language at all.
However, when I mentally try to solve the prime_sequence, I think I succeed and grow the sequence, but of course with imperative programming apriori.
What is plain wrong in my recursive definition, that makes this code not work in Haskell ?
The culprit is this definition:
prime_sequence = [n | n <- [2,3..], is_prime n] where
is_prime n = accumulate_and
(takeWhile (< n) (prime_sequence))
( n `is_not_dividable_by`)
Trying to find the head element of prime_sequence (the first of the 20 to be printed by your main) leads to takeWhile needing to examine prime_sequence's head element. Which leads to a takeWhile call needing to examine prime_sequence's head element. And so it goes, again and again.
That's the black hole, right away. takeWhile can't even start walking along its input, because nothing's there yet.
This is fixed easily enough by priming the sequence:
prime_sequence = 2 : [n | n <- [3,4..], is_prime n] where
is_prime n = accumulate_and
(takeWhile (< n) (prime_sequence))
( n `is_not_dividable_by`)
Now it gets to work, and hits the second problem, described in Rufflewind's answer: takeWhile can't stop walking along its input. The simplest fix is to stop at n/2. But it is much better to stop at the sqrt:
prime_sequence = 2 : [n | n <- [3,4..], is_prime n] where
is_prime n = accumulate_and
(takeWhile ((<= n).(^ 2)) (prime_sequence))
( n `is_not_dividable_by`)
Now it should work.
The reason it's an infinite loop is because of this line:
prime_sequence =
[n | n <- integers, is_prime n]
where is_prime n = accumulate_and (takeWhile (< n) prime_sequence)
(n `is_not_dividable_by`)
In order to compute is_prime n, it needs to take all the prime numbers less than n. However, in order for takeWhile to know when to stop taking it needs need to also check for n, which hasn't been computed yet.
(In a hand-wavy manner, it means your prime_sequence is too lazy so it ends up biting its own tail and becoming an infinite loop.)
Here's how you can generate an infinite list of prime numbers without running into an infinite loop:
-- | An infinite list of prime numbers in ascending order.
prime_sequence :: [Integer]
prime_sequence = find [] integers
where find :: [Integer] -> [Integer] -> [Integer]
find primes [] = []
find primes (n : remaining)
| is_prime = n : find (n : primes) remaining
| otherwise = find primes remaining
where is_prime = accumulate_and primes (n `is_not_dividable_by`)
The important function here is find, which takes an existing list of primes and a list of remaining integers and produces the next remaining integer that is prime, then delays the remaining computation until later by capturing it with (:).
I'm currently stuck on setting upper limits in list comprehensions.
What I'm trying to do is to find all Fibonacci numbers below one million.
For this I had designed a rather simple recursive Fibonacci function
fib :: Int -> Integer
fib n
n == 0 = 0
n == 1 = 1
otherwise = fib (n-1) + fib (n-2)
The thing where I'm stuck on is defining the one million part. What I've got now is:
[ fib x | x <- [0..35], fib x < 1000000 ]
This because I know that the 35th number in the Fibonacci sequence is a high enough number.
However, what I'd like to have is to find that limit via a function and set it that way.
[ fib x | x <- [0..], fib x < 1000000 ]
This does give me the numbers, but it simply doesn't stop. It results in Haskell trying to find Fibonacci numbers below one million further in the sequence, which is rather fruitless.
Could anyone help me out with this? It'd be much appreciated!
The check fib x < 1000000 in the list comprehension filters away the fib x values that are less than 1000000; but the list comprehension has no way of knowing that greater values of x imply greater value of fib x and hence must continue until all x have been checked.
Use takeWhile instead:
takeWhile (< 1000000) [ fib x | x <- [0..35]]
A list comprehension is guaranteed to look at every element of the list. You want takeWhile :: (a -> Bool) -> [a] -> [a]. With it, your list is simply takeWhile (< 1000000) $ map fib [1..]. The takeWhile function simply returns the leading portion of the list which satisfies the given predicate; there's also a similar dropWhile function which drops the leading portion of the list which satisfies the given predicate, as well as span :: (a -> Bool) -> [a] -> ([a], [a]), which is just (takeWhile p xs, dropWhile p xs), and the similar break, which breaks the list in two when the predicate is true (and is equivalent to span (not . p). Thus, for instance:
takeWhile (< 3) [1,2,3,4,5,4,3,2,1] == [1,2]
dropWhile (< 3) [1,2,3,4,5,4,3,2,1] == [3,4,5,4,3,2,1]
span (< 3) [1,2,3,4,5,4,3,2,1] == ([1,2],[3,4,5,4,3,2,1])
break (> 3) [1,2,3,4,5,4,3,2,1] == ([1,2,3],[4,5,4,3,2,1])
It should be mentioned that for such a task the "canonical" (and faster) way is to define the numbers as an infinite stream, e.g.
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
takeWhile (<100) fibs
--[0,1,1,2,3,5,8,13,21,34,55,89]
The recursive definition may look scary (or even "magic") at first, but if you "think lazy", it will make sense.
A "loopy" (and in a sense more "imperative") way to define such an infinite list is:
fibs = map fst $ iterate (\(a,b) -> (b,a+b)) (0,1)
[Edit]
For an efficient direct calculation (without infinite list) you can use matrix multiplication:
fib n = second $ (0,1,1,1) ** n where
p ** 0 = (1,0,0,1)
p ** 1 = p
p ** n | even n = (p `x` p) ** (n `div` 2)
| otherwise = p `x` (p ** (n-1))
(a,b,c,d) `x` (q,r,s,t) = (a*q+b*s, a*r+b*t,c*q+d*s,c*r+d*t)
second (_,f,_,_) = f
(That was really fun to write, but I'm always grateful for suggestions)
The simplest thing I can think of is:
[ fib x | x <- [1..1000000] ]
Since fib n > n for all n > 3.
I am attempting to make an algorithm to solve Project Euler Problem 255
I came up with this solution:
roundedSq n | (roundedSq n) == roundedSq (n-1) = n : roundedSq (n+1)
| rem n 2 == 1 = n : floor ( ((2*10^((d-1) `div` 2)) + ceiling (n `div` (2*10^((d-1) `div` 2)) )) `div` 2 )
| otherwise = n : floor ( ((7*10^((d-2) `div` 2)) + ceiling (n `div` (7*10^((d-2) `div` 2)) )) `div` 2 )
where
d = length (map (digitToInt) (show (n)))
average a = (sum a) `div` (length a)
answer = average [map roundedSq [10E13..10E14]]
main = do
print answer
But every time I try to load, it comes with an error for the answer calculating function.
What have I done wrong and will this even give me a correct solution or will it just get stick in a loop??
answer = average [map roundedSq [10E13..10E14]]
You've put the mapped list into a list of one element here. I think perhaps you mean:
answer = average (map roundedSq [10E13..10E14])
There is a problem with your average.
average a = (sum a) `div` (length a)
sum uses the entire list. length also uses the entire list. This means that the whole list will be generated and held in memory while one of these functions traverses it, and will not be garbage collected until the other function traverses it.
You pass average a very large list, so you will run out of memory.
Solution: rewrite average as a function that only traverses the list once, so that the list can be garbage collected as it is generated.
Example (untested):
average a = sum `div` length
where (sum, length) = foldl' f (0, 0) a
f (sum, length) i = (sum + i, length + 1)
Note that this uses foldl', from Data.List, not foldl. foldl has its own space issues (which someone more knowledgeable than me may wish to comment on).
And as Tobias Wärre has pointed out,
roundedSq n | (roundedSq n) == roundedSq (n-1) = n : roundedSq (n+1)
will result in an endless loop:
we want to evaluate roundedSq n
if (roundedSq n) == roundedSq (n-1) is true, we will return n : roundedSq (n+1) as the answer
we need to evaluate (roundedSq n) == roundedSq (n-1)
we need to evaluate roundedSq n
goto 1.
If you want average to return a Fractional number you'll need to use this definition:
average a = (sum a) / (fromIntegral $ length a)
(/) is the Fractional division operator whereas div is the Integral division operator. Note you also need to use fromIntegral because length returns an Int which is not a part of the Fractional type class.
You'll get stuck in an infinite loop due to
roundedSq n | (roundedSq n) ...
Edit: Sometimes it seems like I have a hole in my head. Of course average is ok.
However, since you don't decrement or increment all recursive calls to roundedSq you will never hit the "bottom" and terminate.