具有任意数量的生成器的Haskell列表理解

Question

So what I have so far is something like this:

combs :: [[Char]]
combs = [[i] ++ [j] ++ [k] ++ [l] | i <- x, j <- x, k <- x, l <- x]
  where x = "abc"

So this is the working function for n = 4, is there any way to make this work for an arbitrary number of generators? I could program in for n = 1, 2, 3 etc.. but ideally need it to work for any given n. For reference, x is just an arbitrary string of unique characters. I'm struggling to think of a way to somehow extract it to work for n generators.

Answer 1

You can use replicateM :

replicateM :: Applicative m => Int -> m a -> m [a]

Eg:

generate :: Num a => Int -> [[a]]
generate = flip replicateM [1,2,3]

to generate all possiible lists of a given length and consisting of elements 1..3 .

Answer 2

As far as I know, you can not construct list comprehension with an arbitrary number of generators, but usually if you do something with arbitrary depth, recursion is the way to do it.

So we have to think of solving this, in terms of itself. If you want all possible strings that can be generated with the characters in x . In case n = 0 , we can generate exactly one string: the empty string.

combs 0 = [""]

so a list with one element [] .

Now in case we want to generate strings with one characters, we can of course simply return x :

combs 1 = x

and now the question is what to do in case n > 1 . In that case we can obtain all the strings with length n-1 , and and for each such string, and each such character in x , produce a new string. Like:

combs n = [ (c:cs) | c <- x, cs <- combs (n-1) ]

Note that this makes the second case ( n = 1 ) redundant. We can pick a character c from x , and prepend that to the empty string. So a basic implementation is:

combs :: Int -> [[Char]]
combs 0 = [""]
combs n = [(c:cs) | c <- x, cs <- combs (n-1)]
    where x = "abc"

Now we can still look for improvements. List comprehensions are basically syntactical sugar for the list monad. So we can use liftA2 here:

combs :: Int -> [[Char]]
combs 0 = [""]
combs n = 
    where x = "abc"

we probably also want to make the set of characters a parameter:

import Control.Applicative(liftA2)

combs ::  Int -> [[Char]]
combs  0 = [""]
combs  n = liftA2 (:) x (combs (n-1))

and we do not have to restrict us to characters, we can produce a certesian power for all possible types:

import Control.Applicative(liftA2)

combs :: [] -> Int -> [[]]
combs _ 0 = []
combs x n = liftA2 (:) x (combs (n-1))

Answer 3

First I would translate the comprehension as a monadic expression.

x >>= \i -> x >>= \j -> x >>= \k -> x >>= \l -> return [i,j,k,l]

With n = 4 we see we have 4 x 's, and generally will have n x 's. Therefore, I am thinking about a list of x 's of length n .

[x,x,x,x] :: [[a]]

How might we go from [x,x,x,x] to the monadic expression? A first good guess is foldr , since we want to do something with each element of the list. Particularly, we want to take an element from each x and form a list with these elements.

foldr :: (a -> b -> b) -> b -> [a] -> b
-- Or more accurately for our scenario:
foldr :: ([a] -> [[a]] -> [[a]]) -> [[a]] -> [[a]] -> [[a]]

There are two terms to come up with for foldr, which I will call f :: [a] -> [[a]] -> [[a]] and z :: [[a]] . We know what foldr fz [x,x,x,x] is:

foldr f z [x,x,x,x] = f x (f x (f x (f x z)))

If we add parentheses to the earlier monadic expression, we have this:

x >>= \i -> (x >>= \j -> (x >>= \k -> (x >>= \l -> return [i,j,k,l])))

You can see how the two expressions are looking similar. We should be able to find an f and z to make them the same. If we choose f = \\xa -> x >>= \\x' -> a >>= \\a' -> return (x' : a') we get:

f x (f x (f x (f x z)))
= (\x a -> a >>= \a' -> x >>= \x' -> return (x' : a')) x (f x (f x (f x z)))
= f x (f x (f x z)) >>= \a' -> x >>= \x' -> return (x' : a')
= f x (f x (f x z)) >>= \a' -> x >>= \l -> return (l : a')
= (f x (f x z) >>= \a' -> x >>= \k -> return (k : a')) >>= \a' -> x >>= \l -> return (l : a')
= f x (f x z) >>= \a' -> x >>= \k -> x >>= \l -> return (l : k : a')

• Note that I have reversed the order of i,j,k,l to l,k,j,i but in context of finding combinations, this should be irrelevant. We could try a' ++ [x'] instead if it was really of concern.

The last step is because (a >>= \\b -> c) >>= \\d -> e is the same as a >>= \\b -> c >>= \\d -> e (when accounting for variable hygiene) and return a >>= \\b -> c is the same as (\\b -> c) a .

If we keep unfolding this expression, eventually we will reach z >>= \\a' -> … on the front. The only choice that makes sense here then is z = [[]] . This means that foldr fz [] = [[]] which may not be desirable (preferring [] instead). Instead, we might use foldr1 (for non-empty lists, and we might use Data.NonEmpty ) or we might add a separate clause for empty lists to combs .

Looking at f = \\xa -> x >>= \\x' -> a >>= \\a' -> return (x' : a') we might realise this helpful equivalence: a >>= \\b -> return (cb) = c <$> a . Therefore, f = \\xa -> x >>= \\x' -> (x' :) <$> a . Then also, a >>= \\b -> c (gb) = g <$> a >>= \\b -> c and so f = (:) <$> x >>= \\x' -> x' <$> a . Finally, a <*> b = a >>= \\x -> x <$> b and so f = (:) <$> x <*> a .

The official implementation of sequenceA for lists is foldr (\\xa -> (:) <$> x <*> a) (pure []) , exactly what we came up with here too. This can be further shortened as foldr (liftA2 (:)) (pure []) but there is possibly some optimisation difference that made the implementors not choose this.

Last step is to merely come up with a list of n x 's. This is just replicate replicate nx . There happens to be a function which does both replication and sequencing, called replicateM replicateM nx .