Church lists in Haskell

问题:

I had to implement the haskell map function to work with church lists which are defined as following:

type Churchlist t u = (t->u->u)->u->u

In lambda calculus, lists are encoded as following:

[] := λc. λn. n
[1,2,3] := λc. λn. c 1 (c 2 (c 3 n))

The sample solution of this exercise is:

mapChurch :: (t->s) -> (Churchlist t u) -> (Churchlist s u)
mapChurch f l = \c n -> l (c.f) n

I have NO idea how this solution works and I don't know how to create such a function. I have already experience with lambda calculus and church numerals, but this exercise has been a big headache for me and I have to be able to understand and solve such problems for my exam next week. Could someone please give me a good source where I could learn to solve such problems or give me a little guidance on how it works?

回答1:

All lambda calculus data structures are, well, functions, because that's all there is in the lambda calculus. That means that the representation for a boolean, tuple, list, number, or anything, has to be some function that represents the active behavior of that thing.

For lists, it is a "fold". Immutable singly-linked lists are usually defined List a = Cons a (List a) | Nil, meaning the only ways you can construct a list is either Nil or Cons anElement anotherList. If you write it out in lisp-style, where c is Cons and n is Nil, then the list [1,2,3] looks like this:

(c 1 (c 2 (c 3 n)))

When you perform a fold over a list, you simply provide your own "Cons" and "Nil" to replace the list ones. In Haskell, the library function for this is foldr

foldr :: (a -> b -> b) -> b -> [a] -> b

Look familiar? Take out the [a] and you have the exact same type as Churchlist a b. Like I said, church encoding represents lists as their folding function.

---

So the example defines map. Notice how l is used as a function: it is the function that folds over some list, after all. \c n -> l (c.f) n basically says "replace every c with c . f and every n with n".

(c 1 (c 2 (c 3 n)))
-- replace `c` with `(c . f)`, and `n` with `n`
((c . f) 1 ((c . f) 2) ((c . f) 3 n)))
-- simplify `(foo . bar) baz` to `foo (bar baz)`
(c (f 1) (c (f 2) (c (f 3) n))

It should be apparent now that this is indeed a mapping function, because it looks just like the original, except 1 turned into (f 1), 2 to (f 2), and 3 to (f 3).

回答2:

So let's start by encoding the two list constructors, using your example as reference:

[] := λc. λn. n
[1,2,3] := λc. λn. c 1 (c 2 (c 3 n))

[] is the end of list constructor, and we can lift that straight from the example. [] already has meaning in haskell, so let's call ours nil:

nil = \c n -> n

The other constructor we need takes an element and an existing list, and creates a new list. Canonically, this is called cons, with the definition:

cons x xs = \c n -> c x (xs c n)

We can check that this is consistent with the example above, since

cons 1 (cons 2 (cons 3 nil))) =
cons 1 (cons 2 (cons 3 (\c n -> n)) = 
cons 1 (cons 2 (\c n -> c 3 ((\c' n' -> n') c n))) =
cons 1 (cons 2 (\c n -> c 3 n)) =
cons 1 (\c n -> c 2 ((\c' n' -> c' 3 n') c n) ) =
cons 1 (\c n -> c 2 (c 3 n)) =
\c n -> c 1 ((\c' n' -> c' 2 (c' 3 n')) c n) =
\c n -> c 1 (c 2 (c 3 n)) =

Now, consider the purpose of the map function - it is to apply the given function to each element of the list. So let's see how that works for each of the constructors.

nil has no elements, so mapChurch f nil should just be nil:

mapChurch f nil
= \c n -> nil (c.f) n
= \c n -> (\c' n' -> n') (c.f) n
= \c n -> n
= nil

cons has an element and a rest of list, so, in order for mapChurch f to work propery, it must apply f to the element and mapChurch f to rest of the list. That is, mapChurch f (cons x xs) should be the same as cons (f x) (mapChurch f xs).

mapChurch f (cons x xs)
= \c n -> (cons x xs) (c.f) n
= \c n -> (\c' n' -> c' x (xs c' n')) (c.f) n
= \c n -> (c.f) x (xs (c.f) n)
-- (c.f) x = c (f x) by definition of (.)
= \c n -> c (f x) (xs (c.f) n)
= \c n -> c (f x) ((\c' n' -> xs (c'.f) n') c n)
= \c n -> c (f x) ((mapChurch f xs) c n)
= cons (f x) (mapChurch f xs)

So since all lists are made from those two constructors, and mapChurch works on both of them as expected, mapChurch must work as expected on all lists.

回答3:

Well, we can comment the Churchlist type this way to clarify it:

-- Tell me...
type Churchlist t u = (t -> u -> u) -- ...how to handle a pair
                    -> u            -- ...and how to handle an empty list
                    -> u            -- ...and then I'll transform a list into 
                                    -- the type you want

Note that this is intimately related to the foldr function:

foldr :: (t -> u -> u) -> u -> [t] -> u
foldr k z [] = z
foldr k z (x:xs) = k x (foldr k z xs)

foldr is a very general function that can implement all sorts of other list functions. A trivial example that will help you is implementing a list copy with foldr:

copyList :: [t] -> [t]
copyList xs = foldr (:) [] xs

Using the commented type above, foldr (:) [] means this: "if you see an empty list return the empty list, and if you see a pair return head:tailResult."

Using Churchlist, you can easily write the counterpart this way:

-- Note that the definitions of nil and cons mirror the two foldr equations!
nil :: Churchlist t u
nil = \k z -> z

cons :: t -> Churchlist t u -> Churchlist t u
cons x xs = \k z -> k x (xs k z)  

copyChurchlist :: ChurchList t u -> Churchlist t u
copyChurchlist xs = xs cons nil

Now, to implement map, you just need to replace cons with a suitable function, like this:

map :: (a -> b) -> [a] -> [b]
map f xs = foldr (\x xs' -> f x:xs') [] xs

Mapping is like copying a list, except that instead of just copying the elements verbatim you apply f to each of them.

Study all of this carefully, and you should be able to write mapChurchlist :: (t -> t') -> Churchlist t u -> Churchlist t' u on your own.

Extra exercise (easy): write these list functions in terms of foldr, and write counterparts for Churchlist:

filter :: (a -> Bool) -> [a] -> [a]
append :: [a] -> [a] -> [a]

-- Return first element of list that satisfies predicate, or Nothing
find :: (a -> Bool) -> [a] -> Maybe a

If you're feeling like tackling something harder, try writing tail for Churchlist. (Start by writing tail for [a] using foldr.)