[lambda.git] / list_monad_as_continuation_monad.mdwn

Rethinking the list monad
-------------------------

To construct a monad, the key element is to settle on a type
constructor, and the monad naturally follows from that.  We'll remind
you of some examples of how monads follow from the type constructor in
a moment.  This will involve some review of familair material, but
it's worth doing for two reasons: it will set up a pattern for the new
discussion further below, and it will tie together some previously
unconnected elements of the course (more specifically, version 3 lists
and monads).

For instance, take the **Reader Monad**.  Once we decide that the type
constructor is

    type 'a reader = env -> 'a

then the choice of unit and bind is natural:

    let r_unit (a : 'a) : 'a reader = fun (e : env) -> a

Since the type of an `'a reader` is `env -> 'a` (by definition),
the type of the `r_unit` function is `'a -> env -> 'a`, which is a
specific case of the type of the *K* combinator.  So it makes sense
that *K* is the unit for the reader monad.

Since the type of the `bind` operator is required to be

    r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)

We can reason our way to the correct `bind` function as follows. We
start by declaring the types determined by the definition of a bind operation:

    let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...

Now we have to open up the `u` box and get out the `'a` object in order to
feed it to `f`.  Since `u` is a function from environments to
objects of type `'a`, the way we open a box in this monad is
by applying it to an environment:

        ... f (u e) ...

This subexpression types to `'b reader`, which is good.  The only
problem is that we invented an environment `e` that we didn't already have ,
so we have to abstract over that variable to balance the books:

        fun e -> f (u e) ...

This types to `env -> 'b reader`, but we want to end up with `env ->
'b`.  Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment.  So we end up as follows:

    r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
                f (u e) e         

And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does.

[The bind we cite here is a condensed version of the careful `let a = u e in ...`
constructions we provided in earlier lectures.  We use the condensed
version here in order to emphasize similarities of structure across
monads.]

The **State Monad** is similar.  Once we've decided to use the following type constructor:

    type 'a state = store -> ('a, store)

Then our unit is naturally:

    let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)

And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box:

    let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = 
                ... f (...) ...

But unlocking the `u` box is a little more complicated.  As before, we
need to posit a state `s` that we can apply `u` to.  Once we do so,
however, we won't have an `'a`, we'll have a pair whose first element
is an `'a`.  So we have to unpack the pair:

        ... let (a, s') = u s in ... (f a) ...

Abstracting over the `s` and adjusting the types gives the result:

        let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = 
                fun (s : store) -> let (a, s') = u s in f a s'

The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
won't pause to explore it here, though conceptually its unit and bind
follow just as naturally from its type constructor.

Our other familiar monad is the **List Monad**, which we were told
looks like this:

    type 'a list = ['a];;
    l_unit (a : 'a) = [a];;
    l_bind u f = List.concat (List.map f u);;

Thinking through the list monad will take a little time, but doing so
will provide a connection with continuations.

Recall that `List.map` takes a function and a list and returns the
result to applying the function to the elements of the list:

    List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]

and List.concat takes a list of lists and erases the embdded list
boundaries:

    List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]

And sure enough, 

    l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]

Now, why this unit, and why this bind?  Well, ideally a unit should
not throw away information, so we can rule out `fun x -> []` as an
ideal unit.  And units should not add more information than required,
so there's no obvious reason to prefer `fun x -> [x,x]`.  In other
words, `fun x -> [x]` is a reasonable choice for a unit.

As for bind, an `'a list` monadic object contains a lot of objects of
type `'a`, and we want to make some use of each of them (rather than
arbitrarily throwing some of them away).  The only
thing we know for sure we can do with an object of type `'a` is apply
the function of type `'a -> 'a list` to them.  Once we've done so, we
have a collection of lists, one for each of the `'a`'s.  One
possibility is that we could gather them all up in a list, so that
`bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`.  But that restricts
the object returned by the second argument of `bind` to always be of
type `'b list list`.  We can elimiate that restriction by flattening
the list of lists into a single list: this is
just List.concat applied to the output of List.map.  So there is some logic to the
choice of unit and bind for the list monad.  

Yet we can still desire to go deeper, and see if the appropriate bind
behavior emerges from the types, as it did for the previously
considered monads.  But we can't do that if we leave the list type 
as a primitive Ocaml type.  However, we know several ways of implementing
lists using just functions.  In what follows, we're going to use type
3 lists (the right fold implementation), though it's important to
wonder how things would change if we used some other strategy for
implementating lists.  These were the lists that made lists look like
Church numerals with extra bits embdded in them:

    empty list:                fun f z -> z
    list with one element:     fun f z -> f 1 z
    list with two elements:    fun f z -> f 2 (f 1 z)
    list with three elements:  fun f z -> f 3 (f 2 (f 1 z))

and so on.  To save time, we'll let the OCaml interpreter infer the
principle types of these functions (rather than inferring what the
types should be ourselves):

        # fun f z -> z;;
        - : 'a -> 'b -> 'b = <fun>
        # fun f z -> f 1 z;;
        - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
        # fun f z -> f 2 (f 1 z);;
        - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
        # fun f z -> f 3 (f 2 (f 1 z))
        - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>

We can see what the consistent, general principle types are at the end, so we
can stop. These types should remind you of the simply-typed lambda calculus
types for Church numerals (`(o -> o) -> o -> o`) with one extra type
thrown in, the type of the element a the head of the list
(in this case, an int).

So here's our type constructor for our hand-rolled lists:

    type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b

Generalizing to lists that contain any kind of element (not just
ints), we have

    type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b

So an `('a, 'b) list'` is a list containing elements of type `'a`,
where `'b` is the type of some part of the plumbing.  This is more
general than an ordinary OCaml list, but we'll see how to map them
into OCaml lists soon.  We don't need to fully grasp the role of the `'b`'s
in order to proceed to build a monad:

    l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z

No problem.  Arriving at bind is a little more complicated, but
exactly the same principles apply, you just have to be careful and
systematic about it.

    l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list'  = ...

Unpacking the types gives:

    l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
            (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
            : ('c -> 'd -> 'd) -> 'd -> 'd = ...

Perhaps a bit intimiating.
But it's a rookie mistake to quail before complicated types. You should
be no more intimiated by complex types than by a linguistic tree with
deeply embedded branches: complex structure created by repeated
application of simple rules.

[This would be a good time to try to build your own term for the types
just given.  Doing so (or attempting to do so) will make the next
paragraph much easier to follow.]

As usual, we need to unpack the `u` box.  Examine the type of `u`.
This time, `u` will only deliver up its contents if we give `u` an
argument that is a function expecting an `'a` and a `'b`. `u` will 
fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:

        ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...

In order for `u` to have the kind of argument it needs, the `... (f a) ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `(f a)`:

        ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...

Now we have an argument `b` of type `'b`, so we can supply that to `(f a) k`, getting a result of type `'b`, as we need:

        ... u (fun (a : 'a) (b : 'b) -> f a k b) ...

Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:

        fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)

This whole expression has type `('c -> 'b -> 'b) -> 'b -> 'b`, which is exactly the type of a `('c, 'b) list'`. So we can hypothesize that we our bind is:

    l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
            (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
            : ('c -> 'b -> 'b) -> 'b -> 'b = 
      fun k -> u (fun a b -> f a k b)

That is a function of the right type for our bind, but to check whether it works, we have to verify it (with the unit we chose) against the monad laws, and reason whether it will have the right behavior.

Here's a way to persuade yourself that it will have the right behavior. First, it will be handy to eta-expand our `fun k -> u (fun a b -> f a k b)` to:

      fun k z -> u (fun a b -> f a k b) z

Now let's think about what this does. It's a wrapper around `u`. In order to behave as the list which is the result of mapping `f` over each element of `u`, and then joining (`concat`ing) the results, this wrapper would have to accept arguments `k` and `z` and fold them in just the same way that the list which is the result of mapping `f` and then joining the results would fold them. Will it?

Suppose we have a list' whose contents are `[1; 2; 4; 8]`---that is, our list' will be `fun f z -> f 1 (f 2 (f 4 (f 8 z)))`. We call that list' `u`. Suppose we also have a function `f` that for each `int` we give it, gives back a list of the divisors of that `int` that are greater than 1. Intuitively, then, binding `u` to `f` should give us:

        concat (map f u) =
        concat [[]; [2]; [2; 4]; [2; 4; 8]] =
        [2; 2; 4; 2; 4; 8]

Or rather, it should give us a list' version of that, which takes a function `k` and value `z` as arguments, and returns the right fold of `k` and `z` over those elements. What does our formula

      fun k z -> u (fun a b -> f a k b) z
        
do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:

        []
        [2]
        [2; 4]
        [2; 4; 8]

(or rather, their list' versions). Then it takes the accumulated result `b` of previous steps in the fold, and it folds `k` and `b` over the list generated by `f a`. The result of doing so is passed on to the next step as the accumulated result so far.

So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:

        0 ==>
        right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
        right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
        right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
        right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0

which indeed is the result of right-folding + and 0 over `[2; 2; 4; 2; 4; 8]`. If you trace through how this works, you should be able to persuade yourself that our formula:

      fun k z -> u (fun a b -> f a k b) z

will deliver just the same folds, for arbitrary choices of `k` and `z` (with the right types), and arbitrary list's `u` and appropriately-typed `f`s, as

        fun k z -> List.fold_right k (concat (map f u)) z

would.

For future reference, we might make two eta-reductions to our formula, so that we have instead:

      let l'_bind = fun k -> u (fun a -> f a k);;

Let's make some more tests:


    l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]

    l'_bind (fun f z -> f 1 (f 2 z)) 
            (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>

Sigh.  OCaml won't show us our own list.  So we have to choose an `f`
and a `z` that will turn our hand-crafted lists into standard OCaml
lists, so that they will print out.

        # let cons h t = h :: t;;  (* OCaml is stupid about :: *)
        # l'_bind (fun f z -> f 1 (f 2 z)) 
                          (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
        - : int list = [1; 2; 2; 3]

Ta da!