from zippers to lists to continuations

[lambda.git] / zipper-lists-continuations.mdwn

Today we're going to encounter continuations.  We're going to come at
them from three different directions, and each time we're going to end
up at the same place: a particular monad, which we'll call the
continuation monad.

The three approches are:

*    Rethinking the list monad;
*    Montague's PTQ treatment of DPs as generalized quantifiers; and
*    Refunctionalizing zippers (Shan: zippers are defunctionalized continuations);

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.  I'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 = fun e:env -> 'a

then we can deduce the unit and the bind:

    runit x:'a -> 'a reader = fun (e:env) -> x

Since the type of an `'a reader` is `fun e:env -> 'a` (by definition),
the type of the `runit` function is `'a -> e: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 deduce the correct `bind` function as follows:

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

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`, we'll have

         .... f (u e) ...

This subexpression types to `'b reader`, which is good.  The only
problem is that we don't have an `e`, so we have to abstract over that
variable:

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

This types to `env -> 'b reader`, but we want to end up with `env ->
'b`.  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.

The **State Monad** is similar.  We somehow intuit that we want to use
the following type constructor:

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

So our unit is naturally

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

And we deduce the bind 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) = 

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:state) -> let (a, s') = u s in f a s'

The **Option 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 (x:'a) = [x];;
    l_bind u f = List.concat (List.map f u);;

Recall that `List.map` take 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]

But where is the reasoning that led us to this unit and bind?
And what is the type `['a]`?  Magic.

So let's take a *completely useless digressing* and see if we can
gain some insight into the details of the List monad.  Let's choose
type constructor that we can peer into, using some of the technology
we built up so laboriously during the first half of the course.  I'm
going to use type 3 lists, partly because I know they'll give the
result I want, but also because they're my favorite.  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 deducing what the
types should be):

<pre>
# 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>
</pre>

Finally, we're getting consistent principle types, 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 bit
thrown in (in this case, and int).

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

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

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 grasp the role of the `'b`'s
in order to proceed to build a monad:

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

Unfortunately, we'll need to spell out the types:

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

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.

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` as an
argument a function expecting an `'a`.  Once that argument is applied
to an object of type `'a`, we'll have what we need.  Thus:

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

In order for `u` to have the kind of argument it needs, we have to
adjust `(f a)` (which has type `('c -> 'd -> 'd) -> 'd -> 'd`) in
order to deliver something of type `'b -> 'b`.  The easiest way is to
alias `'d` to `'b`, and provide `(f a)` with an argument of type `'c
-> 'b -> 'b`.  Thus:

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

[Excercise: can you arrive at a fully general bind for this type
constructor, one that does not collapse `'d`'s with `'b`'s?]

As usual, we have to abstract over `k`, but this time, no further
adjustments are needed:

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

You should carefully check to make sure that this term is consistent
with the typing.

Our theory is that this monad should be capable of exactly
replicating the behavior of the standard List monad.  Let's test:


    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!

Just for mnemonic purposes (sneaking in an instance of eta reduction
to the definition of unit), we can summarize the result as follows:

    type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
    l'_unit x = fun f -> f x
    l'_bind u f = fun k -> u (fun x -> f x k)

To bad this digression, though it ties together various
elements of the course, has *no relevance whatsoever* to the topic of
continuations.