lists-monad tweaks

[lambda.git] / list_monad_as_continuation_monad.mdwn

[[!toc]]

We're going to come at continuations 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.

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

To construct a monad, the key element is to settle on a type
constructor, and the monad more or less 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 familiar
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

The reason this is a fairly natural choice is that because the type of
an `'a reader` is `env -> 'a` (by definition), the type of the
`r_unit` function is `'a -> env -> 'a`, which is an instance 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 traditional reader `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 made use of an environment `e` that we didn't already have,
so we must abstract over that variable to balance the books:

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

[To preview the discussion of the Curry-Howard correspondence, what
we're doing here is constructing an intuitionistic proof of the type,
and using the Curry-Howard labeling of the proof as our bind term.]

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 store `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 embedded 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 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 eliminate 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 and
intriguing to wonder how things would change if we used some other
strategy for implementing lists).  These were the lists that made
lists look like Church numerals with extra bits embedded 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 at 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
`int`s), 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 k z -> k 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 intimidating.
But it's a rookie mistake to quail before complicated types. You should
be no more intimidated 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 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:

        []        ; result of applying f to leftmost a
        [2]
        [2; 4]
        [2; 4; 8] ; result of applying f to rightmost a

(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!


Montague's PTQ treatment of DPs as generalized quantifiers
----------------------------------------------------------

We've hinted that Montague's treatment of DPs as generalized
quantifiers embodies the spirit of continuations (see de Groote 2001,
Barker 2002 for lengthy discussion).  Let's see why.

First, we'll need a type constructor.  As we've said,
Montague replaced individual-denoting determiner phrases (with type `e`)
with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
In particular, the denotation of a proper name like *John*, which
might originally denote a object `j` of type `e`, came to denote a
generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
Let's write a general function that will map individuals into their
corresponding generalized quantifier:

   gqize (a : e) = fun (p : e -> t) -> p a

This function is what Partee 1987 calls LIFT, and it would be
reasonable to use it here, but we will avoid that name, given that we
use that word to refer to other functions.

This function wraps up an individual in a box.  That is to say,
we are in the presence of a monad.  The type constructor, the unit and
the bind follow naturally.  We've done this enough times that we won't
belabor the construction of the bind function, the derivation is
highly similar to the List monad just given:

        type 'a continuation = ('a -> 'b) -> 'b
        c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
        c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
            fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)

Note that `c_unit` is exactly the `gqize` function that Montague used
to lift individuals into the continuation monad.

That last bit in `c_bind` looks familiar---we just saw something like
it in the List monad.  How similar is it to the List monad?  Let's
examine the type constructor and the terms from the list monad derived
above:

        type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
        let l'_unit a = fun k z -> k a z

This can be eta-reduced to:

        let l'_unit a = fun k -> k a

        let l'_bind u f =
            (* we mentioned three versions of this, the fully eta-expanded: *)
            fun k z -> u (fun a b -> f a k b) z
                (* an intermediate version, and the fully eta-reduced: *)
            fun k -> u (fun a -> f a k)

Consider the most eta-reduced versions of `l'_unit` and `l'_bind`. They're the same as the unit and bind for the Montague continuation monad! In other words, the behavior of our v3-list monad and the behavior of the continuations monad are
parallel in a deep sense.

Have we really discovered that lists are secretly continuations?  Or
have we merely found a way of simulating lists using list
continuations?  Well, strictly speaking, what we have done is shown
that one particular implementation of lists---the right fold
implementation---gives rise to a continuation monad fairly naturally,
and that this monad can reproduce the behavior of the standard list
monad.  But what about other list implementations?  Do they give rise
to monads that can be understood in terms of continuations?