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:
[[!toc]]
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. 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 type:
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);;
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 indulge ourselves in a completely useless digression 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. We're going to use type 3 lists, partly because we know
they'll give the result we want, but also because they're the coolest.
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):
# fun f z -> z;;
- : 'a -> 'b -> 'b =
# fun f z -> f 1 z;;
- : (int -> 'a -> 'b) -> 'a -> 'b =
# fun f z -> f 2 (f 1 z);;
- : (int -> 'a -> 'a) -> 'a -> 'a =
# fun f z -> f 3 (f 2 (f 1 z))
- : (int -> 'a -> 'a) -> 'a -> 'a =
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 bit thrown in
(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 = ...
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.
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` 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)) ~~>
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!
To bad this digression, though it ties together various
elements of the course, has *no relevance whatsoever* to the topic of
continuations...
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 you probably know,
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 wraps up an individual in a fancy 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
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)
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
l'_unit a = fun f -> f a
l'_bind u f = fun k -> u (fun a -> f a k)
(We performed a sneaky but valid eta reduction in the unit term.)
The unit and the bind for the Montague continuation monad and the
homemade List monad are the same terms! In other words, the behavior
of the List monad and the behavior of the continuations monad are
parallel in a deep sense. To emphasize the parallel, we can
instantiate the type of the list' monad using the OCaml list type:
type 'a c_list = ('a -> 'a list) -> 'a list
Have we really discovered that lists are secretly continuations? Or
have we merely found a way of simulating lists using list
continuations? Both perspectives are valid, and we can use our
intuitions about the list monad to understand continuations, and vice
versa (not to mention our intuitions about primitive recursion in
Church numerals too). The connections will be expecially relevant
when we consider indefinites and Hamblin semantics on the linguistic
side, and non-determinism on the list monad side.
Refunctionalizing zippers
-------------------------