3 We're going to come at continuations from three different directions, and each
4 time we're going to end up at the same place: a particular monad, which we'll
5 call the continuation monad.
7 Rethinking the list monad
8 -------------------------
10 To construct a monad, the key element is to settle on a type
11 constructor, and the monad more or less naturally follows from that.
12 We'll remind you of some examples of how monads follow from the type
13 constructor in a moment. This will involve some review of familair
14 material, but it's worth doing for two reasons: it will set up a
15 pattern for the new discussion further below, and it will tie together
16 some previously unconnected elements of the course (more specifically,
17 version 3 lists and monads).
19 For instance, take the **Reader Monad**. Once we decide that the type
22 type 'a reader = env -> 'a
24 then the choice of unit and bind is natural:
26 let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
28 The reason this is a fairly natural choice is that because the type of
29 an `'a reader` is `env -> 'a` (by definition), the type of the
30 `r_unit` function is `'a -> env -> 'a`, which is an instance of the
31 type of the *K* combinator. So it makes sense that *K* is the unit
34 Since the type of the `bind` operator is required to be
36 r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
38 We can reason our way to the traditional reader `bind` function as
39 follows. We start by declaring the types determined by the definition
42 let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
44 Now we have to open up the `u` box and get out the `'a` object in order to
45 feed it to `f`. Since `u` is a function from environments to
46 objects of type `'a`, the way we open a box in this monad is
47 by applying it to an environment:
53 This subexpression types to `'b reader`, which is good. The only
54 problem is that we made use of an environment `e` that we didn't already have,
55 so we must abstract over that variable to balance the books:
59 [To preview the discussion of the Curry-Howard correspondence, what
60 we're doing here is constructing an intuitionistic proof of the type,
61 and using the Curry-Howard labeling of the proof as our bind term.]
63 This types to `env -> 'b reader`, but we want to end up with `env ->
64 '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:
67 r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = f (u e) e
70 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.
72 [The bind we cite here is a condensed version of the careful `let a = u e in ...`
73 constructions we provided in earlier lectures. We use the condensed
74 version here in order to emphasize similarities of structure across
77 The **State Monad** is similar. Once we've decided to use the following type constructor:
79 type 'a state = store -> ('a, store)
81 Then our unit is naturally:
83 let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
85 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:
87 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
90 But unlocking the `u` box is a little more complicated. As before, we
91 need to posit a state `s` that we can apply `u` to. Once we do so,
92 however, we won't have an `'a`, we'll have a pair whose first element
93 is an `'a`. So we have to unpack the pair:
95 ... let (a, s') = u s in ... (f a) ...
97 Abstracting over the `s` and adjusting the types gives the result:
99 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
100 fun (s : store) -> let (a, s') = u s in f a s'
102 The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
103 won't pause to explore it here, though conceptually its unit and bind
104 follow just as naturally from its type constructor.
106 Our other familiar monad is the **List Monad**, which we were told
109 type 'a list = ['a];;
110 l_unit (a : 'a) = [a];;
111 l_bind u f = List.concat (List.map f u);;
113 Thinking through the list monad will take a little time, but doing so
114 will provide a connection with continuations.
116 Recall that `List.map` takes a function and a list and returns the
117 result to applying the function to the elements of the list:
119 List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
121 and List.concat takes a list of lists and erases the embdded list
124 List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
128 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
130 Now, why this unit, and why this bind? Well, ideally a unit should
131 not throw away information, so we can rule out `fun x -> []` as an
132 ideal unit. And units should not add more information than required,
133 so there's no obvious reason to prefer `fun x -> [x,x]`. In other
134 words, `fun x -> [x]` is a reasonable choice for a unit.
136 As for bind, an `'a list` monadic object contains a lot of objects of
137 type `'a`, and we want to make use of each of them (rather than
138 arbitrarily throwing some of them away). The only
139 thing we know for sure we can do with an object of type `'a` is apply
140 the function of type `'a -> 'a list` to them. Once we've done so, we
141 have a collection of lists, one for each of the `'a`'s. One
142 possibility is that we could gather them all up in a list, so that
143 `bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts
144 the object returned by the second argument of `bind` to always be of
145 type `'b list list`. We can elimiate that restriction by flattening
146 the list of lists into a single list: this is
147 just List.concat applied to the output of List.map. So there is some logic to the
148 choice of unit and bind for the list monad.
150 Yet we can still desire to go deeper, and see if the appropriate bind
151 behavior emerges from the types, as it did for the previously
152 considered monads. But we can't do that if we leave the list type as
153 a primitive Ocaml type. However, we know several ways of implementing
154 lists using just functions. In what follows, we're going to use type
155 3 lists, the right fold implementation (though it's important and
156 intriguing to wonder how things would change if we used some other
157 strategy for implementating lists). These were the lists that made
158 lists look like Church numerals with extra bits embdded in them:
160 empty list: fun f z -> z
161 list with one element: fun f z -> f 1 z
162 list with two elements: fun f z -> f 2 (f 1 z)
163 list with three elements: fun f z -> f 3 (f 2 (f 1 z))
165 and so on. To save time, we'll let the OCaml interpreter infer the
166 principle types of these functions (rather than inferring what the
167 types should be ourselves):
170 - : 'a -> 'b -> 'b = <fun>
172 - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
173 # fun f z -> f 2 (f 1 z);;
174 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
175 # fun f z -> f 3 (f 2 (f 1 z))
176 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
178 We can see what the consistent, general principle types are at the end, so we
179 can stop. These types should remind you of the simply-typed lambda calculus
180 types for Church numerals (`(o -> o) -> o -> o`) with one extra type
181 thrown in, the type of the element a the head of the list
182 (in this case, an int).
184 So here's our type constructor for our hand-rolled lists:
186 type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
188 Generalizing to lists that contain any kind of element (not just
191 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
193 So an `('a, 'b) list'` is a list containing elements of type `'a`,
194 where `'b` is the type of some part of the plumbing. This is more
195 general than an ordinary OCaml list, but we'll see how to map them
196 into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
197 in order to proceed to build a monad:
199 l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
201 No problem. Arriving at bind is a little more complicated, but
202 exactly the same principles apply, you just have to be careful and
205 l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
207 Unpacking the types gives:
209 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
210 (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
211 : ('c -> 'd -> 'd) -> 'd -> 'd = ...
213 Perhaps a bit intimiating.
214 But it's a rookie mistake to quail before complicated types. You should
215 be no more intimiated by complex types than by a linguistic tree with
216 deeply embedded branches: complex structure created by repeated
217 application of simple rules.
219 [This would be a good time to try to build your own term for the types
220 just given. Doing so (or attempting to do so) will make the next
221 paragraph much easier to follow.]
223 As usual, we need to unpack the `u` box. Examine the type of `u`.
224 This time, `u` will only deliver up its contents if we give `u` an
225 argument that is a function expecting an `'a` and a `'b`. `u` will
226 fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:
228 ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
230 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)`:
232 ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
234 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:
236 ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
238 Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
240 fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
242 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:
244 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
245 (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
246 : ('c -> 'b -> 'b) -> 'b -> 'b =
247 fun k -> u (fun a b -> f a k b)
249 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.
251 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:
253 fun k z -> u (fun a b -> f a k b) z
255 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?
257 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:
260 concat [[]; [2]; [2; 4]; [2; 4; 8]] =
263 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
265 fun k z -> u (fun a b -> f a k b) z
267 do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
274 (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.
276 So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
279 right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
280 right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
281 right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
282 right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
284 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:
286 fun k z -> u (fun a b -> f a k b) z
288 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
290 fun k z -> List.fold_right k (concat (map f u)) z
294 For future reference, we might make two eta-reductions to our formula, so that we have instead:
296 let l'_bind = fun k -> u (fun a -> f a k);;
298 Let's make some more tests:
301 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
303 l'_bind (fun f z -> f 1 (f 2 z))
304 (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
306 Sigh. OCaml won't show us our own list. So we have to choose an `f`
307 and a `z` that will turn our hand-crafted lists into standard OCaml
308 lists, so that they will print out.
310 # let cons h t = h :: t;; (* OCaml is stupid about :: *)
311 # l'_bind (fun f z -> f 1 (f 2 z))
312 (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
313 - : int list = [1; 2; 2; 3]
318 Montague's PTQ treatment of DPs as generalized quantifiers
319 ----------------------------------------------------------
321 We've hinted that Montague's treatment of DPs as generalized
322 quantifiers embodies the spirit of continuations (see de Groote 2001,
323 Barker 2002 for lengthy discussion). Let's see why.
325 First, we'll need a type constructor. As you probably know,
326 Montague replaced individual-denoting determiner phrases (with type `e`)
327 with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
328 In particular, the denotation of a proper name like *John*, which
329 might originally denote a object `j` of type `e`, came to denote a
330 generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
331 Let's write a general function that will map individuals into their
332 corresponding generalized quantifier:
334 gqize (a : e) = fun (p : e -> t) -> p a
336 This function is what Partee 1987 calls LIFT, and it would be
337 reasonable to use it here, but we will avoid that name, given that we
338 use that word to refer to other functions.
340 This function wraps up an individual in a box. That is to say,
341 we are in the presence of a monad. The type constructor, the unit and
342 the bind follow naturally. We've done this enough times that we won't
343 belabor the construction of the bind function, the derivation is
344 highly similar to the List monad just given:
346 type 'a continuation = ('a -> 'b) -> 'b
347 c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
348 c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
349 fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
351 Note that `c_unit` is exactly the `gqize` function that Montague used
352 to lift individuals into the continuation monad.
354 That last bit in `c_bind` looks familiar---we just saw something like
355 it in the List monad. How similar is it to the List monad? Let's
356 examine the type constructor and the terms from the list monad derived
359 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
360 l'_unit a = fun f -> f a
361 l'_bind u f = fun k -> u (fun a -> f a k)
363 (We performed a sneaky but valid eta reduction in the unit term.)
365 The unit and the bind for the Montague continuation monad and the
366 homemade List monad are the same terms! In other words, the behavior
367 of the List monad and the behavior of the continuations monad are
368 parallel in a deep sense.
370 Have we really discovered that lists are secretly continuations? Or
371 have we merely found a way of simulating lists using list
372 continuations? Well, strictly speaking, what we have done is shown
373 that one particular implementation of lists---the right fold
374 implementation---gives rise to a continuation monad fairly naturally,
375 and that this monad can reproduce the behavior of the standard list
376 monad. But what about other list implementations? Do they give rise
377 to monads that can be understood in terms of continuations?