4 Rethinking the list monad
5 -------------------------
7 To construct a monad, the key element is to settle on a type
8 constructor, and the monad naturally follows from that. We'll remind
9 you of some examples of how monads follow from the type constructor in
10 a moment. This will involve some review of familair material, but
11 it's worth doing for two reasons: it will set up a pattern for the new
12 discussion further below, and it will tie together some previously
13 unconnected elements of the course (more specifically, version 3 lists
16 For instance, take the **Reader Monad**. Once we decide that the type
19 type 'a reader = env -> 'a
21 then the choice of unit and bind is natural:
23 let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
25 Since the type of an `'a reader` is `env -> 'a` (by definition),
26 the type of the `r_unit` function is `'a -> env -> 'a`, which is a
27 specific case of the type of the *K* combinator. So it makes sense
28 that *K* is the unit for the reader monad.
30 Since the type of the `bind` operator is required to be
32 r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
34 We can reason our way to the correct `bind` function as follows. We
35 start by declaring the types determined by the definition of a bind operation:
37 let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
39 Now we have to open up the `u` box and get out the `'a` object in order to
40 feed it to `f`. Since `u` is a function from environments to
41 objects of type `'a`, the way we open a box in this monad is
42 by applying it to an environment:
46 This subexpression types to `'b reader`, which is good. The only
47 problem is that we invented an environment `e` that we didn't already have ,
48 so we have to abstract over that variable to balance the books:
52 This types to `env -> 'b reader`, but we want to end up with `env ->
53 '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:
55 r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
58 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.
60 [The bind we cite here is a condensed version of the careful `let a = u e in ...`
61 constructions we provided in earlier lectures. We use the condensed
62 version here in order to emphasize similarities of structure across
65 The **State Monad** is similar. Once we've decided to use the following type constructor:
67 type 'a state = store -> ('a, store)
69 Then our unit is naturally:
71 let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
73 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:
75 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
78 But unlocking the `u` box is a little more complicated. As before, we
79 need to posit a state `s` that we can apply `u` to. Once we do so,
80 however, we won't have an `'a`, we'll have a pair whose first element
81 is an `'a`. So we have to unpack the pair:
83 ... let (a, s') = u s in ... (f a) ...
85 Abstracting over the `s` and adjusting the types gives the result:
87 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
88 fun (s : store) -> let (a, s') = u s in f a s'
90 The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
91 won't pause to explore it here, though conceptually its unit and bind
92 follow just as naturally from its type constructor.
94 Our other familiar monad is the **List Monad**, which we were told
98 l_unit (a : 'a) = [a];;
99 l_bind u f = List.concat (List.map f u);;
101 Thinking through the list monad will take a little time, but doing so
102 will provide a connection with continuations.
104 Recall that `List.map` takes a function and a list and returns the
105 result to applying the function to the elements of the list:
107 List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
109 and List.concat takes a list of lists and erases the embdded list
112 List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
116 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
118 Now, why this unit, and why this bind? Well, ideally a unit should
119 not throw away information, so we can rule out `fun x -> []` as an
120 ideal unit. And units should not add more information than required,
121 so there's no obvious reason to prefer `fun x -> [x,x]`. In other
122 words, `fun x -> [x]` is a reasonable choice for a unit.
124 As for bind, an `'a list` monadic object contains a lot of objects of
125 type `'a`, and we want to make some use of each of them (rather than
126 arbitrarily throwing some of them away). The only
127 thing we know for sure we can do with an object of type `'a` is apply
128 the function of type `'a -> 'a list` to them. Once we've done so, we
129 have a collection of lists, one for each of the `'a`'s. One
130 possibility is that we could gather them all up in a list, so that
131 `bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts
132 the object returned by the second argument of `bind` to always be of
133 type `'b list list`. We can elimiate that restriction by flattening
134 the list of lists into a single list: this is
135 just List.concat applied to the output of List.map. So there is some logic to the
136 choice of unit and bind for the list monad.
138 Yet we can still desire to go deeper, and see if the appropriate bind
139 behavior emerges from the types, as it did for the previously
140 considered monads. But we can't do that if we leave the list type
141 as a primitive Ocaml type. However, we know several ways of implementing
142 lists using just functions. In what follows, we're going to use type
143 3 lists (the right fold implementation), though it's important to
144 wonder how things would change if we used some other strategy for
145 implementating lists. These were the lists that made lists look like
146 Church numerals with extra bits embdded in them:
148 empty list: fun f z -> z
149 list with one element: fun f z -> f 1 z
150 list with two elements: fun f z -> f 2 (f 1 z)
151 list with three elements: fun f z -> f 3 (f 2 (f 1 z))
153 and so on. To save time, we'll let the OCaml interpreter infer the
154 principle types of these functions (rather than inferring what the
155 types should be ourselves):
158 - : 'a -> 'b -> 'b = <fun>
160 - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
161 # fun f z -> f 2 (f 1 z);;
162 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
163 # fun f z -> f 3 (f 2 (f 1 z))
164 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
166 We can see what the consistent, general principle types are at the end, so we
167 can stop. These types should remind you of the simply-typed lambda calculus
168 types for Church numerals (`(o -> o) -> o -> o`) with one extra type
169 thrown in, the type of the element a the head of the list
170 (in this case, an int).
172 So here's our type constructor for our hand-rolled lists:
174 type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
176 Generalizing to lists that contain any kind of element (not just
179 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
181 So an `('a, 'b) list'` is a list containing elements of type `'a`,
182 where `'b` is the type of some part of the plumbing. This is more
183 general than an ordinary OCaml list, but we'll see how to map them
184 into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
185 in order to proceed to build a monad:
187 l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
189 No problem. Arriving at bind is a little more complicated, but
190 exactly the same principles apply, you just have to be careful and
193 l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
195 Unpacking the types gives:
197 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
198 (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
199 : ('c -> 'd -> 'd) -> 'd -> 'd = ...
201 Perhaps a bit intimiating.
202 But it's a rookie mistake to quail before complicated types. You should
203 be no more intimiated by complex types than by a linguistic tree with
204 deeply embedded branches: complex structure created by repeated
205 application of simple rules.
207 [This would be a good time to try to build your own term for the types
208 just given. Doing so (or attempting to do so) will make the next
209 paragraph much easier to follow.]
211 As usual, we need to unpack the `u` box. Examine the type of `u`.
212 This time, `u` will only deliver up its contents if we give `u` an
213 argument that is a function expecting an `'a` and a `'b`. `u` will
214 fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:
216 ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
218 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)`:
220 ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
222 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:
224 ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
226 Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
228 fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
230 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:
232 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
233 (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
234 : ('c -> 'b -> 'b) -> 'b -> 'b =
235 fun k -> u (fun a b -> f a k b)
237 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.
239 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:
241 fun k z -> u (fun a b -> f a k b) z
243 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?
245 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:
248 concat [[]; [2]; [2; 4]; [2; 4; 8]] =
251 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
253 fun k z -> u (fun a b -> f a k b) z
255 do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
262 (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.
264 So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
267 right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
268 right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
269 right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
270 right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
272 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:
274 fun k z -> u (fun a b -> f a k b) z
276 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
278 fun k z -> List.fold_right k (concat (map f u)) z
282 For future reference, we might make two eta-reductions to our formula, so that we have instead:
284 let l'_bind = fun k -> u (fun a -> f a k);;
286 Let's make some more tests:
289 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
291 l'_bind (fun f z -> f 1 (f 2 z))
292 (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
294 Sigh. OCaml won't show us our own list. So we have to choose an `f`
295 and a `z` that will turn our hand-crafted lists into standard OCaml
296 lists, so that they will print out.
298 # let cons h t = h :: t;; (* OCaml is stupid about :: *)
299 # l'_bind (fun f z -> f 1 (f 2 z))
300 (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
301 - : int list = [1; 2; 2; 3]
306 Montague's PTQ treatment of DPs as generalized quantifiers
307 ----------------------------------------------------------
309 We've hinted that Montague's treatment of DPs as generalized
310 quantifiers embodies the spirit of continuations (see de Groote 2001,
311 Barker 2002 for lengthy discussion). Let's see why.
313 First, we'll need a type constructor. As you probably know,
314 Montague replaced individual-denoting determiner phrases (with type `e`)
315 with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
316 In particular, the denotation of a proper name like *John*, which
317 might originally denote a object `j` of type `e`, came to denote a
318 generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
319 Let's write a general function that will map individuals into their
320 corresponding generalized quantifier:
322 gqize (a : e) = fun (p : e -> t) -> p a
324 This function is what Partee 1987 calls LIFT, and it would be
325 reasonable to use it here, but we will avoid that name, given that we
326 use that word to refer to other functions.
328 This function wraps up an individual in a box. That is to say,
329 we are in the presence of a monad. The type constructor, the unit and
330 the bind follow naturally. We've done this enough times that we won't
331 belabor the construction of the bind function, the derivation is
332 highly similar to the List monad just given:
334 type 'a continuation = ('a -> 'b) -> 'b
335 c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
336 c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
337 fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
339 Note that `c_bind` is exactly the `gqize` function that Montague used
340 to lift individuals into the continuation monad.
342 That last bit in `c_bind` looks familiar---we just saw something like
343 it in the List monad. How similar is it to the List monad? Let's
344 examine the type constructor and the terms from the list monad derived
347 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
348 l'_unit a = fun f -> f a
349 l'_bind u f = fun k -> u (fun a -> f a k)
351 (We performed a sneaky but valid eta reduction in the unit term.)
353 The unit and the bind for the Montague continuation monad and the
354 homemade List monad are the same terms! In other words, the behavior
355 of the List monad and the behavior of the continuations monad are
356 parallel in a deep sense.
358 Have we really discovered that lists are secretly continuations? Or
359 have we merely found a way of simulating lists using list
360 continuations? Well, strictly speaking, what we have done is shown
361 that one particular implementation of lists---the right fold
362 implementation---gives rise to a continuation monad fairly naturally,
363 and that this monad can reproduce the behavior of the standard list
364 monad. But what about other list implementations? Do they give rise
365 to monads that can be understood in terms of continuations?