4 Today we're going to encounter continuations. We're going to come at
5 them from three different directions, and each time we're going to end
6 up at the same place: a particular monad, which we'll call the
9 Much of this discussion benefited from detailed comments and
10 suggestions from Ken Shan.
13 Rethinking the list monad
14 -------------------------
16 To construct a monad, the key element is to settle on a type
17 constructor, and the monad naturally follows from that. We'll remind
18 you of some examples of how monads follow from the type constructor in
19 a moment. This will involve some review of familair material, but
20 it's worth doing for two reasons: it will set up a pattern for the new
21 discussion further below, and it will tie together some previously
22 unconnected elements of the course (more specifically, version 3 lists
25 For instance, take the **Reader Monad**. Once we decide that the type
28 type 'a reader = env -> 'a
30 then the choice of unit and bind is natural:
32 let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
34 Since the type of an `'a reader` is `env -> 'a` (by definition),
35 the type of the `r_unit` function is `'a -> env -> 'a`, which is a
36 specific case of the type of the *K* combinator. So it makes sense
37 that *K* is the unit for the reader monad.
39 Since the type of the `bind` operator is required to be
41 r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
43 We can reason our way to the correct `bind` function as follows. We
44 start by declaring the types determined by the definition of a bind operation:
46 let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
48 Now we have to open up the `u` box and get out the `'a` object in order to
49 feed it to `f`. Since `u` is a function from environments to
50 objects of type `'a`, the way we open a box in this monad is
51 by applying it to an environment:
55 This subexpression types to `'b reader`, which is good. The only
56 problem is that we invented an environment `e` that we didn't already have ,
57 so we have to abstract over that variable to balance the books:
61 This types to `env -> 'b reader`, but we want to end up with `env ->
62 '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:
64 r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
67 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.
69 [The bind we cite here is a condensed version of the careful `let a = u e in ...`
70 constructions we provided in earlier lectures. We use the condensed
71 version here in order to emphasize similarities of structure across
74 The **State Monad** is similar. Once we've decided to use the following type constructor:
76 type 'a state = store -> ('a, store)
78 Then our unit is naturally:
80 let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
82 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:
84 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
87 But unlocking the `u` box is a little more complicated. As before, we
88 need to posit a state `s` that we can apply `u` to. Once we do so,
89 however, we won't have an `'a`, we'll have a pair whose first element
90 is an `'a`. So we have to unpack the pair:
92 ... let (a, s') = u s in ... (f a) ...
94 Abstracting over the `s` and adjusting the types gives the result:
96 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
97 fun (s : store) -> let (a, s') = u s in f a s'
99 The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
100 won't pause to explore it here, though conceptually its unit and bind
101 follow just as naturally from its type constructor.
103 Our other familiar monad is the **List Monad**, which we were told
106 type 'a list = ['a];;
107 l_unit (a : 'a) = [a];;
108 l_bind u f = List.concat (List.map f u);;
110 Thinking through the list monad will take a little time, but doing so
111 will provide a connection with continuations.
113 Recall that `List.map` takes a function and a list and returns the
114 result to applying the function to the elements of the list:
116 List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
118 and List.concat takes a list of lists and erases the embdded list
121 List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
125 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
127 Now, why this unit, and why this bind? Well, ideally a unit should
128 not throw away information, so we can rule out `fun x -> []` as an
129 ideal unit. And units should not add more information than required,
130 so there's no obvious reason to prefer `fun x -> [x,x]`. In other
131 words, `fun x -> [x]` is a reasonable choice for a unit.
133 As for bind, an `'a list` monadic object contains a lot of objects of
134 type `'a`, and we want to make some use of each of them (rather than
135 arbitrarily throwing some of them away). The only
136 thing we know for sure we can do with an object of type `'a` is apply
137 the function of type `'a -> 'a list` to them. Once we've done so, we
138 have a collection of lists, one for each of the `'a`'s. One
139 possibility is that we could gather them all up in a list, so that
140 `bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts
141 the object returned by the second argument of `bind` to always be of
142 type `'b list list`. We can elimiate that restriction by flattening
143 the list of lists into a single list: this is
144 just List.concat applied to the output of List.map. So there is some logic to the
145 choice of unit and bind for the list monad.
147 Yet we can still desire to go deeper, and see if the appropriate bind
148 behavior emerges from the types, as it did for the previously
149 considered monads. But we can't do that if we leave the list type
150 as a primitive Ocaml type. However, we know several ways of implementing
151 lists using just functions. In what follows, we're going to use type
152 3 lists (the right fold implementation), though it's important to
153 wonder how things would change if we used some other strategy for
154 implementating lists. These were the lists that made lists look like
155 Church numerals with extra bits embdded in them:
157 empty list: fun f z -> z
158 list with one element: fun f z -> f 1 z
159 list with two elements: fun f z -> f 2 (f 1 z)
160 list with three elements: fun f z -> f 3 (f 2 (f 1 z))
162 and so on. To save time, we'll let the OCaml interpreter infer the
163 principle types of these functions (rather than inferring what the
164 types should be ourselves):
167 - : 'a -> 'b -> 'b = <fun>
169 - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
170 # fun f z -> f 2 (f 1 z);;
171 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
172 # fun f z -> f 3 (f 2 (f 1 z))
173 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
175 We can see what the consistent, general principle types are at the end, so we
176 can stop. These types should remind you of the simply-typed lambda calculus
177 types for Church numerals (`(o -> o) -> o -> o`) with one extra type
178 thrown in, the type of the element a the head of the list
179 (in this case, an int).
181 So here's our type constructor for our hand-rolled lists:
183 type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
185 Generalizing to lists that contain any kind of element (not just
188 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
190 So an `('a, 'b) list'` is a list containing elements of type `'a`,
191 where `'b` is the type of some part of the plumbing. This is more
192 general than an ordinary OCaml list, but we'll see how to map them
193 into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
194 in order to proceed to build a monad:
196 l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
198 No problem. Arriving at bind is a little more complicated, but
199 exactly the same principles apply, you just have to be careful and
202 l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
204 Unpacking the types gives:
206 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
207 (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
208 : ('c -> 'd -> 'd) -> 'd -> 'd = ...
210 Perhaps a bit intimiating.
211 But it's a rookie mistake to quail before complicated types. You should
212 be no more intimiated by complex types than by a linguistic tree with
213 deeply embedded branches: complex structure created by repeated
214 application of simple rules.
216 [This would be a good time to try to build your own term for the types
217 just given. Doing so (or attempting to do so) will make the next
218 paragraph much easier to follow.]
220 As usual, we need to unpack the `u` box. Examine the type of `u`.
221 This time, `u` will only deliver up its contents if we give `u` an
222 argument that is a function expecting an `'a` and a `'b`. `u` will
223 fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:
225 ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
227 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)`:
229 ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
231 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:
233 ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
235 Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
237 fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
239 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:
241 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
242 (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
243 : ('c -> 'b -> 'b) -> 'b -> 'b =
244 fun k -> u (fun a b -> f a k b)
246 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.
248 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:
250 fun k z -> u (fun a b -> f a k b) z
252 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?
254 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:
257 concat [[]; [2]; [2; 4]; [2; 4; 8]] =
260 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
262 fun k z -> u (fun a b -> f a k b) z
264 do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
271 (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.
273 So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
276 right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
277 right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
278 right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
279 right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
281 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:
283 fun k z -> u (fun a b -> f a k b) z
285 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
287 fun k z -> List.fold_right k (concat (map f u)) z
291 For future reference, we might make two eta-reductions to our formula, so that we have instead:
293 let l'_bind = fun k -> u (fun a -> f a k);;
295 Let's make some more tests:
298 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
300 l'_bind (fun f z -> f 1 (f 2 z))
301 (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
303 Sigh. OCaml won't show us our own list. So we have to choose an `f`
304 and a `z` that will turn our hand-crafted lists into standard OCaml
305 lists, so that they will print out.
307 # let cons h t = h :: t;; (* OCaml is stupid about :: *)
308 # l'_bind (fun f z -> f 1 (f 2 z))
309 (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
310 - : int list = [1; 2; 2; 3]
315 Montague's PTQ treatment of DPs as generalized quantifiers
316 ----------------------------------------------------------
318 We've hinted that Montague's treatment of DPs as generalized
319 quantifiers embodies the spirit of continuations (see de Groote 2001,
320 Barker 2002 for lengthy discussion). Let's see why.
322 First, we'll need a type constructor. As you probably know,
323 Montague replaced individual-denoting determiner phrases (with type `e`)
324 with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
325 In particular, the denotation of a proper name like *John*, which
326 might originally denote a object `j` of type `e`, came to denote a
327 generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
328 Let's write a general function that will map individuals into their
329 corresponding generalized quantifier:
331 gqize (a : e) = fun (p : e -> t) -> p a
333 This function is what Partee 1987 calls LIFT, and it would be
334 reasonable to use it here, but we will avoid that name, given that we
335 use that word to refer to other functions.
337 This function wraps up an individual in a box. That is to say,
338 we are in the presence of a monad. The type constructor, the unit and
339 the bind follow naturally. We've done this enough times that we won't
340 belabor the construction of the bind function, the derivation is
341 highly similar to the List monad just given:
343 type 'a continuation = ('a -> 'b) -> 'b
344 c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
345 c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
346 fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
348 Note that `c_bind` is exactly the `gqize` function that Montague used
349 to lift individuals into the continuation monad.
351 That last bit in `c_bind` looks familiar---we just saw something like
352 it in the List monad. How similar is it to the List monad? Let's
353 examine the type constructor and the terms from the list monad derived
356 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
357 l'_unit a = fun f -> f a
358 l'_bind u f = fun k -> u (fun a -> f a k)
360 (We performed a sneaky but valid eta reduction in the unit term.)
362 The unit and the bind for the Montague continuation monad and the
363 homemade List monad are the same terms! In other words, the behavior
364 of the List monad and the behavior of the continuations monad are
365 parallel in a deep sense.
367 Have we really discovered that lists are secretly continuations? Or
368 have we merely found a way of simulating lists using list
369 continuations? Well, strictly speaking, what we have done is shown
370 that one particular implementation of lists---the right fold
371 implementation---gives rise to a continuation monad fairly naturally,
372 and that this monad can reproduce the behavior of the standard list
373 monad. But what about other list implementations? Do they give rise
374 to monads that can be understood in terms of continuations?
376 Refunctionalizing zippers
377 -------------------------
379 Manipulating trees with monads
380 ------------------------------
382 This thread develops an idea based on a detailed suggestion of Ken
383 Shan's. We'll build a series of functions that operate on trees,
384 doing various things, including replacing leaves, counting nodes, and
385 converting a tree to a list of leaves. The end result will be an
386 application for continuations.
388 From an engineering standpoint, we'll build a tree transformer that
389 deals in monads. We can modify the behavior of the system by swapping
390 one monad for another. (We've already seen how adding a monad can add
391 a layer of funtionality without disturbing the underlying system, for
392 instance, in the way that the reader monad allowed us to add a layer
393 of intensionality to an extensional grammar, but we have not yet seen
394 the utility of replacing one monad with other.)
396 First, we'll be needing a lot of trees during the remainder of the
397 course. Here's a type constructor for binary trees:
399 type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
401 These are trees in which the internal nodes do not have labels. [How
402 would you adjust the type constructor to allow for labels on the
405 We'll be using trees where the nodes are integers, e.g.,
409 let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
410 (Node ((Leaf 5),(Node ((Leaf 7),
425 Our first task will be to replace each leaf with its double:
428 let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
429 match t with Leaf x -> Leaf (newleaf x)
430 | Node (l, r) -> Node ((treemap newleaf l),
431 (treemap newleaf r));;
433 `treemap` takes a function that transforms old leaves into new leaves,
434 and maps that function over all the leaves in the tree, leaving the
435 structure of the tree unchanged. For instance:
438 let double i = i + i;;
441 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
455 We could have built the doubling operation right into the `treemap`
456 code. However, because what to do to each leaf is a parameter, we can
457 decide to do something else to the leaves without needing to rewrite
458 `treemap`. For instance, we can easily square each leaf instead by
459 supplying the appropriate `int -> int` operation in place of `double`:
462 let square x = x * x;;
465 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
468 Note that what `treemap` does is take some global, contextual
469 information---what to do to each leaf---and supplies that information
470 to each subpart of the computation. In other words, `treemap` has the
471 behavior of a reader monad. Let's make that explicit.
473 In general, we're on a journey of making our treemap function more and
474 more flexible. So the next step---combining the tree transducer with
475 a reader monad---is to have the treemap function return a (monadized)
476 tree that is ready to accept any `int->int` function and produce the
479 \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))
493 That is, we want to transform the ordinary tree `t1` (of type `int
494 tree`) into a reader object of type `(int->int)-> int tree`: something
495 that, when you apply it to an `int->int` function returns an `int
496 tree` in which each leaf `x` has been replaced with `(f x)`.
498 With previous readers, we always knew which kind of environment to
499 expect: either an assignment function (the original calculator
500 simulation), a world (the intensionality monad), an integer (the
501 Jacobson-inspired link monad), etc. In this situation, it will be
502 enough for now to expect that our reader will expect a function of
506 type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *)
507 let reader_unit (x:'a): 'a reader = fun _ -> x;;
508 let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;
511 It's easy to figure out how to turn an `int` into an `int reader`:
514 let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
515 int2int_reader 2 (fun i -> i + i);;
519 But what do we do when the integers are scattered over the leaves of a
520 tree? A binary tree is not the kind of thing that we can apply a
521 function of type `int->int` to.
524 let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
525 match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
526 | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
527 reader_bind (treemonadizer f r) (fun y ->
528 reader_unit (Node (x, y))));;
531 This function says: give me a function `f` that knows how to turn
532 something of type `'a` into an `'b reader`, and I'll show you how to
533 turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
534 the `treemonadizer` function builds plumbing that connects all of the
535 leaves of a tree into one connected monadic network; it threads the
536 monad through the leaves.
539 # treemonadizer int2int_reader t1 (fun i -> i + i);;
541 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
544 Here, our environment is the doubling function (`fun i -> i + i`). If
545 we apply the very same `int tree reader` (namely, `treemonadizer
546 int2int_reader t1`) to a different `int->int` function---say, the
547 squaring function, `fun i -> i * i`---we get an entirely different
551 # treemonadizer int2int_reader t1 (fun i -> i * i);;
553 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
556 Now that we have a tree transducer that accepts a monad as a
557 parameter, we can see what it would take to swap in a different monad.
558 For instance, we can use a state monad to count the number of nodes in
562 type 'a state = int -> 'a * int;;
563 let state_unit x i = (x, i+.5);;
564 let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
567 Gratifyingly, we can use the `treemonadizer` function without any
568 modification whatsoever, except for replacing the (parametric) type
569 `reader` with `state`:
572 let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
573 match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
574 | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
575 state_bind (treemonadizer f r) (fun y ->
576 state_unit (Node (x, y))));;
579 Then we can count the number of nodes in the tree:
582 # treemonadizer state_unit t1 0;;
584 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
598 Notice that we've counted each internal node twice---it's a good
599 exercise to adjust the code to count each node once.
601 One more revealing example before getting down to business: replacing
602 `state` everywhere in `treemonadizer` with `list` gives us
605 # treemonadizer (fun x -> [ [x; square x] ]) t1;;
606 - : int list tree list =
608 (Node (Leaf [2; 4], Leaf [3; 9]),
609 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
612 Unlike the previous cases, instead of turning a tree into a function
613 from some input to a result, this transformer replaces each `int` with
616 Now for the main point. What if we wanted to convert a tree to a list
620 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
621 let continuation_unit x c = c x;;
622 let continuation_bind u f c = u (fun a -> f a c);;
624 let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
625 match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
626 | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
627 continuation_bind (treemonadizer f r) (fun y ->
628 continuation_unit (Node (x, y))));;
631 We use the continuation monad described above, and insert the
632 `continuation` type in the appropriate place in the `treemonadizer` code.
636 # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
637 - : int list = [2; 3; 5; 7; 11]
640 We have found a way of collapsing a tree into a list of its leaves.
642 The continuation monad is amazingly flexible; we can use it to
643 simulate some of the computations performed above. To see how, first
644 note that an interestingly uninteresting thing happens if we use the
645 continuation unit as our first argument to `treemonadizer`, and then
646 apply the result to the identity function:
649 # treemonadizer continuation_unit t1 (fun x -> x);;
651 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
654 That is, nothing happens. But we can begin to substitute more
655 interesting functions for the first argument of `treemonadizer`:
658 (* Simulating the tree reader: distributing a operation over the leaves *)
659 # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
661 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
663 (* Simulating the int list tree list *)
664 # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
667 (Node (Leaf [2; 4], Leaf [3; 9]),
668 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
670 (* Counting leaves *)
671 # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
675 We could simulate the tree state example too, but it would require
676 generalizing the type of the continuation monad to
678 type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
680 The binary tree monad
681 ---------------------
683 Of course, by now you may have realized that we have discovered a new
684 monad, the binary tree monad:
687 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
688 let tree_unit (x:'a) = Leaf x;;
689 let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
690 match u with Leaf x -> f x
691 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
694 For once, let's check the Monad laws. The left identity law is easy:
696 Left identity: bind (unit a) f = bind (Leaf a) f = fa
698 To check the other two laws, we need to make the following
699 observation: it is easy to prove based on `tree_bind` by a simple
700 induction on the structure of the first argument that the tree
701 resulting from `bind u f` is a tree with the same strucure as `u`,
702 except that each leaf `a` has been replaced with `fa`:
704 \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
721 Given this equivalence, the right identity law
723 Right identity: bind u unit = u
725 falls out once we realize that
727 bind (Leaf a) unit = unit a = Leaf a
729 As for the associative law,
731 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
733 we'll give an example that will show how an inductive proof would
734 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
736 \tree (. (. (. (. (a1)(a2)))))
737 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
742 bind __|__ f = __|_ = . .
744 a1 a2 fa1 fa2 | | | |
748 Now when we bind this tree to `g`, we get
760 At this point, it should be easy to convince yourself that
761 using the recipe on the right hand side of the associative law will
762 built the exact same final tree.
764 So binary trees are a monad.
766 Haskell combines this monad with the Option monad to provide a monad
768 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
770 represent non-deterministic computations as a tree.
773 Lists, zippers, continuations
774 -----------------------------
776 Let's work with lists of chars for a change. To maximize readability, we'll
777 indulge in an abbreviatory convention that "abc" abbreviates the
778 list `['a'; 'b'; 'c']`.
780 Task 1: replace each occurrence of 'S' with a copy of the string up to
786 t1 "abSe" ~~> "ababe"
790 In linguistic terms, this is a kind of anaphora
791 resolution, where `'S'` is functioning like an anaphoric element, and
792 the preceding string portion is the antecedent.
794 This deceptively simple task gives rise to some mind-bending complexity.
795 Note that it matters which 'S' you target first (the position of the *
796 indicates the targeted 'S'):
827 ~~> t1 "aSbaaaSbaabab"
832 Aparently, this task, as simple as it is, is a form of computation,
833 and the order in which the `'S'`s get evaluated can lead to divergent
836 For now, as usual, we'll agree to always ``evaluate'' the leftmost `'S'`.
838 This is a task well-suited to using a zipper.
841 type 'a list_zipper = ('a list) * ('a list);;
843 let rec t1 (z:char list_zipper) =
844 match z with (sofar, []) -> List.rev(sofar)
845 | (sofar, 'S'::rest) -> t1 ((List.append sofar sofar), rest)
846 | (sofar, fst::rest) -> t1 (fst::sofar, rest);;
848 # t1 ([], ['a'; 'b'; 'S'; 'e']);;
849 - : char list = ['a'; 'b'; 'a'; 'b'; 'e']
851 # t1 ([], ['a'; 'S'; 'b'; 'S']);;
852 - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
855 Note that this implementation enforces the evaluate-leftmost rule.