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. 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 start by declaring the type:
45 let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
47 Now we have to open up the `u` box and get out the `'a` object in order to
48 feed it to `f`. Since `u` is a function from environments to
49 objects of type `'a`, the way we open a box in this monad is
50 by applying it to an environment:
54 This subexpression types to `'b reader`, which is good. The only
55 problem is that we invented an environment `e` that we didn't already have ,
56 so we have to abstract over that variable to balance the books:
60 This types to `env -> 'b reader`, but we want to end up with `env ->
61 '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:
63 r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
66 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.
68 [The bind we cite here is a condensed version of the careful `let a = u e in ...`
69 constructions we provided in earlier lectures. We use the condensed
70 version here in order to emphasize similarities of structure across
73 The **State Monad** is similar. Once we've decided to use the following type constructor:
75 type 'a state = store -> ('a, store)
77 Then our unit is naturally:
79 let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
81 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:
83 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
86 But unlocking the `u` box is a little more complicated. As before, we
87 need to posit a state `s` that we can apply `u` to. Once we do so,
88 however, we won't have an `'a`, we'll have a pair whose first element
89 is an `'a`. So we have to unpack the pair:
91 ... let (a, s') = u s in ... (f a) ...
93 Abstracting over the `s` and adjusting the types gives the result:
95 let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
96 fun (s : store) -> let (a, s') = u s in f a s'
98 The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
99 won't pause to explore it here, though conceptually its unit and bind
100 follow just as naturally from its type constructor.
102 Our other familiar monad is the **List Monad**, which we were told
105 type 'a list = ['a];;
106 l_unit (a : 'a) = [a];;
107 l_bind u f = List.concat (List.map f u);;
109 Thinking through the list monad will take a little time, but doing so
110 will provide a connection with continuations.
112 Recall that `List.map` takes a function and a list and returns the
113 result to applying the function to the elements of the list:
115 List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]
117 and List.concat takes a list of lists and erases the embdded list
120 List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]
124 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
126 Now, why this unit, and why this bind? Well, ideally a unit should
127 not throw away information, so we can rule out `fun x -> []` as an
128 ideal unit. And units should not add more information than required,
129 so there's no obvious reason to prefer `fun x -> [x,x]`. In other
130 words, `fun x -> [x]` is a reasonable guess for a unit.
132 As for bind, an `'a list` monadic object contains a lot of objects of
133 type `'a`, and we want to make some use of each of them. The only
134 thing we know for sure we can do with an object of type `'a` is apply
135 the function of type `'a -> 'a list` to them. Once we've done so, we
136 have a collection of lists, one for each of the `'a`'s. One
137 possibility is that we could gather them all up in a list, so that
138 `bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts
139 the object returned by the second argument of `bind` to always be of
140 type `'b list list`. We can elimiate that restriction by flattening
141 the list of lists into a single list. So there is some logic to the
142 choice of unit and bind for the list monad.
144 Yet we can still desire to go deeper, and see if the appropriate bind
145 behavior emerges from the types, as it did for the previously
146 considered monads. But we can't do that if we leave the list type is
147 a primitive Ocaml type. However, we know several ways of implementing
148 lists using just functions. In what follows, we're going to use type
149 3 lists (the right fold implementation), though it's important to
150 wonder how things would change if we used some other strategy for
151 implementating lists. These were the lists that made lists look like
152 Church numerals with extra bits embdded in them:
154 empty list: fun f z -> z
155 list with one element: fun f z -> f 1 z
156 list with two elements: fun f z -> f 2 (f 1 z)
157 list with three elements: fun f z -> f 3 (f 2 (f 1 z))
159 and so on. To save time, we'll let the OCaml interpreter infer the
160 principle types of these functions (rather than deducing what the
164 - : 'a -> 'b -> 'b = <fun>
166 - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
167 # fun f z -> f 2 (f 1 z);;
168 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
169 # fun f z -> f 3 (f 2 (f 1 z))
170 - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
172 We can see what the consistent, general principle types are at the end, so we
173 can stop. These types should remind you of the simply-typed lambda calculus
174 types for Church numerals (`(o -> o) -> o -> o`) with one extra bit thrown in
175 (in this case, an int).
177 So here's our type constructor for our hand-rolled lists:
179 type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b
181 Generalizing to lists that contain any kind of element (not just
184 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
186 So an `('a, 'b) list'` is a list containing elements of type `'a`,
187 where `'b` is the type of some part of the plumbing. This is more
188 general than an ordinary OCaml list, but we'll see how to map them
189 into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
190 in order to proceed to build a monad:
192 l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
194 No problem. Arriving at bind is a little more complicated, but
195 exactly the same principles apply, you just have to be careful and
198 l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ...
200 Unpacking the types gives:
202 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
203 (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
204 : ('c -> 'd -> 'd) -> 'd -> 'd = ...
206 But it's a rookie mistake to quail before complicated types. You should
207 be no more intimiated by complex types than by a linguistic tree with
208 deeply embedded branches: complex structure created by repeated
209 application of simple rules.
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` as an
213 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:
215 ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
217 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)`:
219 ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
221 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:
223 ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
225 Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
227 fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
229 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:
231 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
232 (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
233 : ('c -> 'b -> 'b) -> 'b -> 'b =
234 fun k -> u (fun a b -> f a k b)
236 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.
238 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:
240 fun k z -> u (fun a b -> f a k b) z
242 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?
244 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:
247 concat [[]; [2]; [2; 4]; [2; 4; 8]] =
250 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
252 fun k z -> u (fun a b -> f a k b) z
254 do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
261 (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.
263 So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
266 right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
267 right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
268 right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
269 right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
271 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:
273 fun k z -> u (fun a b -> f a k b) z
275 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
277 fun k z -> List.fold_right k (concat (map f u)) z
281 For future reference, we might make two eta-reductions to our formula, so that we have instead:
283 let l'_bind = fun k -> u (fun a -> f a k);;
285 Let's make some more tests:
288 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
290 l'_bind (fun f z -> f 1 (f 2 z))
291 (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
293 Sigh. OCaml won't show us our own list. So we have to choose an `f`
294 and a `z` that will turn our hand-crafted lists into standard OCaml
295 lists, so that they will print out.
297 # let cons h t = h :: t;; (* OCaml is stupid about :: *)
298 # l'_bind (fun f z -> f 1 (f 2 z))
299 (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
300 - : int list = [1; 2; 2; 3]
305 Montague's PTQ treatment of DPs as generalized quantifiers
306 ----------------------------------------------------------
308 We've hinted that Montague's treatment of DPs as generalized
309 quantifiers embodies the spirit of continuations (see de Groote 2001,
310 Barker 2002 for lengthy discussion). Let's see why.
312 First, we'll need a type constructor. As you probably know,
313 Montague replaced individual-denoting determiner phrases (with type `e`)
314 with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
315 In particular, the denotation of a proper name like *John*, which
316 might originally denote a object `j` of type `e`, came to denote a
317 generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
318 Let's write a general function that will map individuals into their
319 corresponding generalized quantifier:
321 gqize (a : e) = fun (p : e -> t) -> p a
323 This function wraps up an individual in a fancy box. That is to say,
324 we are in the presence of a monad. The type constructor, the unit and
325 the bind follow naturally. We've done this enough times that we won't
326 belabor the construction of the bind function, the derivation is
327 similar to the List monad just given:
329 type 'a continuation = ('a -> 'b) -> 'b
330 c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
331 c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
332 fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
334 How similar is it to the List monad? Let's examine the type
335 constructor and the terms from the list monad derived above:
337 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
338 l'_unit a = fun f -> f a
339 l'_bind u f = fun k -> u (fun a -> f a k)
341 (We performed a sneaky but valid eta reduction in the unit term.)
343 The unit and the bind for the Montague continuation monad and the
344 homemade List monad are the same terms! In other words, the behavior
345 of the List monad and the behavior of the continuations monad are
346 parallel in a deep sense.
348 Have we really discovered that lists are secretly continuations? Or
349 have we merely found a way of simulating lists using list
350 continuations? Well, strictly speaking, what we have done is shown
351 that one particular implementation of lists---the left fold
352 implementation---gives rise to a continuation monad fairly naturally,
353 and that this monad can reproduce the behavior of the standard list
354 monad. But what about other list implementations? Do they give rise
355 to monads that can be understood in terms of continuations?
357 Refunctionalizing zippers
358 -------------------------
360 Manipulating trees with monads
361 ------------------------------
363 This thread develops an idea based on a detailed suggestion of Ken
364 Shan's. We'll build a series of functions that operate on trees,
365 doing various things, including replacing leaves, counting nodes, and
366 converting a tree to a list of leaves. The end result will be an
367 application for continuations.
369 From an engineering standpoint, we'll build a tree transformer that
370 deals in monads. We can modify the behavior of the system by swapping
371 one monad for another. (We've already seen how adding a monad can add
372 a layer of funtionality without disturbing the underlying system, for
373 instance, in the way that the reader monad allowed us to add a layer
374 of intensionality to an extensional grammar, but we have not yet seen
375 the utility of replacing one monad with other.)
377 First, we'll be needing a lot of trees during the remainder of the
378 course. Here's a type constructor for binary trees:
380 type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
382 These are trees in which the internal nodes do not have labels. [How
383 would you adjust the type constructor to allow for labels on the
386 We'll be using trees where the nodes are integers, e.g.,
390 let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
391 (Node ((Leaf 5),(Node ((Leaf 7),
406 Our first task will be to replace each leaf with its double:
409 let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
410 match t with Leaf x -> Leaf (newleaf x)
411 | Node (l, r) -> Node ((treemap newleaf l),
412 (treemap newleaf r));;
414 `treemap` takes a function that transforms old leaves into new leaves,
415 and maps that function over all the leaves in the tree, leaving the
416 structure of the tree unchanged. For instance:
419 let double i = i + i;;
422 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
436 We could have built the doubling operation right into the `treemap`
437 code. However, because what to do to each leaf is a parameter, we can
438 decide to do something else to the leaves without needing to rewrite
439 `treemap`. For instance, we can easily square each leaf instead by
440 supplying the appropriate `int -> int` operation in place of `double`:
443 let square x = x * x;;
446 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
449 Note that what `treemap` does is take some global, contextual
450 information---what to do to each leaf---and supplies that information
451 to each subpart of the computation. In other words, `treemap` has the
452 behavior of a reader monad. Let's make that explicit.
454 In general, we're on a journey of making our treemap function more and
455 more flexible. So the next step---combining the tree transducer with
456 a reader monad---is to have the treemap function return a (monadized)
457 tree that is ready to accept any `int->int` function and produce the
460 \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))
474 That is, we want to transform the ordinary tree `t1` (of type `int
475 tree`) into a reader object of type `(int->int)-> int tree`: something
476 that, when you apply it to an `int->int` function returns an `int
477 tree` in which each leaf `x` has been replaced with `(f x)`.
479 With previous readers, we always knew which kind of environment to
480 expect: either an assignment function (the original calculator
481 simulation), a world (the intensionality monad), an integer (the
482 Jacobson-inspired link monad), etc. In this situation, it will be
483 enough for now to expect that our reader will expect a function of
487 type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *)
488 let reader_unit (x:'a): 'a reader = fun _ -> x;;
489 let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;
492 It's easy to figure out how to turn an `int` into an `int reader`:
495 let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
496 int2int_reader 2 (fun i -> i + i);;
500 But what do we do when the integers are scattered over the leaves of a
501 tree? A binary tree is not the kind of thing that we can apply a
502 function of type `int->int` to.
505 let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
506 match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
507 | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
508 reader_bind (treemonadizer f r) (fun y ->
509 reader_unit (Node (x, y))));;
512 This function says: give me a function `f` that knows how to turn
513 something of type `'a` into an `'b reader`, and I'll show you how to
514 turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
515 the `treemonadizer` function builds plumbing that connects all of the
516 leaves of a tree into one connected monadic network; it threads the
517 monad through the leaves.
520 # treemonadizer int2int_reader t1 (fun i -> i + i);;
522 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
525 Here, our environment is the doubling function (`fun i -> i + i`). If
526 we apply the very same `int tree reader` (namely, `treemonadizer
527 int2int_reader t1`) to a different `int->int` function---say, the
528 squaring function, `fun i -> i * i`---we get an entirely different
532 # treemonadizer int2int_reader t1 (fun i -> i * i);;
534 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
537 Now that we have a tree transducer that accepts a monad as a
538 parameter, we can see what it would take to swap in a different monad.
539 For instance, we can use a state monad to count the number of nodes in
543 type 'a state = int -> 'a * int;;
544 let state_unit x i = (x, i+.5);;
545 let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
548 Gratifyingly, we can use the `treemonadizer` function without any
549 modification whatsoever, except for replacing the (parametric) type
550 `reader` with `state`:
553 let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
554 match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
555 | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
556 state_bind (treemonadizer f r) (fun y ->
557 state_unit (Node (x, y))));;
560 Then we can count the number of nodes in the tree:
563 # treemonadizer state_unit t1 0;;
565 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
579 Notice that we've counted each internal node twice---it's a good
580 exercise to adjust the code to count each node once.
582 One more revealing example before getting down to business: replacing
583 `state` everywhere in `treemonadizer` with `list` gives us
586 # treemonadizer (fun x -> [ [x; square x] ]) t1;;
587 - : int list tree list =
589 (Node (Leaf [2; 4], Leaf [3; 9]),
590 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
593 Unlike the previous cases, instead of turning a tree into a function
594 from some input to a result, this transformer replaces each `int` with
597 Now for the main point. What if we wanted to convert a tree to a list
601 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
602 let continuation_unit x c = c x;;
603 let continuation_bind u f c = u (fun a -> f a c);;
605 let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
606 match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
607 | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
608 continuation_bind (treemonadizer f r) (fun y ->
609 continuation_unit (Node (x, y))));;
612 We use the continuation monad described above, and insert the
613 `continuation` type in the appropriate place in the `treemonadizer` code.
617 # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
618 - : int list = [2; 3; 5; 7; 11]
621 We have found a way of collapsing a tree into a list of its leaves.
623 The continuation monad is amazingly flexible; we can use it to
624 simulate some of the computations performed above. To see how, first
625 note that an interestingly uninteresting thing happens if we use the
626 continuation unit as our first argument to `treemonadizer`, and then
627 apply the result to the identity function:
630 # treemonadizer continuation_unit t1 (fun x -> x);;
632 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
635 That is, nothing happens. But we can begin to substitute more
636 interesting functions for the first argument of `treemonadizer`:
639 (* Simulating the tree reader: distributing a operation over the leaves *)
640 # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
642 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
644 (* Simulating the int list tree list *)
645 # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
648 (Node (Leaf [2; 4], Leaf [3; 9]),
649 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
651 (* Counting leaves *)
652 # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
656 We could simulate the tree state example too, but it would require
657 generalizing the type of the continuation monad to
659 type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
661 The binary tree monad
662 ---------------------
664 Of course, by now you may have realized that we have discovered a new
665 monad, the binary tree monad:
668 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
669 let tree_unit (x:'a) = Leaf x;;
670 let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
671 match u with Leaf x -> f x
672 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
675 For once, let's check the Monad laws. The left identity law is easy:
677 Left identity: bind (unit a) f = bind (Leaf a) f = fa
679 To check the other two laws, we need to make the following
680 observation: it is easy to prove based on `tree_bind` by a simple
681 induction on the structure of the first argument that the tree
682 resulting from `bind u f` is a tree with the same strucure as `u`,
683 except that each leaf `a` has been replaced with `fa`:
685 \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
702 Given this equivalence, the right identity law
704 Right identity: bind u unit = u
706 falls out once we realize that
708 bind (Leaf a) unit = unit a = Leaf a
710 As for the associative law,
712 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
714 we'll give an example that will show how an inductive proof would
715 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
717 \tree (. (. (. (. (a1)(a2)))))
718 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
723 bind __|__ f = __|_ = . .
725 a1 a2 fa1 fa2 | | | |
729 Now when we bind this tree to `g`, we get
741 At this point, it should be easy to convince yourself that
742 using the recipe on the right hand side of the associative law will
743 built the exact same final tree.
745 So binary trees are a monad.
747 Haskell combines this monad with the Option monad to provide a monad
749 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
751 represent non-deterministic computations as a tree.