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 The best way to follow the next long, somewhat intricate paragraph
217 immediately following is to take this type and try to construct a term
218 for it, just as we did for the monads above. If you suceed, the
219 discussion will just make brilliant sense. If you get stuck, the
220 discussion will explain how to proceed.
222 As usual, we need to unpack the `u` box. Examine the type of `u`.
223 This time, `u` will only deliver up its contents if we give `u` an
224 argument that is 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:
226 ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...
228 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)`:
230 ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...
232 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:
234 ... u (fun (a : 'a) (b : 'b) -> f a k b) ...
236 Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:
238 fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)
240 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:
242 l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
243 (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
244 : ('c -> 'b -> 'b) -> 'b -> 'b =
245 fun k -> u (fun a b -> f a k b)
247 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.
249 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:
251 fun k z -> u (fun a b -> f a k b) z
253 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?
255 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:
258 concat [[]; [2]; [2; 4]; [2; 4; 8]] =
261 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
263 fun k z -> u (fun a b -> f a k b) z
265 do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:
272 (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.
274 So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:
277 right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
278 right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
279 right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
280 right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0
282 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:
284 fun k z -> u (fun a b -> f a k b) z
286 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
288 fun k z -> List.fold_right k (concat (map f u)) z
292 For future reference, we might make two eta-reductions to our formula, so that we have instead:
294 let l'_bind = fun k -> u (fun a -> f a k);;
296 Let's make some more tests:
299 l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]
301 l'_bind (fun f z -> f 1 (f 2 z))
302 (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>
304 Sigh. OCaml won't show us our own list. So we have to choose an `f`
305 and a `z` that will turn our hand-crafted lists into standard OCaml
306 lists, so that they will print out.
308 # let cons h t = h :: t;; (* OCaml is stupid about :: *)
309 # l'_bind (fun f z -> f 1 (f 2 z))
310 (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
311 - : int list = [1; 2; 2; 3]
316 Montague's PTQ treatment of DPs as generalized quantifiers
317 ----------------------------------------------------------
319 We've hinted that Montague's treatment of DPs as generalized
320 quantifiers embodies the spirit of continuations (see de Groote 2001,
321 Barker 2002 for lengthy discussion). Let's see why.
323 First, we'll need a type constructor. As you probably know,
324 Montague replaced individual-denoting determiner phrases (with type `e`)
325 with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
326 In particular, the denotation of a proper name like *John*, which
327 might originally denote a object `j` of type `e`, came to denote a
328 generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
329 Let's write a general function that will map individuals into their
330 corresponding generalized quantifier:
332 gqize (a : e) = fun (p : e -> t) -> p a
334 This function wraps up an individual in a fancy box. That is to say,
335 we are in the presence of a monad. The type constructor, the unit and
336 the bind follow naturally. We've done this enough times that we won't
337 belabor the construction of the bind function, the derivation is
338 similar to the List monad just given:
340 type 'a continuation = ('a -> 'b) -> 'b
341 c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
342 c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
343 fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
345 How similar is it to the List monad? Let's examine the type
346 constructor and the terms from the list monad derived above:
348 type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
349 l'_unit a = fun f -> f a
350 l'_bind u f = fun k -> u (fun a -> f a k)
352 (We performed a sneaky but valid eta reduction in the unit term.)
354 The unit and the bind for the Montague continuation monad and the
355 homemade List monad are the same terms! In other words, the behavior
356 of the List monad and the behavior of the continuations monad are
357 parallel in a deep sense.
359 Have we really discovered that lists are secretly continuations? Or
360 have we merely found a way of simulating lists using list
361 continuations? Well, strictly speaking, what we have done is shown
362 that one particular implementation of lists---the left fold
363 implementation---gives rise to a continuation monad fairly naturally,
364 and that this monad can reproduce the behavior of the standard list
365 monad. But what about other list implementations? Do they give rise
366 to monads that can be understood in terms of continuations?
368 Refunctionalizing zippers
369 -------------------------
371 Manipulating trees with monads
372 ------------------------------
374 This thread develops an idea based on a detailed suggestion of Ken
375 Shan's. We'll build a series of functions that operate on trees,
376 doing various things, including replacing leaves, counting nodes, and
377 converting a tree to a list of leaves. The end result will be an
378 application for continuations.
380 From an engineering standpoint, we'll build a tree transformer that
381 deals in monads. We can modify the behavior of the system by swapping
382 one monad for another. (We've already seen how adding a monad can add
383 a layer of funtionality without disturbing the underlying system, for
384 instance, in the way that the reader monad allowed us to add a layer
385 of intensionality to an extensional grammar, but we have not yet seen
386 the utility of replacing one monad with other.)
388 First, we'll be needing a lot of trees during the remainder of the
389 course. Here's a type constructor for binary trees:
391 type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
393 These are trees in which the internal nodes do not have labels. [How
394 would you adjust the type constructor to allow for labels on the
397 We'll be using trees where the nodes are integers, e.g.,
401 let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
402 (Node ((Leaf 5),(Node ((Leaf 7),
417 Our first task will be to replace each leaf with its double:
420 let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
421 match t with Leaf x -> Leaf (newleaf x)
422 | Node (l, r) -> Node ((treemap newleaf l),
423 (treemap newleaf r));;
425 `treemap` takes a function that transforms old leaves into new leaves,
426 and maps that function over all the leaves in the tree, leaving the
427 structure of the tree unchanged. For instance:
430 let double i = i + i;;
433 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
447 We could have built the doubling operation right into the `treemap`
448 code. However, because what to do to each leaf is a parameter, we can
449 decide to do something else to the leaves without needing to rewrite
450 `treemap`. For instance, we can easily square each leaf instead by
451 supplying the appropriate `int -> int` operation in place of `double`:
454 let square x = x * x;;
457 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
460 Note that what `treemap` does is take some global, contextual
461 information---what to do to each leaf---and supplies that information
462 to each subpart of the computation. In other words, `treemap` has the
463 behavior of a reader monad. Let's make that explicit.
465 In general, we're on a journey of making our treemap function more and
466 more flexible. So the next step---combining the tree transducer with
467 a reader monad---is to have the treemap function return a (monadized)
468 tree that is ready to accept any `int->int` function and produce the
471 \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))
485 That is, we want to transform the ordinary tree `t1` (of type `int
486 tree`) into a reader object of type `(int->int)-> int tree`: something
487 that, when you apply it to an `int->int` function returns an `int
488 tree` in which each leaf `x` has been replaced with `(f x)`.
490 With previous readers, we always knew which kind of environment to
491 expect: either an assignment function (the original calculator
492 simulation), a world (the intensionality monad), an integer (the
493 Jacobson-inspired link monad), etc. In this situation, it will be
494 enough for now to expect that our reader will expect a function of
498 type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *)
499 let reader_unit (x:'a): 'a reader = fun _ -> x;;
500 let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;
503 It's easy to figure out how to turn an `int` into an `int reader`:
506 let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
507 int2int_reader 2 (fun i -> i + i);;
511 But what do we do when the integers are scattered over the leaves of a
512 tree? A binary tree is not the kind of thing that we can apply a
513 function of type `int->int` to.
516 let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
517 match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
518 | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
519 reader_bind (treemonadizer f r) (fun y ->
520 reader_unit (Node (x, y))));;
523 This function says: give me a function `f` that knows how to turn
524 something of type `'a` into an `'b reader`, and I'll show you how to
525 turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
526 the `treemonadizer` function builds plumbing that connects all of the
527 leaves of a tree into one connected monadic network; it threads the
528 monad through the leaves.
531 # treemonadizer int2int_reader t1 (fun i -> i + i);;
533 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
536 Here, our environment is the doubling function (`fun i -> i + i`). If
537 we apply the very same `int tree reader` (namely, `treemonadizer
538 int2int_reader t1`) to a different `int->int` function---say, the
539 squaring function, `fun i -> i * i`---we get an entirely different
543 # treemonadizer int2int_reader t1 (fun i -> i * i);;
545 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
548 Now that we have a tree transducer that accepts a monad as a
549 parameter, we can see what it would take to swap in a different monad.
550 For instance, we can use a state monad to count the number of nodes in
554 type 'a state = int -> 'a * int;;
555 let state_unit x i = (x, i+.5);;
556 let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
559 Gratifyingly, we can use the `treemonadizer` function without any
560 modification whatsoever, except for replacing the (parametric) type
561 `reader` with `state`:
564 let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
565 match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
566 | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
567 state_bind (treemonadizer f r) (fun y ->
568 state_unit (Node (x, y))));;
571 Then we can count the number of nodes in the tree:
574 # treemonadizer state_unit t1 0;;
576 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
590 Notice that we've counted each internal node twice---it's a good
591 exercise to adjust the code to count each node once.
593 One more revealing example before getting down to business: replacing
594 `state` everywhere in `treemonadizer` with `list` gives us
597 # treemonadizer (fun x -> [ [x; square x] ]) t1;;
598 - : int list tree list =
600 (Node (Leaf [2; 4], Leaf [3; 9]),
601 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
604 Unlike the previous cases, instead of turning a tree into a function
605 from some input to a result, this transformer replaces each `int` with
608 Now for the main point. What if we wanted to convert a tree to a list
612 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
613 let continuation_unit x c = c x;;
614 let continuation_bind u f c = u (fun a -> f a c);;
616 let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
617 match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
618 | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
619 continuation_bind (treemonadizer f r) (fun y ->
620 continuation_unit (Node (x, y))));;
623 We use the continuation monad described above, and insert the
624 `continuation` type in the appropriate place in the `treemonadizer` code.
628 # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
629 - : int list = [2; 3; 5; 7; 11]
632 We have found a way of collapsing a tree into a list of its leaves.
634 The continuation monad is amazingly flexible; we can use it to
635 simulate some of the computations performed above. To see how, first
636 note that an interestingly uninteresting thing happens if we use the
637 continuation unit as our first argument to `treemonadizer`, and then
638 apply the result to the identity function:
641 # treemonadizer continuation_unit t1 (fun x -> x);;
643 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
646 That is, nothing happens. But we can begin to substitute more
647 interesting functions for the first argument of `treemonadizer`:
650 (* Simulating the tree reader: distributing a operation over the leaves *)
651 # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
653 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
655 (* Simulating the int list tree list *)
656 # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
659 (Node (Leaf [2; 4], Leaf [3; 9]),
660 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
662 (* Counting leaves *)
663 # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
667 We could simulate the tree state example too, but it would require
668 generalizing the type of the continuation monad to
670 type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
672 The binary tree monad
673 ---------------------
675 Of course, by now you may have realized that we have discovered a new
676 monad, the binary tree monad:
679 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
680 let tree_unit (x:'a) = Leaf x;;
681 let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
682 match u with Leaf x -> f x
683 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
686 For once, let's check the Monad laws. The left identity law is easy:
688 Left identity: bind (unit a) f = bind (Leaf a) f = fa
690 To check the other two laws, we need to make the following
691 observation: it is easy to prove based on `tree_bind` by a simple
692 induction on the structure of the first argument that the tree
693 resulting from `bind u f` is a tree with the same strucure as `u`,
694 except that each leaf `a` has been replaced with `fa`:
696 \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
713 Given this equivalence, the right identity law
715 Right identity: bind u unit = u
717 falls out once we realize that
719 bind (Leaf a) unit = unit a = Leaf a
721 As for the associative law,
723 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
725 we'll give an example that will show how an inductive proof would
726 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
728 \tree (. (. (. (. (a1)(a2)))))
729 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
734 bind __|__ f = __|_ = . .
736 a1 a2 fa1 fa2 | | | |
740 Now when we bind this tree to `g`, we get
752 At this point, it should be easy to convince yourself that
753 using the recipe on the right hand side of the associative law will
754 built the exact same final tree.
756 So binary trees are a monad.
758 Haskell combines this monad with the Option monad to provide a monad
760 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
762 represent non-deterministic computations as a tree.