3 Manipulating trees with monads
4 ------------------------------
6 This topic develops an idea based on a suggestion of Ken Shan's.
7 We'll build a series of functions that operate on trees, doing various
8 things, including updating leaves with a Reader monad, counting nodes
9 with a State monad, replacing leaves with a List monad, and converting
10 a tree into a list of leaves with a Continuation monad. It will turn
11 out that the continuation monad can simulate the behavior of each of
14 From an engineering standpoint, we'll build a tree transformer that
15 deals in monads. We can modify the behavior of the system by swapping
16 one monad for another. We've already seen how adding a monad can add
17 a layer of funtionality without disturbing the underlying system, for
18 instance, in the way that the Reader monad allowed us to add a layer
19 of intensionality to an extensional grammar, but we have not yet seen
20 the utility of replacing one monad with other.
22 First, we'll be needing a lot of trees for the remainder of the
23 course. Here again is a type constructor for leaf-labeled, binary trees:
25 type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree);;
27 [How would you adjust the type constructor to allow for labels on the
30 We'll be using trees where the nodes are integers, e.g.,
33 let t1 = Node (Node (Leaf 2, Leaf 3),
34 Node (Leaf 5, Node (Leaf 7,
47 Our first task will be to replace each leaf with its double:
49 let rec tree_map (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree =
51 | Leaf i -> Leaf (leaf_modifier i)
52 | Node (l, r) -> Node (tree_map leaf_modifier l,
53 tree_map leaf_modifier r);;
55 `tree_map` takes a function that transforms old leaves into new leaves,
56 and maps that function over all the leaves in the tree, leaving the
57 structure of the tree unchanged. For instance:
59 let double i = i + i;;
62 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
75 We could have built the doubling operation right into the `tree_map`
76 code. However, because we've made what to do to each leaf a
77 parameter, we can decide to do something else to the leaves without
78 needing to rewrite `tree_map`. For instance, we can easily square
79 each leaf instead by supplying the appropriate `int -> int` operation
82 let square i = i * i;;
85 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
87 Note that what `tree_map` does is take some unchanging contextual
88 information---what to do to each leaf---and supplies that information
89 to each subpart of the computation. In other words, `tree_map` has the
90 behavior of a Reader monad. Let's make that explicit.
92 In general, we're on a journey of making our `tree_map` function more and
93 more flexible. So the next step---combining the tree transformer with
94 a Reader monad---is to have the `tree_map` function return a (monadized)
95 tree that is ready to accept any `int -> int` function and produce the
109 That is, we want to transform the ordinary tree `t1` (of type `int
110 tree`) into a reader monadic object of type `(int -> int) -> int
111 tree`: something that, when you apply it to an `int -> int` function
112 `f` returns an `int tree` in which each leaf `i` has been replaced
115 [Application note: this kind of reader object could provide a model
116 for Kaplan's characters. It turns an ordinary tree into one that
117 expects contextual information (here, the `λ f`) that can be
118 used to compute the content of indexicals embedded arbitrarily deeply
121 With our previous applications of the Reader monad, we always knew
122 which kind of environment to expect: either an assignment function, as
123 in the original calculator simulation; a world, as in the
124 intensionality monad; an individual, as in the Jacobson-inspired link
125 monad; etc. In the present case, we expect that our "environment"
126 will be some function of type `int -> int`. "Looking up" some `int` in
127 the environment will return us the `int` that comes out the other side
130 type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *)
131 let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
132 let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;
134 It would be a simple matter to turn an *integer* into an `int reader`:
136 let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;;
137 int_readerize 2 (fun i -> i + i);;
140 But how do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader?
141 A tree is not the kind of thing that we can apply a
142 function of type `int -> int` to.
146 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
148 | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
149 | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->
150 reader_bind (tree_monadize f r) (fun r' ->
151 reader_unit (Node (l', r'))));;
153 This function says: give me a function `f` that knows how to turn
154 something of type `'a` into an `'b reader`---this is a function of the same type that you could bind an `'a reader` to---and I'll show you how to
155 turn an `'a tree` into an `'b tree reader`. That is, if you show me how to do this:
161 then I'll give you back the ability to do this:
165 __|___ ---> | __|___ |
170 And how will that boxed tree behave? Whatever actions you perform on it will be transmitted down to corresponding operations on its leaves. For instance, our `int reader` expects an `int -> int` environment. If supplying environment `e` to our `int reader` doubles the contained `int`:
173 1 ---> | 1 | applied to e ~~> 2
176 Then we can expect that supplying it to our `int tree reader` will double all the leaves:
180 __|___ ---> | __|___ | applied to e ~~> __|___
185 In more fanciful terms, the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the
186 `'b reader` monad through the original tree's leaves.
188 # tree_monadize int_readerize t1 double;;
190 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
192 Here, our environment is the doubling function (`fun i -> i + i`). If
193 we apply the very same `int tree reader` (namely, `tree_monadize
194 int_readerize t1`) to a different `int -> int` function---say, the
195 squaring function, `fun i -> i * i`---we get an entirely different
198 # tree_monadize int_readerize t1 square;;
200 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
202 Now that we have a tree transformer that accepts a *reader* monad as a
203 parameter, we can see what it would take to swap in a different monad.
205 For instance, we can use a State monad to count the number of leaves in
208 type 'a state = int -> 'a * int;;
209 let state_unit a = fun s -> (a, s);;
210 let state_bind u f = fun s -> let (a, s') = u s in f a s';;
212 Gratifyingly, we can use the `tree_monadize` function without any
213 modification whatsoever, except for replacing the (parametric) type
214 `'b reader` with `'b state`, and substituting in the appropriate unit and bind:
216 let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
218 | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
219 | Node (l, r) -> state_bind (tree_monadize f l) (fun l' ->
220 state_bind (tree_monadize f r) (fun r' ->
221 state_unit (Node (l', r'))));;
223 Then we can count the number of leaves in the tree:
225 # tree_monadize (fun a -> fun s -> (a, s+1)) t1 0;;
227 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
240 Note that the value returned is a pair consisting of a tree and an
241 integer, 5, which represents the count of the leaves in the tree.
243 Why does this work? Because the operation `fun a -> fun s -> (a, s+1)`
244 takes an `int` and wraps it in an `int state` monadic box that
245 increments the state. When we give that same operations to our
246 `tree_monadize` function, it then wraps an `int tree` in a box, one
247 that does the same state-incrementing for each of its leaves.
249 We can use the state monad to replace leaves with a number
250 corresponding to that leave's ordinal position. When we do so, we
251 reveal the order in which the monadic tree forces evaluation:
253 # tree_monadize (fun a -> fun s -> (s+1, s+1)) t1 0;;
255 (Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5)
257 The key thing to notice is that instead of copying `a` into the
258 monadic box, we throw away the `a` and put a copy of the state in
261 Reversing the order requires reversing the order of the state_bind
262 operations. It's not obvious that this will type correctly, so think
265 let rec tree_monadize_rev (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
267 | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
268 | Node (l, r) -> state_bind (tree_monadize f r) (fun r' -> (* R first *)
269 state_bind (tree_monadize f l) (fun l'-> (* Then L *)
270 state_unit (Node (l', r'))));;
272 # tree_monadize_rev (fun a -> fun s -> (s+1, s+1)) t1 0;;
274 (Node (Node (Leaf 5, Leaf 4), Node (Leaf 3, Node (Leaf 2, Leaf 1))), 5)
276 We will need below to depend on controlling the order in which nodes
277 are visited when we use the continuation monad to solve the
280 One more revealing example before getting down to business: replacing
281 `state` everywhere in `tree_monadize` with `list` gives us
283 # tree_monadize (fun i -> [ [i; square i] ]) t1;;
284 - : int list tree list =
286 (Node (Leaf [2; 4], Leaf [3; 9]),
287 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
289 Unlike the previous cases, instead of turning a tree into a function
290 from some input to a result, this transformer replaces each `int` with
291 a list of `int`'s. We might also have done this with a Reader monad, though then our environments would need to be of type `int -> int list`. Experiment with what happens if you supply the `tree_monadize` based on the List monad an operation like `fun i -> [2*i; 3*i]`. Use small trees for your experiment.
293 [Why is the argument to `tree_monadize` `int -> int list list` instead
294 of `int -> int list`? Well, as usual, the List monad bind operation
295 will erase the outer list box, so if we want to replace the leaves
296 with lists, we have to nest the replacement lists inside a disposable
299 Now for the main point. What if we wanted to convert a tree to a list
302 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
303 let continuation_unit a = fun k -> k a;;
304 let continuation_bind u f = fun k -> u (fun a -> f a k);;
306 let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
308 | Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b))
309 | Node (l, r) -> continuation_bind (tree_monadize f l) (fun l' ->
310 continuation_bind (tree_monadize f r) (fun r' ->
311 continuation_unit (Node (l', r'))));;
313 We use the Continuation monad described above, and insert the
314 `continuation` type in the appropriate place in the `tree_monadize` code. Then if we give the `tree_monadize` function an operation that converts `int`s into `'b`-wrapping Continuation monads, it will give us back a way to turn `int tree`s into corresponding `'b tree`-wrapping Continuation monads.
316 So for example, we compute:
318 # tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;
319 - : int list = [2; 3; 5; 7; 11]
321 We have found a way of collapsing a tree into a list of its
322 leaves. Can you trace how this is working? Think first about what the
323 operation `fun a -> fun k -> a :: k a` does when you apply it to a
324 plain `int`, and the continuation `fun _ -> []`. Then given what we've
325 said about `tree_monadize`, what should we expect `tree_monadize (fun
326 a -> fun k -> a :: k a` to do?
328 In a moment, we'll return to the same-fringe problem. Since the
329 simple but inefficient way to solve it is to map each tree to a list
330 of its leaves, this transformation is on the path to a more efficient
331 solution. We'll just have to figure out how to postpone computing the
332 tail of the list until it's needed...
334 The Continuation monad is amazingly flexible; we can use it to
335 simulate some of the computations performed above. To see how, first
336 note that an interestingly uninteresting thing happens if we use
337 `continuation_unit` as our first argument to `tree_monadize`, and then
338 apply the result to the identity function:
340 # tree_monadize continuation_unit t1 (fun t -> t);;
342 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
344 That is, nothing happens. But we can begin to substitute more
345 interesting functions for the first argument of `tree_monadize`:
347 (* Simulating the tree reader: distributing a operation over the leaves *)
348 # tree_monadize (fun a -> fun k -> k (square a)) t1 (fun t -> t);;
350 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
352 (* Simulating the int list tree list *)
353 # tree_monadize (fun a -> fun k -> k [a; square a]) t1 (fun t -> t);;
356 (Node (Leaf [2; 4], Leaf [3; 9]),
357 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
359 (* Counting leaves *)
360 # tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;
363 [To be fixed: exactly which kind of monad each of these computations simulates.]
365 We could simulate the tree state example too by setting the relevant
366 type to `('a, 'state -> 'result) continuation`.
367 In fact, Andre Filinsky has suggested that the continuation monad is
368 able to simulate any other monad (Google for "mother of all monads").
370 We would eventually want to generalize the continuation type to
372 type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
374 If you want to see how to parameterize the definition of the `tree_monadize` function, so that you don't have to keep rewriting it for each new monad, see [this code](/code/tree_monadize.ml).
376 Using continuations to solve the same fringe problem
377 ----------------------------------------------------
379 We've seen two solutions to the same fringe problem so far.
380 The problem, recall, is to take two trees and decide whether they have
381 the same leaves in the same order.
393 let ta = Node (Leaf 1, Node (Leaf 2, Leaf 3));;
394 let tb = Node (Node (Leaf 1, Leaf 2), Leaf 3);;
395 let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));;
398 So `ta` and `tb` are different trees that have the same fringe, but
399 `ta` and `tc` are not.
401 The simplest solution is to map each tree to a list of its leaves,
402 then compare the lists. But because we will have computed the entire
403 fringe before starting the comparison, if the fringes differ in an
404 early position, we've wasted our time examining the rest of the trees.
406 The second solution was to use tree zippers and mutable state to
407 simulate coroutines (see [[coroutines and aborts]]). In that
408 solution, we pulled the zipper on the first tree until we found the
409 next leaf, then stored the zipper structure in the mutable variable
410 while we turned our attention to the other tree. Because we stopped
411 as soon as we find the first mismatched leaf, this solution does not
412 have the flaw just mentioned of the solution that maps both trees to a
413 list of leaves before beginning comparison.
415 Since zippers are just continuations reified, we expect that the
416 solution in terms of zippers can be reworked using continuations, and
417 this is indeed the case. Before we can arrive at a solution, however,
418 we must define a data structure called a stream:
420 type 'a stream = End | Next of 'a * (unit -> 'a stream);;
422 A stream is like a list in that it contains a series of objects (all
423 of the same type, here, type `'a`). The first object in the stream
424 corresponds to the head of a list, which we pair with a stream
425 representing the rest of a the list. There is a special stream called
426 `End` that represents a stream that contains no (more) elements,
427 analogous to the empty list `[]`.
429 Actually, we pair each element not with a stream, but with a thunked
430 stream, that is, a function from the unit type to streams. The idea
431 is that the next element in the stream is not computed until we forced
432 the thunk by applying it to the unit:
435 # let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));;
436 val make_int_stream : int -> int stream = <fun>
437 # let int_stream = make_int_stream 1;;
438 val int_stream : int stream = Next (1, <fun>) (* First element: 1 *)
439 # match int_stream with Next (i, rest) -> rest;;
440 - : unit -> int stream = <fun> (* Rest: a thunk *)
442 (* Force the thunk to compute the second element *)
443 # (match int_stream with Next (i, rest) -> rest) ();;
444 - : int stream = Next (2, <fun>)
447 You can think of `int_stream` as a functional object that provides
448 access to an infinite sequence of integers, one at a time. It's as if
449 we had written `[1;2;...]` where `...` meant "continue indefinitely".
451 So, with streams in hand, we need only rewrite our continuation tree
452 monadizer so that instead of mapping trees to lists, it maps them to
455 # tree_monadize (fun a k -> a :: k a) t1 (fun t -> []);;
456 - : int list = [2; 3; 5; 7; 11]
460 # tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End);;
461 - : int stream = Next (2, <fun>)
463 We can see the first element in the stream, the first leaf (namely,
464 2), but in order to see the next, we'll have to force a thunk.
466 Then to complete the same-fringe function, we simply convert both
467 trees into leaf-streams, then compare the streams element by element.
468 The code is enitrely routine, but for the sake of completeness, here it is:
471 let rec compare_streams stream1 stream2 =
472 match stream1, stream2 with
473 | End, End -> true (* Done! Fringes match. *)
474 | Next (next1, rest1), Next (next2, rest2) when next1 = next2 -> compare_streams (rest1 ()) (rest2 ())
477 let same_fringe t1 t2 =
478 let stream1 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End) in
479 let stream2 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t2 (fun _ -> End) in
480 compare_streams stream1 stream2;;
483 Notice the forcing of the thunks in the recursive call to
484 `compare_streams`. So indeed:
487 # same_fringe ta tb;;
489 # same_fringe ta tc;;
493 Now, this implementation is a bit silly, since in order to convert the
494 trees to leaf streams, our tree_monadizer function has to visit every
495 node in the tree. But if we needed to compare each tree to a large
496 set of other trees, we could arrange to monadize each tree only once,
497 and then run compare_streams on the monadized trees.
499 By the way, what if you have reason to believe that the fringes of
500 your trees are more likely to differ near the right edge than the left
501 edge? If we reverse evaluation order in the tree_monadizer function,
502 as shown above when we replaced leaves with their ordinal position,
503 then the resulting streams would produce leaves from the right to the
506 The idea of using continuations to characterize natural language meaning
507 ------------------------------------------------------------------------
509 We might a philosopher or a linguist be interested in continuations,
510 especially if efficiency of computation is usually not an issue?
511 Well, the application of continuations to the same-fringe problem
512 shows that continuations can manage order of evaluation in a
513 well-controlled manner. In a series of papers, one of us (Barker) and
514 Ken Shan have argued that a number of phenomena in natural langauge
515 semantics are sensitive to the order of evaluation. We can't
516 reproduce all of the intricate arguments here, but we can give a sense
517 of how the analyses use continuations to achieve an analysis of
518 natural language meaning.
520 **Quantification and default quantifier scope construal**.
522 We saw in the copy-string example and in the same-fringe example that
523 local properties of a tree (whether a character is `S` or not, which
524 integer occurs at some leaf position) can control global properties of
525 the computation (whether the preceeding string is copied or not,
526 whether the computation halts or proceeds). Local control of
527 surrounding context is a reasonable description of in-situ
530 (1) John saw everyone yesterday.
532 This sentence means (roughly)
534 forall x . yesterday(saw x) john
536 That is, the quantifier *everyone* contributes a variable in the
537 direct object position, and a universal quantifier that takes scope
538 over the whole sentence. If we have a lexical meaning function like
542 let lex (s:string) k = match s with
543 | "everyone" -> Node (Leaf "forall x", k "x")
544 | "someone" -> Node (Leaf "exists y", k "y")
547 let sentence1 = Node (Leaf "John",
548 Node (Node (Leaf "saw",
553 Then we can crudely approximate quantification as follows:
556 # tree_monadize lex sentence1 (fun x -> x);;
560 Node (Leaf "John", Node (Node (Leaf "saw", Leaf "x"), Leaf "yesterday")))
563 In order to see the effects of evaluation order,
564 observe what happens when we combine two quantifiers in the same
568 # let sentence2 = Node (Leaf "everyone", Node (Leaf "saw", Leaf "someone"));;
569 # tree_monadize lex sentence2 (fun x -> x);;
573 Node (Leaf "exists y", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
576 The universal takes scope over the existential. If, however, we
577 replace the usual tree_monadizer with tree_monadizer_rev, we get
581 # tree_monadize_rev lex sentence2 (fun x -> x);;
585 Node (Leaf "forall x", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
588 There are many crucially important details about quantification that
589 are being simplified here, and the continuation treatment here is not
590 scalable for a number of reasons. Nevertheless, it will serve to give
591 an idea of how continuations can provide insight into the behavior of
595 The Binary Tree monad
596 ---------------------
598 Of course, by now you may have realized that we have discovered a new
599 monad, the Binary Tree monad. Just as mere lists are in fact a monad,
600 so are trees. Here is the type constructor, unit, and bind:
602 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
603 let tree_unit (a: 'a) : 'a tree = Leaf a;;
604 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
607 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
609 For once, let's check the Monad laws. The left identity law is easy:
611 Left identity: bind (unit a) f = bind (Leaf a) f = f a
613 To check the other two laws, we need to make the following
614 observation: it is easy to prove based on `tree_bind` by a simple
615 induction on the structure of the first argument that the tree
616 resulting from `bind u f` is a tree with the same strucure as `u`,
617 except that each leaf `a` has been replaced with `f a`:
633 Given this equivalence, the right identity law
635 Right identity: bind u unit = u
637 falls out once we realize that
639 bind (Leaf a) unit = unit a = Leaf a
641 As for the associative law,
643 Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
645 we'll give an example that will show how an inductive proof would
646 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
651 bind __|__ f = __|_ = . .
653 a1 a2 f a1 f a2 | | | |
656 Now when we bind this tree to `g`, we get
666 At this point, it should be easy to convince yourself that
667 using the recipe on the right hand side of the associative law will
668 built the exact same final tree.
670 So binary trees are a monad.
672 Haskell combines this monad with the Option monad to provide a monad
674 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
675 that is intended to represent non-deterministic computations as a tree.
678 What's this have to do with tree\_monadize?
679 --------------------------------------------
681 So we've defined a Tree monad:
683 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
684 let tree_unit (a: 'a) : 'a tree = Leaf a;;
685 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
688 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
690 What's this have to do with the `tree_monadize` functions we defined earlier?
692 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
694 | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
695 | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->
696 reader_bind (tree_monadize f r) (fun r' ->
697 reader_unit (Node (l', r'))));;
699 ... and so on for different monads?
701 The answer is that each of those `tree_monadize` functions is adding a Tree monad *layer* to a pre-existing Reader (and so on) monad. We discuss that further here: [[Monad Transformers]].