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
98 \tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))
111 That is, we want to transform the ordinary tree `t1` (of type `int
112 tree`) into a reader monadic object of type `(int -> int) -> int
113 tree`: something that, when you apply it to an `int -> int` function
114 `f` returns an `int tree` in which each leaf `i` has been replaced
117 [Application note: this kind of reader object could provide a model
118 for Kaplan's characters. It turns an ordinary tree into one that
119 expects contextual information (here, the `λ f`) that can be
120 used to compute the content of indexicals embedded arbitrarily deeply
123 With our previous applications of the Reader monad, we always knew
124 which kind of environment to expect: either an assignment function, as
125 in the original calculator simulation; a world, as in the
126 intensionality monad; an individual, as in the Jacobson-inspired link
127 monad; etc. In the present case, we expect that our "environment"
128 will be some function of type `int -> int`. "Looking up" some `int` in
129 the environment will return us the `int` that comes out the other side
132 type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *)
133 let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
134 let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;
136 It would be a simple matter to turn an *integer* into an `int reader`:
138 let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;;
139 int_readerize 2 (fun i -> i + i);;
142 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?
143 A tree is not the kind of thing that we can apply a
144 function of type `int -> int` to.
148 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
150 | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
151 | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->
152 reader_bind (tree_monadize f r) (fun r' ->
153 reader_unit (Node (l', r'))));;
155 This function says: give me a function `f` that knows how to turn
156 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
157 turn an `'a tree` into an `'b tree reader`. That is, if you show me how to do this:
163 then I'll give you back the ability to do this:
167 __|___ ---> | __|___ |
172 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`:
175 1 ---> | 1 | applied to e ~~> 2
178 Then we can expect that supplying it to our `int tree reader` will double all the leaves:
182 __|___ ---> | __|___ | applied to e ~~> __|___
187 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
188 `'b reader` monad through the original tree's leaves.
190 # tree_monadize int_readerize t1 double;;
192 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
194 Here, our environment is the doubling function (`fun i -> i + i`). If
195 we apply the very same `int tree reader` (namely, `tree_monadize
196 int_readerize t1`) to a different `int -> int` function---say, the
197 squaring function, `fun i -> i * i`---we get an entirely different
200 # tree_monadize int_readerize t1 square;;
202 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
204 Now that we have a tree transformer that accepts a *reader* monad as a
205 parameter, we can see what it would take to swap in a different monad.
207 For instance, we can use a State monad to count the number of leaves in
210 type 'a state = int -> 'a * int;;
211 let state_unit a = fun s -> (a, s);;
212 let state_bind u f = fun s -> let (a, s') = u s in f a s';;
214 Gratifyingly, we can use the `tree_monadize` function without any
215 modification whatsoever, except for replacing the (parametric) type
216 `'b reader` with `'b state`, and substituting in the appropriate unit and bind:
218 let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
220 | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
221 | Node (l, r) -> state_bind (tree_monadize f l) (fun l' ->
222 state_bind (tree_monadize f r) (fun r' ->
223 state_unit (Node (l', r'))));;
225 Then we can count the number of leaves in the tree:
227 # tree_monadize (fun a -> fun s -> (a, s+1)) t1 0;;
229 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
242 Note that the value returned is a pair consisting of a tree and an
243 integer, 5, which represents the count of the leaves in the tree.
245 Why does this work? Because the operation `fun a -> fun s -> (a, s+1)`
246 takes an `int` and wraps it in an `int state` monadic box that
247 increments the state. When we give that same operations to our
248 `tree_monadize` function, it then wraps an `int tree` in a box, one
249 that does the same state-incrementing for each of its leaves.
251 We can use the state monad to replace leaves with a number
252 corresponding to that leave's ordinal position. When we do so, we
253 reveal the order in which the monadic tree forces evaluation:
255 # tree_monadize (fun a -> fun s -> (s+1, s+1)) t1 0;;
257 (Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5)
259 The key thing to notice is that instead of copying `a` into the
260 monadic box, we throw away the `a` and put a copy of the state in
263 Reversing the order requires reversing the order of the state_bind
264 operations. It's not obvious that this will type correctly, so think
267 let rec tree_monadize_rev (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
269 | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
270 | Node (l, r) -> state_bind (tree_monadize f r) (fun r' ->
271 state_bind (tree_monadize f l) (fun l' ->
272 state_unit (Node (l', r'))));;
274 # tree_monadize_rev (fun a -> fun s -> (s+1, s+1)) t1 0;;
276 (Node (Node (Leaf 5, Leaf 4), Node (Leaf 3, Node (Leaf 2, Leaf 1))), 5)
278 We will need below to depend on controlling the order in which nodes
279 are visited when we use the continuation monad to solve the
282 One more revealing example before getting down to business: replacing
283 `state` everywhere in `tree_monadize` with `list` gives us
285 # tree_monadize (fun i -> [ [i; square i] ]) t1;;
286 - : int list tree list =
288 (Node (Leaf [2; 4], Leaf [3; 9]),
289 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
291 Unlike the previous cases, instead of turning a tree into a function
292 from some input to a result, this transformer replaces each `int` with
293 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.
295 [Why is the argument to `tree_monadize` `int -> int list list` instead
296 of `int -> int list`? Well, as usual, the List monad bind operation
297 will erase the outer list box, so if we want to replace the leaves
298 with lists, we have to nest the replacement lists inside a disposable
301 Now for the main point. What if we wanted to convert a tree to a list
304 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
305 let continuation_unit a = fun k -> k a;;
306 let continuation_bind u f = fun k -> u (fun a -> f a k);;
308 let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
310 | Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b))
311 | Node (l, r) -> continuation_bind (tree_monadize f l) (fun l' ->
312 continuation_bind (tree_monadize f r) (fun r' ->
313 continuation_unit (Node (l', r'))));;
315 We use the Continuation monad described above, and insert the
316 `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.
318 So for example, we compute:
320 # tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;
321 - : int list = [2; 3; 5; 7; 11]
323 We have found a way of collapsing a tree into a list of its leaves. Can you trace how this is working? Think first about what the operation `fun a -> fun k -> a :: k a` does when you apply it to a plain `int`, and the continuation `fun _ -> []`. Then given what we've said about `tree_monadize`, what should we expect `tree_monadize (fun a -> fun k -> a :: k a` to do?
325 The Continuation monad is amazingly flexible; we can use it to
326 simulate some of the computations performed above. To see how, first
327 note that an interestingly uninteresting thing happens if we use
328 `continuation_unit` as our first argument to `tree_monadize`, and then
329 apply the result to the identity function:
331 # tree_monadize continuation_unit t1 (fun t -> t);;
333 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
335 That is, nothing happens. But we can begin to substitute more
336 interesting functions for the first argument of `tree_monadize`:
338 (* Simulating the tree reader: distributing a operation over the leaves *)
339 # tree_monadize (fun a -> fun k -> k (square a)) t1 (fun t -> t);;
341 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
343 (* Simulating the int list tree list *)
344 # tree_monadize (fun a -> fun k -> k [a; square a]) t1 (fun t -> t);;
347 (Node (Leaf [2; 4], Leaf [3; 9]),
348 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
350 (* Counting leaves *)
351 # tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;
354 We could simulate the tree state example too, but it would require
355 generalizing the type of the Continuation monad to
357 type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
359 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).
361 Using continuations to solve the same fringe problem
362 ----------------------------------------------------
364 We've seen two solutions to the same fringe problem so far.
365 The simplest is to map each tree to a list of its leaves, then compare
366 the lists. But if the fringes differ in an early position, we've
367 wasted our time visiting the rest of the tree.
369 The second solution was to use tree zippers and mutable state to
370 simulate coroutines. We would unzip the first tree until we found the
371 next leaf, then store the zipper structure in the mutable variable
372 while we turned our attention to the other tree. Because we stop as
373 soon as we find the first mismatched leaf, this solution does not have
374 the flaw just mentioned of the solution that maps both trees to a list
375 of leaves before beginning comparison.
377 Since zippers are just continuations reified, we expect that the
378 solution in terms of zippers can be reworked using continuations, and
379 this is indeed the case. To make this work in the most convenient
380 way, we need to use the fully general type for continuations just mentioned.
382 tree_monadize (fun a k -> a, k a) t1 (fun t -> 0);;
386 The Binary Tree monad
387 ---------------------
389 Of course, by now you may have realized that we have discovered a new
390 monad, the Binary Tree monad. Just as mere lists are in fact a monad,
391 so are trees. Here is the type constructor, unit, and bind:
393 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
394 let tree_unit (a: 'a) : 'a tree = Leaf a;;
395 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
398 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
400 For once, let's check the Monad laws. The left identity law is easy:
402 Left identity: bind (unit a) f = bind (Leaf a) f = f a
404 To check the other two laws, we need to make the following
405 observation: it is easy to prove based on `tree_bind` by a simple
406 induction on the structure of the first argument that the tree
407 resulting from `bind u f` is a tree with the same strucure as `u`,
408 except that each leaf `a` has been replaced with `f a`:
410 \tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
426 Given this equivalence, the right identity law
428 Right identity: bind u unit = u
430 falls out once we realize that
432 bind (Leaf a) unit = unit a = Leaf a
434 As for the associative law,
436 Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
438 we'll give an example that will show how an inductive proof would
439 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
441 \tree (. (. (. (. (a1) (a2)))))
442 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))
447 bind __|__ f = __|_ = . .
449 a1 a2 f a1 f a2 | | | |
452 Now when we bind this tree to `g`, we get
462 At this point, it should be easy to convince yourself that
463 using the recipe on the right hand side of the associative law will
464 built the exact same final tree.
466 So binary trees are a monad.
468 Haskell combines this monad with the Option monad to provide a monad
470 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
471 that is intended to represent non-deterministic computations as a tree.
474 What's this have to do with tree\_mondadize?
475 --------------------------------------------
477 So we've defined a Tree monad:
479 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
480 let tree_unit (a: 'a) : 'a tree = Leaf a;;
481 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
484 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
486 What's this have to do with the `tree_monadize` functions we defined earlier?
488 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
490 | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
491 | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->
492 reader_bind (tree_monadize f r) (fun r' ->
493 reader_unit (Node (l', r'))));;
495 ... and so on for different monads?
497 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]].