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 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?
323 The Continuation monad is amazingly flexible; we can use it to
324 simulate some of the computations performed above. To see how, first
325 note that an interestingly uninteresting thing happens if we use
326 `continuation_unit` as our first argument to `tree_monadize`, and then
327 apply the result to the identity function:
329 # tree_monadize continuation_unit t1 (fun t -> t);;
331 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
333 That is, nothing happens. But we can begin to substitute more
334 interesting functions for the first argument of `tree_monadize`:
336 (* Simulating the tree reader: distributing a operation over the leaves *)
337 # tree_monadize (fun a -> fun k -> k (square a)) t1 (fun t -> t);;
339 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
341 (* Simulating the int list tree list *)
342 # tree_monadize (fun a -> fun k -> k [a; square a]) t1 (fun t -> t);;
345 (Node (Leaf [2; 4], Leaf [3; 9]),
346 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
348 (* Counting leaves *)
349 # tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;
352 We could simulate the tree state example too, but it would require
353 generalizing the type of the Continuation monad to
355 type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
357 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).
359 Using continuations to solve the same fringe problem
360 ----------------------------------------------------
362 We've seen two solutions to the same fringe problem so far.
363 The simplest is to map each tree to a list of its leaves, then compare
364 the lists. But if the fringes differ in an early position, we've
365 wasted our time visiting the rest of the tree.
367 The second solution was to use tree zippers and mutable state to
368 simulate coroutines. We would unzip the first tree until we found the
369 next leaf, then store the zipper structure in the mutable variable
370 while we turned our attention to the other tree. Because we stop as
371 soon as we find the first mismatched leaf, this solution does not have
372 the flaw just mentioned of the solution that maps both trees to a list
373 of leaves before beginning comparison.
375 Since zippers are just continuations reified, we expect that the
376 solution in terms of zippers can be reworked using continuations, and
377 this is indeed the case. To make this work in the most convenient
378 way, we need to use the fully general type for continuations just mentioned.
380 tree_monadize (fun a k -> a, k a) t1 (fun t -> 0);;
384 The Binary Tree monad
385 ---------------------
387 Of course, by now you may have realized that we have discovered a new
388 monad, the Binary Tree monad. Just as mere lists are in fact a monad,
389 so are trees. Here is the type constructor, unit, and bind:
391 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
392 let tree_unit (a: 'a) : 'a tree = Leaf a;;
393 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
396 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
398 For once, let's check the Monad laws. The left identity law is easy:
400 Left identity: bind (unit a) f = bind (Leaf a) f = f a
402 To check the other two laws, we need to make the following
403 observation: it is easy to prove based on `tree_bind` by a simple
404 induction on the structure of the first argument that the tree
405 resulting from `bind u f` is a tree with the same strucure as `u`,
406 except that each leaf `a` has been replaced with `f a`:
422 Given this equivalence, the right identity law
424 Right identity: bind u unit = u
426 falls out once we realize that
428 bind (Leaf a) unit = unit a = Leaf a
430 As for the associative law,
432 Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
434 we'll give an example that will show how an inductive proof would
435 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
440 bind __|__ f = __|_ = . .
442 a1 a2 f a1 f a2 | | | |
445 Now when we bind this tree to `g`, we get
455 At this point, it should be easy to convince yourself that
456 using the recipe on the right hand side of the associative law will
457 built the exact same final tree.
459 So binary trees are a monad.
461 Haskell combines this monad with the Option monad to provide a monad
463 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
464 that is intended to represent non-deterministic computations as a tree.
467 What's this have to do with tree\_mondadize?
468 --------------------------------------------
470 So we've defined a Tree monad:
472 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
473 let tree_unit (a: 'a) : 'a tree = Leaf a;;
474 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
477 | Node (l, r) -> Node (tree_bind l f, tree_bind r f);;
479 What's this have to do with the `tree_monadize` functions we defined earlier?
481 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
483 | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
484 | Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->
485 reader_bind (tree_monadize f r) (fun r' ->
486 reader_unit (Node (l', r'))));;
488 ... and so on for different monads?
490 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]].