3 Manipulating trees with monads
4 ------------------------------
6 This topic develops an idea based on a detailed suggestion of Ken
7 Shan's. We'll build a series of functions that operate on trees,
8 doing various things, including replacing leaves, counting nodes, and
9 converting a tree to a list of leaves. The end result will be an
10 application for continuations.
12 From an engineering standpoint, we'll build a tree transformer that
13 deals in monads. We can modify the behavior of the system by swapping
14 one monad for another. We've already seen how adding a monad can add
15 a layer of funtionality without disturbing the underlying system, for
16 instance, in the way that the reader monad allowed us to add a layer
17 of intensionality to an extensional grammar, but we have not yet seen
18 the utility of replacing one monad with other.
20 First, we'll be needing a lot of trees for the remainder of the
21 course. Here again is a type constructor for leaf-labeled, binary trees:
23 type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
25 [How would you adjust the type constructor to allow for labels on the
28 We'll be using trees where the nodes are integers, e.g.,
31 let t1 = Node (Node (Leaf 2, Leaf 3),
32 Node (Leaf 5, Node (Leaf 7,
45 Our first task will be to replace each leaf with its double:
47 let rec tree_map (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree =
49 | Leaf i -> Leaf (leaf_modifier i)
50 | Node (l, r) -> Node (tree_map leaf_modifier l,
51 tree_map leaf_modifier r);;
53 `tree_map` takes a function that transforms old leaves into new leaves,
54 and maps that function over all the leaves in the tree, leaving the
55 structure of the tree unchanged. For instance:
57 let double i = i + i;;
60 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
73 We could have built the doubling operation right into the `tree_map`
74 code. However, because we've left what to do to each leaf as a parameter, we can
75 decide to do something else to the leaves without needing to rewrite
76 `tree_map`. For instance, we can easily square each leaf instead by
77 supplying the appropriate `int -> int` operation in place of `double`:
79 let square i = i * i;;
82 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
84 Note that what `tree_map` does is take some global, contextual
85 information---what to do to each leaf---and supplies that information
86 to each subpart of the computation. In other words, `tree_map` has the
87 behavior of a reader monad. Let's make that explicit.
89 In general, we're on a journey of making our `tree_map` function more and
90 more flexible. So the next step---combining the tree transformer with
91 a reader monad---is to have the `tree_map` function return a (monadized)
92 tree that is ready to accept any `int -> int` function and produce the
95 \tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))
108 That is, we want to transform the ordinary tree `t1` (of type `int
109 tree`) into a reader object of type `(int -> int) -> int tree`: something
110 that, when you apply it to an `int -> int` function `f` returns an `int
111 tree` in which each leaf `i` has been replaced with `f i`.
113 With previous readers, we always knew which kind of environment to
114 expect: either an assignment function (the original calculator
115 simulation), a world (the intensionality monad), an integer (the
116 Jacobson-inspired link monad), etc. In the present case, it will be
117 enough to expect that our "environment" will be some function of type
120 type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *)
121 let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
122 let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;
124 It would be a simple matter to turn an *integer* into an `int reader`:
126 let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;;
127 int_readerize 2 (fun i -> i + i);;
130 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?
131 A tree is not the kind of thing that we can apply a
132 function of type `int -> int` to.
136 let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
138 | Leaf i -> reader_bind (f i) (fun i' -> reader_unit (Leaf i'))
139 | Node (l, r) -> reader_bind (tree_monadize f l) (fun x ->
140 reader_bind (tree_monadize f r) (fun y ->
141 reader_unit (Node (x, y))));;
143 This function says: give me a function `f` that knows how to turn
144 something of type `'a` into an `'b reader`, and I'll show you how to
145 turn an `'a tree` into an `'b tree reader`. In more fanciful terms,
146 the `tree_monadize` function builds plumbing that connects all of the
147 leaves of a tree into one connected monadic network; it threads the
148 `'b reader` monad through the original tree's leaves.
150 # tree_monadize int_readerize t1 double;;
152 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
154 Here, our environment is the doubling function (`fun i -> i + i`). If
155 we apply the very same `int tree reader` (namely, `tree_monadize
156 int_readerize t1`) to a different `int -> int` function---say, the
157 squaring function, `fun i -> i * i`---we get an entirely different
160 # tree_monadize int_readerize t1 square;;
162 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
164 Now that we have a tree transformer that accepts a reader monad as a
165 parameter, we can see what it would take to swap in a different monad.
167 For instance, we can use a state monad to count the number of leaves in
170 type 'a state = int -> 'a * int;;
171 let state_unit a = fun s -> (a, s);;
172 let state_bind u f = fun s -> let (a, s') = u s in f a s';;
174 Gratifyingly, we can use the `tree_monadize` function without any
175 modification whatsoever, except for replacing the (parametric) type
176 `'b reader` with `'b state`, and substituting in the appropriate unit and bind:
178 let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
180 | Leaf i -> state_bind (f i) (fun i' -> state_unit (Leaf i'))
181 | Node (l, r) -> state_bind (tree_monadize f l) (fun x ->
182 state_bind (tree_monadize f r) (fun y ->
183 state_unit (Node (x, y))));;
185 Then we can count the number of leaves in the tree:
187 # tree_monadize (fun a s -> (a, s+1)) t1 0;;
189 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
203 One more revealing example before getting down to business: replacing
204 `state` everywhere in `tree_monadize` with `list` gives us
206 # tree_monadize (fun i -> [ [i; square i] ]) t1;;
207 - : int list tree list =
209 (Node (Leaf [2; 4], Leaf [3; 9]),
210 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
212 Unlike the previous cases, instead of turning a tree into a function
213 from some input to a result, this transformer replaces each `int` with
217 FIXME: We don't make it clear why the fun has to be int -> int list list, instead of int -> int list
221 Now for the main point. What if we wanted to convert a tree to a list
224 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
225 let continuation_unit a = fun k -> k a;;
226 let continuation_bind u f = fun k -> u (fun a -> f a k);;
228 let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
230 | Leaf i -> continuation_bind (f i) (fun i' -> continuation_unit (Leaf i'))
231 | Node (l, r) -> continuation_bind (tree_monadize f l) (fun x ->
232 continuation_bind (tree_monadize f r) (fun y ->
233 continuation_unit (Node (x, y))));;
235 We use the continuation monad described above, and insert the
236 `continuation` type in the appropriate place in the `tree_monadize` code.
239 # tree_monadize (fun a k -> a :: (k a)) t1 (fun t -> []);;
240 - : int list = [2; 3; 5; 7; 11]
242 We have found a way of collapsing a tree into a list of its leaves.
244 The continuation monad is amazingly flexible; we can use it to
245 simulate some of the computations performed above. To see how, first
246 note that an interestingly uninteresting thing happens if we use
247 `continuation_unit` as our first argument to `tree_monadize`, and then
248 apply the result to the identity function:
250 # tree_monadize continuation_unit t1 (fun i -> i);;
252 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
254 That is, nothing happens. But we can begin to substitute more
255 interesting functions for the first argument of `tree_monadize`:
257 (* Simulating the tree reader: distributing a operation over the leaves *)
258 # tree_monadize (fun a k -> k (square a)) t1 (fun i -> i);;
260 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
262 (* Simulating the int list tree list *)
263 # tree_monadize (fun a k -> k [a; square a]) t1 (fun i -> i);;
266 (Node (Leaf [2; 4], Leaf [3; 9]),
267 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
269 (* Counting leaves *)
270 # tree_monadize (fun a k -> 1 + k a) t1 (fun i -> 0);;
273 We could simulate the tree state example too, but it would require
274 generalizing the type of the continuation monad to
276 type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
278 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).
281 The binary tree monad
282 ---------------------
284 Of course, by now you may have realized that we have discovered a new
285 monad, the binary tree monad:
287 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
288 let tree_unit (a: 'a) = Leaf a;;
289 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
292 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
294 For once, let's check the Monad laws. The left identity law is easy:
296 Left identity: bind (unit a) f = bind (Leaf a) f = f a
298 To check the other two laws, we need to make the following
299 observation: it is easy to prove based on `tree_bind` by a simple
300 induction on the structure of the first argument that the tree
301 resulting from `bind u f` is a tree with the same strucure as `u`,
302 except that each leaf `a` has been replaced with `f a`:
304 \tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
320 Given this equivalence, the right identity law
322 Right identity: bind u unit = u
324 falls out once we realize that
326 bind (Leaf a) unit = unit a = Leaf a
328 As for the associative law,
330 Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
332 we'll give an example that will show how an inductive proof would
333 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
335 \tree (. (. (. (. (a1) (a2)))))
336 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))
341 bind __|__ f = __|_ = . .
343 a1 a2 f a1 f a2 | | | |
346 Now when we bind this tree to `g`, we get
356 At this point, it should be easy to convince yourself that
357 using the recipe on the right hand side of the associative law will
358 built the exact same final tree.
360 So binary trees are a monad.
362 Haskell combines this monad with the Option monad to provide a monad
364 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
365 that is intended to represent non-deterministic computations as a tree.