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 treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
49 | Leaf x -> Leaf (newleaf x)
50 | Node (l, r) -> Node ((treemap newleaf l),
51 (treemap newleaf r));;
53 `treemap` 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 `treemap`
74 code. However, because what to do to each leaf is a parameter, we can
75 decide to do something else to the leaves without needing to rewrite
76 `treemap`. For instance, we can easily square each leaf instead by
77 supplying the appropriate `int -> int` operation in place of `double`:
79 let square x = x * x;;
82 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
84 Note that what `treemap` 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, `treemap` has the
87 behavior of a reader monad. Let's make that explicit.
89 In general, we're on a journey of making our treemap function more and
90 more flexible. So the next step---combining the tree transducer with
91 a reader monad---is to have the treemap function return a (monadized)
92 tree that is ready to accept any `int->int` function and produce the
95 \tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))
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 returns an `int
111 tree` in which each leaf `x` has been replaced with `(f x)`.
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 this situation, it will be
117 enough for now to expect that our reader will expect a function of
120 type 'a reader = (int->int) -> 'a;; (* mnemonic: e for environment *)
121 let reader_unit (x:'a): 'a reader = fun _ -> x;;
122 let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;
124 It's easy to figure out how to turn an `int` into an `int reader`:
126 let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
127 int2int_reader 2 (fun i -> i + i);;
130 But what do we do when the integers are scattered over the leaves of a
131 tree? A binary tree is not the kind of thing that we can apply a
132 function of type `int->int` to.
134 let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
136 | Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
137 | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
138 reader_bind (treemonadizer f r) (fun y ->
139 reader_unit (Node (x, y))));;
141 This function says: give me a function `f` that knows how to turn
142 something of type `'a` into an `'b reader`, and I'll show you how to
143 turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
144 the `treemonadizer` function builds plumbing that connects all of the
145 leaves of a tree into one connected monadic network; it threads the
146 monad through the leaves.
148 # treemonadizer int2int_reader t1 (fun i -> i + i);;
150 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
152 Here, our environment is the doubling function (`fun i -> i + i`). If
153 we apply the very same `int tree reader` (namely, `treemonadizer
154 int2int_reader t1`) to a different `int->int` function---say, the
155 squaring function, `fun i -> i * i`---we get an entirely different
158 # treemonadizer int2int_reader t1 (fun i -> i * i);;
160 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
162 Now that we have a tree transducer that accepts a monad as a
163 parameter, we can see what it would take to swap in a different monad.
164 For instance, we can use a state monad to count the number of nodes in
167 type 'a state = int -> 'a * int;;
168 let state_unit x i = (x, i+.5);;
169 let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
171 Gratifyingly, we can use the `treemonadizer` function without any
172 modification whatsoever, except for replacing the (parametric) type
173 `reader` with `state`:
175 let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
177 | Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
178 | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
179 state_bind (treemonadizer f r) (fun y ->
180 state_unit (Node (x, y))));;
182 Then we can count the number of nodes in the tree:
184 # treemonadizer state_unit t1 0;;
186 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
199 Notice that we've counted each internal node twice---it's a good
200 exercise to adjust the code to count each node once.
202 One more revealing example before getting down to business: replacing
203 `state` everywhere in `treemonadizer` with `list` gives us
205 # treemonadizer (fun x -> [ [x; square x] ]) t1;;
206 - : int list tree list =
208 (Node (Leaf [2; 4], Leaf [3; 9]),
209 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
211 Unlike the previous cases, instead of turning a tree into a function
212 from some input to a result, this transformer replaces each `int` with
215 Now for the main point. What if we wanted to convert a tree to a list
218 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
219 let continuation_unit x c = c x;;
220 let continuation_bind u f c = u (fun a -> f a c);;
222 let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
224 | Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
225 | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
226 continuation_bind (treemonadizer f r) (fun y ->
227 continuation_unit (Node (x, y))));;
229 We use the continuation monad described above, and insert the
230 `continuation` type in the appropriate place in the `treemonadizer` code.
233 # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
234 - : int list = [2; 3; 5; 7; 11]
236 We have found a way of collapsing a tree into a list of its leaves.
238 The continuation monad is amazingly flexible; we can use it to
239 simulate some of the computations performed above. To see how, first
240 note that an interestingly uninteresting thing happens if we use the
241 continuation unit as our first argument to `treemonadizer`, and then
242 apply the result to the identity function:
244 # treemonadizer continuation_unit t1 (fun x -> x);;
246 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
248 That is, nothing happens. But we can begin to substitute more
249 interesting functions for the first argument of `treemonadizer`:
251 (* Simulating the tree reader: distributing a operation over the leaves *)
252 # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
254 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
256 (* Simulating the int list tree list *)
257 # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
260 (Node (Leaf [2; 4], Leaf [3; 9]),
261 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
263 (* Counting leaves *)
264 # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
267 We could simulate the tree state example too, but it would require
268 generalizing the type of the continuation monad to
270 type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
272 The binary tree monad
273 ---------------------
275 Of course, by now you may have realized that we have discovered a new
276 monad, the binary tree monad:
278 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
279 let tree_unit (x:'a) = Leaf x;;
280 let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
283 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
285 For once, let's check the Monad laws. The left identity law is easy:
287 Left identity: bind (unit a) f = bind (Leaf a) f = fa
289 To check the other two laws, we need to make the following
290 observation: it is easy to prove based on `tree_bind` by a simple
291 induction on the structure of the first argument that the tree
292 resulting from `bind u f` is a tree with the same strucure as `u`,
293 except that each leaf `a` has been replaced with `fa`:
295 \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
311 Given this equivalence, the right identity law
313 Right identity: bind u unit = u
315 falls out once we realize that
317 bind (Leaf a) unit = unit a = Leaf a
319 As for the associative law,
321 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
323 we'll give an example that will show how an inductive proof would
324 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
326 \tree (. (. (. (. (a1)(a2)))))
327 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
332 bind __|__ f = __|_ = . .
334 a1 a2 fa1 fa2 | | | |
337 Now when we bind this tree to `g`, we get
347 At this point, it should be easy to convince yourself that
348 using the recipe on the right hand side of the associative law will
349 built the exact same final tree.
351 So binary trees are a monad.
353 Haskell combines this monad with the Option monad to provide a monad
355 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
356 that is intended to represent non-deterministic computations as a tree.