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,
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 transformer 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
107 That is, we want to transform the ordinary tree `t1` (of type `int
108 tree`) into a reader object of type `(int -> int) -> int tree`: something
109 that, when you apply it to an `int -> int` function `f` returns an `int
110 tree` in which each leaf `x` has been replaced with `f x`.
112 With previous readers, we always knew which kind of environment to
113 expect: either an assignment function (the original calculator
114 simulation), a world (the intensionality monad), an integer (the
115 Jacobson-inspired link monad), etc. In this situation, it will be
116 enough for now to expect that our reader will expect a function of
119 type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *)
120 let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
121 let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;
123 It's easy to figure out how to turn an `int` into an `int reader`:
125 let int2int_reader : 'a -> 'b reader = fun (a : 'a) -> fun (op : 'a -> 'b) -> op a;;
126 int2int_reader 2 (fun i -> i + i);;
129 But what do we do when the integers are scattered over the leaves of a
130 tree? A binary tree is not the kind of thing that we can apply a
131 function of type `int -> int` to.
133 let rec treemonadizer (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
135 | Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
136 | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
137 reader_bind (treemonadizer f r) (fun y ->
138 reader_unit (Node (x, y))));;
140 This function says: give me a function `f` that knows how to turn
141 something of type `'a` into an `'b reader`, and I'll show you how to
142 turn an `'a tree` into an `'a tree reader`. In more fanciful terms,
143 the `treemonadizer` function builds plumbing that connects all of the
144 leaves of a tree into one connected monadic network; it threads the
145 `'b reader` monad through the leaves.
147 # treemonadizer int2int_reader t1 (fun i -> i + i);;
149 Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
151 Here, our environment is the doubling function (`fun i -> i + i`). If
152 we apply the very same `int tree reader` (namely, `treemonadizer
153 int2int_reader t1`) to a different `int -> int` function---say, the
154 squaring function, `fun i -> i * i`---we get an entirely different
157 # treemonadizer int2int_reader t1 (fun i -> i * i);;
159 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
161 Now that we have a tree transformer that accepts a reader monad as a
162 parameter, we can see what it would take to swap in a different monad.
163 For instance, we can use a state monad to count the number of nodes in
166 type 'a state = int -> 'a * int;;
167 let state_unit a = fun i -> (a, i);;
168 let state_bind u f = fun i -> let (a, i') = u i in f a (i' + 1);;
170 Gratifyingly, we can use the `treemonadizer` function without any
171 modification whatsoever, except for replacing the (parametric) type
172 `'b reader` with `'b state`, and substituting in the appropriate unit and bind:
174 let rec treemonadizer (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
176 | Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
177 | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
178 state_bind (treemonadizer f r) (fun y ->
179 state_unit (Node (x, y))));;
181 Then we can count the number of nodes in the tree:
183 # treemonadizer state_unit t1 0;;
185 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
198 Notice that we've counted each internal node twice---it's a good
199 exercise to adjust the code to count each node once.
202 A tree with n leaves has 2n - 1 nodes.
203 This function will currently return n*1 + (n-1)*2 = 3n - 2.
204 To convert b = 3n - 2 into 2n - 1, we can use: let n = (b + 2)/3 in 2*n -1
206 But I assume Chris means here, adjust the code so that no corrections of this sort have to be applied.
210 One more revealing example before getting down to business: replacing
211 `state` everywhere in `treemonadizer` with `list` gives us
213 # treemonadizer (fun x -> [ [x; square x] ]) t1;;
214 - : int list tree list =
216 (Node (Leaf [2; 4], Leaf [3; 9]),
217 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
219 Unlike the previous cases, instead of turning a tree into a function
220 from some input to a result, this transformer replaces each `int` with
223 Now for the main point. What if we wanted to convert a tree to a list
226 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
227 let continuation_unit x c = c x;;
228 let continuation_bind u f c = u (fun a -> f a c);;
230 let rec treemonadizer (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
232 | Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
233 | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
234 continuation_bind (treemonadizer f r) (fun y ->
235 continuation_unit (Node (x, y))));;
237 We use the continuation monad described above, and insert the
238 `continuation` type in the appropriate place in the `treemonadizer` code.
241 # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
242 - : int list = [2; 3; 5; 7; 11]
244 We have found a way of collapsing a tree into a list of its leaves.
246 The continuation monad is amazingly flexible; we can use it to
247 simulate some of the computations performed above. To see how, first
248 note that an interestingly uninteresting thing happens if we use the
249 continuation unit as our first argument to `treemonadizer`, and then
250 apply the result to the identity function:
252 # treemonadizer continuation_unit t1 (fun x -> x);;
254 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
256 That is, nothing happens. But we can begin to substitute more
257 interesting functions for the first argument of `treemonadizer`:
259 (* Simulating the tree reader: distributing a operation over the leaves *)
260 # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
262 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
264 (* Simulating the int list tree list *)
265 # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
268 (Node (Leaf [2; 4], Leaf [3; 9]),
269 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
271 (* Counting leaves *)
272 # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
275 We could simulate the tree state example too, but it would require
276 generalizing the type of the continuation monad to
278 type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;
280 The binary tree monad
281 ---------------------
283 Of course, by now you may have realized that we have discovered a new
284 monad, the binary tree monad:
286 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
287 let tree_unit (x: 'a) = Leaf x;;
288 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
291 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
293 For once, let's check the Monad laws. The left identity law is easy:
295 Left identity: bind (unit a) f = bind (Leaf a) f = fa
297 To check the other two laws, we need to make the following
298 observation: it is easy to prove based on `tree_bind` by a simple
299 induction on the structure of the first argument that the tree
300 resulting from `bind u f` is a tree with the same strucure as `u`,
301 except that each leaf `a` has been replaced with `fa`:
303 \tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
319 Given this equivalence, the right identity law
321 Right identity: bind u unit = u
323 falls out once we realize that
325 bind (Leaf a) unit = unit a = Leaf a
327 As for the associative law,
329 Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
331 we'll give an example that will show how an inductive proof would
332 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
334 \tree (. (. (. (. (a1)(a2)))))
335 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
340 bind __|__ f = __|_ = . .
342 a1 a2 fa1 fa2 | | | |
345 Now when we bind this tree to `g`, we get
355 At this point, it should be easy to convince yourself that
356 using the recipe on the right hand side of the associative law will
357 built the exact same final tree.
359 So binary trees are a monad.
361 Haskell combines this monad with the Option monad to provide a monad
363 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
364 that is intended to represent non-deterministic computations as a tree.