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.
169 For instance, we can use a state monad to count the number of nodes in
172 type 'a state = int -> 'a * int;;
173 let state_unit a = fun s -> (a, s);;
174 let state_bind_and_count u f = fun s -> let (a, s') = u s in f a (s' + 1);;
176 Gratifyingly, we can use the `tree_monadize` function without any
177 modification whatsoever, except for replacing the (parametric) type
178 `'b reader` with `'b state`, and substituting in the appropriate unit and bind:
180 let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
182 | Leaf i -> state_bind_and_count (f i) (fun i' -> state_unit (Leaf i'))
183 | Node (l, r) -> state_bind_and_count (tree_monadize f l) (fun x ->
184 state_bind_and_count (tree_monadize f r) (fun y ->
185 state_unit (Node (x, y))));;
187 Then we can count the number of nodes in the tree:
189 # tree_monadize state_unit t1 0;;
191 (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
204 Notice that we've counted each internal node twice---it's a good
205 exercise to adjust the code to count each node once.
208 A tree with n leaves has 2n - 1 nodes.
209 This function will currently return n*1 + (n-1)*2 = 3n - 2.
210 To convert b = 3n - 2 into 2n - 1, we can use: let n = (b + 2)/3 in 2*n -1
212 But I assume Chris means here, adjust the code so that no corrections of this sort have to be applied.
216 One more revealing example before getting down to business: replacing
217 `state` everywhere in `tree_monadize` with `list` gives us
219 # tree_monadize (fun i -> [ [i; square i] ]) t1;;
220 - : int list tree list =
222 (Node (Leaf [2; 4], Leaf [3; 9]),
223 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
225 Unlike the previous cases, instead of turning a tree into a function
226 from some input to a result, this transformer replaces each `int` with
230 We don't make it clear why the fun has to be int -> int list list, instead of int -> int list
234 Now for the main point. What if we wanted to convert a tree to a list
237 type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
238 let continuation_unit a = fun k -> k a;;
239 let continuation_bind u f = fun k -> u (fun a -> f a k);;
241 let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
243 | Leaf i -> continuation_bind (f i) (fun i' -> continuation_unit (Leaf i'))
244 | Node (l, r) -> continuation_bind (tree_monadize f l) (fun x ->
245 continuation_bind (tree_monadize f r) (fun y ->
246 continuation_unit (Node (x, y))));;
248 We use the continuation monad described above, and insert the
249 `continuation` type in the appropriate place in the `tree_monadize` code.
252 # tree_monadize (fun a k -> a :: (k a)) t1 (fun t -> []);;
253 - : int list = [2; 3; 5; 7; 11]
255 <!-- FIXME: what if we had fun t -> [-t]? why `t`? -->
257 We have found a way of collapsing a tree into a list of its leaves.
259 The continuation monad is amazingly flexible; we can use it to
260 simulate some of the computations performed above. To see how, first
261 note that an interestingly uninteresting thing happens if we use
262 `continuation_unit` as our first argument to `tree_monadize`, and then
263 apply the result to the identity function:
265 # tree_monadize continuation_unit t1 (fun i -> i);;
267 Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
269 That is, nothing happens. But we can begin to substitute more
270 interesting functions for the first argument of `tree_monadize`:
272 (* Simulating the tree reader: distributing a operation over the leaves *)
273 # tree_monadize (fun a k -> k (square a)) t1 (fun i -> i);;
275 Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
277 (* Simulating the int list tree list *)
278 # tree_monadize (fun a k -> k [a; square a]) t1 (fun i -> i);;
281 (Node (Leaf [2; 4], Leaf [3; 9]),
282 Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
284 (* Counting leaves *)
285 # tree_monadize (fun a k -> 1 + k a) t1 (fun i -> 0);;
288 We could simulate the tree state example too, but it would require
289 generalizing the type of the continuation monad to
291 type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
293 The binary tree monad
294 ---------------------
296 Of course, by now you may have realized that we have discovered a new
297 monad, the binary tree monad:
299 type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
300 let tree_unit (a: 'a) = Leaf a;;
301 let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
304 | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
306 For once, let's check the Monad laws. The left identity law is easy:
308 Left identity: bind (unit a) f = bind (Leaf a) f = f a
310 To check the other two laws, we need to make the following
311 observation: it is easy to prove based on `tree_bind` by a simple
312 induction on the structure of the first argument that the tree
313 resulting from `bind u f` is a tree with the same strucure as `u`,
314 except that each leaf `a` has been replaced with `f a`:
316 \tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
332 Given this equivalence, the right identity law
334 Right identity: bind u unit = u
336 falls out once we realize that
338 bind (Leaf a) unit = unit a = Leaf a
340 As for the associative law,
342 Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)
344 we'll give an example that will show how an inductive proof would
345 proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
347 \tree (. (. (. (. (a1) (a2)))))
348 \tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))
353 bind __|__ f = __|_ = . .
355 a1 a2 f a1 f a2 | | | |
358 Now when we bind this tree to `g`, we get
368 At this point, it should be easy to convince yourself that
369 using the recipe on the right hand side of the associative law will
370 built the exact same final tree.
372 So binary trees are a monad.
374 Haskell combines this monad with the Option monad to provide a monad
376 [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
377 that is intended to represent non-deterministic computations as a tree.