This topic develops an idea based on a suggestion of Ken Shan's. We'll build a series of functions that operate on trees, doing various things, including updating leaves with a Reader monad, counting nodes with a State monad, copying the tree with a List monad, and converting a tree into a list of leaves with a Continuation monad. It will turn out that the continuation monad can simulate the behavior of each of the other monads.

From an engineering standpoint, we'll build a tree machine that deals in monads. We can modify the behavior of the system by swapping one monad for another. We've already seen how adding a monad can add a layer of funtionality without disturbing the underlying system, for instance, in the way that the Reader monad allowed us to add a layer of intensionality to an extensional grammar. But we have not yet seen the utility of replacing one monad with other.

First, we'll be needing a lot of trees for the remainder of the course. Here again is a type constructor for leaf-labeled, binary trees:

type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree);;

[How would you adjust the type constructor to allow for labels on the internal nodes?]

We'll be using trees where the nodes are integers, e.g.,

let t1 = Node (Node (Leaf 2, Leaf 3),
Node (Leaf 5, Node (Leaf 7,
Leaf 11)))
.
___|___
|     |
.     .
_|_   _|__
|  |  |  |
2  3  5  .
_|__
|  |
7  11

Our first task will be to replace each leaf with its double:

let rec tree_map (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree =
match t with
| Leaf i -> Leaf (leaf_modifier i)
| Node (l, r) -> Node (tree_map leaf_modifier l,
tree_map leaf_modifier r);;

tree_map takes a tree and a function that transforms old leaves into new leaves, and maps that function over all the leaves in the tree, leaving the structure of the tree unchanged. For instance:

let double i = i + i;;
tree_map double t1;;
- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))

.
___|____
|      |
.      .
_|__  __|__
|  |  |   |
4  6  10  .
__|___
|    |
14   22

We could have built the doubling operation right into the tree_map code. However, because we've made what to do to each leaf a parameter, we can decide to do something else to the leaves without needing to rewrite tree_map. For instance, we can easily square each leaf instead, by supplying the appropriate int -> int operation in place of double:

let square i = i * i;;
tree_map square t1;;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))

Note that what tree_map does is take some unchanging contextual information---what to do to each leaf---and supplies that information to each subpart of the computation. In other words, tree_map has the behavior of a Reader monad. Let's make that explicit.

In general, we're on a journey of making our tree_map function more and more flexible. So the next step---combining the tree transformer with a Reader monad---is to have the tree_map function return a (monadized) tree that is ready to accept any int -> int function and produce the updated tree.

fun e ->    .
_____|____
|        |
.        .
__|___   __|___
|    |   |    |
e 2  e 3  e 5  .
__|___
|    |
e 7  e 11

That is, we want to transform the ordinary tree t1 (of type int tree) into a reader monadic object of type (int -> int) -> int tree: something that, when you apply it to an int -> int function e returns an int tree in which each leaf i has been replaced with e i.

[Application note: this kind of reader object could provide a model for Kaplan's characters. It turns an ordinary tree into one that expects contextual information (here, the e) that can be used to compute the content of indexicals embedded arbitrarily deeply in the tree.]

With our previous applications of the Reader monad, we always knew which kind of environment to expect: either an assignment function, as in the original calculator simulation; a world, as in the intensionality monad; an individual, as in the Jacobson-inspired link monad; etc. In the present case, we expect that our "environment" will be some function of type int -> int. "Looking up" some int in the environment will return us the int that comes out the other side of that function.

type 'a reader = (int -> int) -> 'a;;
let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
fun e -> f (u e) e;;

It would be a simple matter to turn an integer into an int reader:

fun (a : int) ->
fun (modifier : int -> int) -> modifier a;;
asker 2 (fun i -> i + i);;
- : int = 4

asker a is a monadic box that waits for an an environment (here, the argument modifier) and returns what that environment maps a to.

How do we do the analagous transformation when our ints are scattered over the leaves of a tree? How do we turn an int tree into a reader? A tree is not the kind of thing that we can apply a function of type int -> int to.

But we can do this:

let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
match t with
| Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b))
| Node (l, r) -> reader_bind (tree_monadize f l) (fun l' ->

This function says: give me a function f that knows how to turn something of type 'a into an 'b reader---this is a function of the same type that you could bind an 'a reader to, such as asker or reader_unit---and I'll show you how to turn an 'a tree into an 'b tree reader. That is, if you show me how to do this:

------------
1     --->  |    1     |
------------

then I'll give you back the ability to do this:

____________
.           |    .     |
__|___  --->  |  __|___  |
|    |        |  |    |  |
1    2        |  1    2  |
------------

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:

------------
1     --->  |    1     |  applied to e  ~~>  2
------------

Then we can expect that supplying it to our int tree reader will double all the leaves:

____________
.           |    .     |                      .
__|___  --->  |  __|___  | applied to e  ~~>  __|___
|    |        |  |    |  |                    |    |
1    2        |  1    2  |                    2    4
------------

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 'b reader monad through the original tree's leaves.

- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))

Here, our environment is the doubling function (fun i -> i + i). If we apply the very same int tree reader (namely, tree_monadize asker t1) to a different int -> int function---say, the squaring function, fun i -> i * i---we get an entirely different result:

- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))

Now that we have a tree transformer that accepts a reader monad as a parameter, we can see what it would take to swap in a different monad.

For instance, we can use a State monad to count the number of leaves in the tree.

type 'a state = int -> 'a * int;;
let state_unit a = fun s -> (a, s);;
let state_bind u f = fun s -> let (a, s') = u s in f a s';;

Gratifyingly, we can use the tree_monadize function without any modification whatsoever, except for replacing the (parametric) type 'b reader with 'b state, and substituting in the appropriate unit and bind:

let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
match t with
| Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
| Node (l, r) -> state_bind (tree_monadize f l) (fun l' ->
state_bind (tree_monadize f r) (fun r' ->
state_unit (Node (l', r'))));;

Then we can count the number of leaves in the tree:

# let incrementer = fun a ->
fun s -> (a, s+1);;

- : int tree * int =
(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)

.
___|___
|     |
.     .
(  _|__  _|__     ,   5 )
|  |  |  |
2  3  5  .
_|__
|  |
7  11

Note that the value returned is a pair consisting of a tree and an integer, 5, which represents the count of the leaves in the tree.

Why does this work? Because the operation incrementer takes an argument a and wraps it in an State monadic box that increments the store and leaves behind a wrapped a. When we give that same operations to our tree_monadize function, it then wraps an int tree in a box, one that does the same store-incrementing for each of its leaves.

We can use the state monad to annotate leaves with a number corresponding to that leave's ordinal position. When we do so, we reveal the order in which the monadic tree forces evaluation:

# tree_monadize (fun a -> fun s -> ((a,s+1), s+1)) t1 0;;
- : int tree * int =
(Node
(Node (Leaf (2, 1), Leaf (3, 2)),
Node
(Leaf (5, 3),
Node (Leaf (7, 4), Leaf (11, 5)))),
5)

The key thing to notice is that instead of just wrapping a in the monadic box, we wrap a pair of a and the current store.

Reversing the annotation order requires reversing the order of the state_bind operations. It's not obvious that this will type correctly, so think it through:

let rec tree_monadize_rev (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
match t with
| Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b))
| Node (l, r) -> state_bind (tree_monadize f r) (fun r' ->     (* R first *)
state_bind (tree_monadize f l) (fun l'->    (* Then L  *)
state_unit (Node (l', r'))));;

# tree_monadize_rev (fun a -> fun s -> ((a,s+1), s+1)) t1 0;;
- : int tree * int =
(Node
(Node (Leaf (2, 5), Leaf (3, 4)),
Node
(Leaf (5, 3),
Node (Leaf (7, 2), Leaf (11, 1)))),
5)

Later, we will talk more about controlling the order in which nodes are visited.

One more revealing example before getting down to business: replacing state everywhere in tree_monadize with list lets us do:

# let decider i = if i = 2 then [20; 21] else [i];;
- : int tree List_monad.m =
[
Node (Node (Leaf 20, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)));
Node (Node (Leaf 21, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
]

Unlike the previous cases, instead of turning a tree into a function from some input to a result, this monadized tree gives us back a list of trees, one for each choice of ints for its leaves.

Now for the main point. What if we wanted to convert a tree to a list of leaves?

type ('r,'a) continuation = ('a -> 'r) -> 'r;;
let continuation_unit a = fun k -> k a;;
let continuation_bind u f = fun k -> u (fun a -> f a k);;

let rec tree_monadize (f : 'a -> ('r,'b) continuation) (t : 'a tree) : ('r,'b tree) continuation =
match t with
| Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b))
| Node (l, r) -> continuation_bind (tree_monadize f l) (fun l' ->
continuation_bind (tree_monadize f r) (fun r' ->
continuation_unit (Node (l', r'))));;

We use the Continuation monad described above, and insert the continuation type in the appropriate place in the tree_monadize code. Then if we give the tree_monadize function an operation that converts ints into 'b-wrapping Continuation monads, it will give us back a way to turn int trees into corresponding 'b tree-wrapping Continuation monads.

So for example, we compute:

# tree_monadize (fun a k -> a :: k ()) t1 (fun _ -> []);;
- : int list = [2; 3; 5; 7; 11]

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 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?

Soon we'll return to the same-fringe problem. Since the simple but inefficient way to solve it is to map each tree to a list of its leaves, this transformation is on the path to a more efficient solution. We'll just have to figure out how to postpone computing the tail of the list until it's needed...

The Continuation monad is amazingly flexible; we can use it to simulate some of the computations performed above. To see how, first note that an interestingly uninteresting thing happens if we use continuation_unit as our first argument to tree_monadize, and then apply the result to the identity function:

# tree_monadize continuation_unit t1 (fun t -> t);;
- : int tree =
Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))

That is, nothing happens. But we can begin to substitute more interesting functions for the first argument of tree_monadize:

(* Simulating the tree reader: distributing a operation over the leaves *)
# tree_monadize (fun a -> fun k -> k (square a)) t1 (fun t -> t);;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))

(* Counting leaves *)
# tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;
- : int = 5

It's not immediately obvious to us how to simulate the List monadization of the tree using this technique.

We could simulate the tree annotating example by setting the relevant type to (store -> 'result, 'a) continuation.

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.

Of course, by now you may have realized that we are working with a new monad, the binary, leaf-labeled Tree monad. Just as mere lists are in fact a monad, so are trees. Here is the type constructor, unit, and bind:

type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
let tree_unit (a: 'a) : 'a tree = Leaf a;;
let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
match u with
| Leaf a -> f a
| Node (l, r) -> Node (tree_bind l f, tree_bind r f);;

For once, let's check the Monad laws. The left identity law is easy:

Left identity: bind (unit a) f = bind (Leaf a) f = f a

To check the other two laws, we need to make the following observation: it is easy to prove based on tree_bind by a simple induction on the structure of the first argument that the tree resulting from bind u f is a tree with the same strucure as u, except that each leaf a has been replaced with the tree returned by f a:

.                         .
__|__                     __|__
|   |                    /\   |
a1  .                   f a1  .
_|__                     __|__
|  |                     |   /\
.  a5                    .  f a5
bind         _|__       f   =        __|__
|  |                    |   /\
.  a4                   .  f a4
__|__                   __|___
|   |                  /\    /\
a2  a3                f a2  f a3

Given this equivalence, the right identity law

Right identity: bind u unit = u

falls out once we realize that

bind (Leaf a) unit = unit a = Leaf a

As for the associative law,

Associativity: bind (bind u f) g = bind u (\a. bind (f a) g)

we'll give an example that will show how an inductive proof would proceed. Let f a = Node (Leaf a, Leaf a). Then

.
____|____
.               .            |       |
bind    __|__   f  =    __|_    =      .       .
|   |           |   |        __|__   __|__
a1  a2        f a1 f a2      |   |   |   |
a1  a1  a1  a1

Now when we bind this tree to g, we get

.
_____|______
|          |
.          .
__|__      __|__
|   |      |   |
g a1 g a1  g a1 g a1

At this point, it should be easy to convince yourself that using the recipe on the right hand side of the associative law will build the exact same final tree.

So binary trees are a monad.

Haskell combines this monad with the Option monad to provide a monad called a SearchTree that is intended to represent non-deterministic computations as a tree.

## What's this have to do with tree_monadize?

Our different implementations of tree_monadize above were different layerings of the Tree monad with other monads (Reader, State, List, and Continuation). We'll explore that further here: ?Monad Transformers.