tree monads

[lambda.git] / zipper-lists-continuations.mdwn

Today we're going to encounter continuations.  We're going to come at
them from three different directions, and each time we're going to end
up at the same place: a particular monad, which we'll call the
continuation monad.

The three approches are:

[[!toc]]

Rethinking the list monad
-------------------------

To construct a monad, the key element is to settle on a type
constructor, and the monad naturally follows from that.  We'll remind
you of some examples of how monads follow from the type constructor in
a moment.  This will involve some review of familair material, but
it's worth doing for two reasons: it will set up a pattern for the new
discussion further below, and it will tie together some previously
unconnected elements of the course (more specifically, version 3 lists
and monads).

For instance, take the **Reader Monad**.  Once we decide that the type
constructor is

    type 'a reader = env -> 'a

then the choice of unit and bind is natural:

    let r_unit (a : 'a) : 'a reader = fun (e : env) -> a

Since the type of an `'a reader` is `env -> 'a` (by definition),
the type of the `r_unit` function is `'a -> env -> 'a`, which is a
specific case of the type of the *K* combinator.  It makes sense
that *K* is the unit for the reader monad.

Since the type of the `bind` operator is required to be

    r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)

We can reason our way to the correct `bind` function as follows. We start by declaring the type:

    let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =

Now we have to open up the `u` box and get out the `'a` object in order to
feed it to `f`.  Since `u` is a function from environments to
objects of type `'a`, the way we open a box in this monad is
by applying it to an environment:

        ... f (u e) ...

This subexpression types to `'b reader`, which is good.  The only
problem is that we invented an environment `e` that we didn't already have ,
so we have to abstract over that variable to balance the books:

        fun e -> f (u e) ...

This types to `env -> 'b reader`, but we want to end up with `env ->
'b`.  Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment.  So we end up as follows:

    r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
                f (u e) e         

And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does.

[The bind we cite here is a condensed version of the careful `let a = u e in ...`
constructions we provided in earlier lectures.  We use the condensed
version here in order to emphasize similarities of structure across
monads.]

The **State Monad** is similar.  Once we've decided to use the following type constructor:

    type 'a state = store -> ('a, store)

Then our unit is naturally:

    let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)

And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box:

    let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = 
                ... f (...) ...

But unlocking the `u` box is a little more complicated.  As before, we
need to posit a state `s` that we can apply `u` to.  Once we do so,
however, we won't have an `'a`, we'll have a pair whose first element
is an `'a`.  So we have to unpack the pair:

        ... let (a, s') = u s in ... (f a) ...

Abstracting over the `s` and adjusting the types gives the result:

        let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = 
                fun (s : store) -> let (a, s') = u s in f a s'

The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
won't pause to explore it here, though conceptually its unit and bind
follow just as naturally from its type constructor.

Our other familiar monad is the **List Monad**, which we were told
looks like this:

    type 'a list = ['a];;
    l_unit (a : 'a) = [a];;
    l_bind u f = List.concat (List.map f u);;

Recall that `List.map` take a function and a list and returns the
result to applying the function to the elements of the list:

    List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]]

and List.concat takes a list of lists and erases the embdded list
boundaries:

    List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3]

And sure enough, 

    l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]

But where is the reasoning that led us to this unit and bind?
And what is the type `['a]`?  Magic.

So let's indulge ourselves in a completely useless digression and see
if we can gain some insight into the details of the List monad.  Let's
choose type constructor that we can peer into, using some of the
technology we built up so laboriously during the first half of the
course.  We're going to use type 3 lists, partly because we know
they'll give the result we want, but also because they're the coolest.
These were the lists that made lists look like Church numerals with
extra bits embdded in them:

    empty list:                fun f z -> z
    list with one element:     fun f z -> f 1 z
    list with two elements:    fun f z -> f 2 (f 1 z)
    list with three elements:  fun f z -> f 3 (f 2 (f 1 z))

and so on.  To save time, we'll let the OCaml interpreter infer the
principle types of these functions (rather than deducing what the
types should be):

        # fun f z -> z;;
        - : 'a -> 'b -> 'b = <fun>
        # fun f z -> f 1 z;;
        - : (int -> 'a -> 'b) -> 'a -> 'b = <fun>
        # fun f z -> f 2 (f 1 z);;
        - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>
        # fun f z -> f 3 (f 2 (f 1 z))
        - : (int -> 'a -> 'a) -> 'a -> 'a = <fun>

We can see what the consistent, general principle types are at the end, so we
can stop. These types should remind you of the simply-typed lambda calculus
types for Church numerals (`(o -> o) -> o -> o`) with one extra bit thrown in
(in this case, an int).

So here's our type constructor for our hand-rolled lists:

    type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b

Generalizing to lists that contain any kind of element (not just
ints), we have

    type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b

So an `('a, 'b) list'` is a list containing elements of type `'a`,
where `'b` is the type of some part of the plumbing.  This is more
general than an ordinary OCaml list, but we'll see how to map them
into OCaml lists soon.  We don't need to fully grasp the role of the `'b`'s
in order to proceed to build a monad:

    l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z

No problem.  Arriving at bind is a little more complicated, but
exactly the same principles apply, you just have to be careful and
systematic about it.

    l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list'  = ...

Unpacking the types gives:

    l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
            (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd)
            : ('c -> 'd -> 'd) -> 'd -> 'd = ...

But it's a rookie mistake to quail before complicated types. You should
be no more intimiated by complex types than by a linguistic tree with
deeply embedded branches: complex structure created by repeated
application of simple rules.

As usual, we need to unpack the `u` box.  Examine the type of `u`.
This time, `u` will only deliver up its contents if we give `u` as an
argument a function expecting an `'a` and a `'b`. `u` will fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus:

        ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ...

In order for `u` to have the kind of argument it needs, the `... (f a) ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `(f a)`:

        ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ...

Now we have an argument `b` of type `'b`, so we can supply that to `(f a) k`, getting a result of type `'b`, as we need:

        ... u (fun (a : 'a) (b : 'b) -> f a k b) ...

Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it:

        fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b)

This whole expression has type `('c -> 'b -> 'b) -> 'b -> 'b`, which is exactly the type of a `('c, 'b) list'`. So we can hypothesize that we our bind is:

    l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b)
            (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b)
            : ('c -> 'b -> 'b) -> 'b -> 'b = 
      fun k -> u (fun a b -> f a k b)

That is a function of the right type for our bind, but to check whether it works, we have to verify it (with the unit we chose) against the monad laws, and reason whether it will have the right behavior.

Here's a way to persuade yourself that it will have the right behavior. First, it will be handy to eta-expand our `fun k -> u (fun a b -> f a k b)` to:

      fun k z -> u (fun a b -> f a k b) z

Now let's think about what this does. It's a wrapper around `u`. In order to behave as the list which is the result of mapping `f` over each element of `u`, and then joining (`concat`ing) the results, this wrapper would have to accept arguments `k` and `z` and fold them in just the same way that the list which is the result of mapping `f` and then joining the results would fold them. Will it?

Suppose we have a list' whose contents are `[1; 2; 4; 8]`---that is, our list' will be `fun f z -> f 1 (f 2 (f 4 (f 8 z)))`. We call that list' `u`. Suppose we also have a function `f` that for each `int` we give it, gives back a list of the divisors of that `int` that are greater than 1. Intuitively, then, binding `u` to `f` should give us:

        concat (map f u) =
        concat [[]; [2]; [2; 4]; [2; 4; 8]] =
        [2; 2; 4; 2; 4; 8]

Or rather, it should give us a list' version of that, which takes a function `k` and value `z` as arguments, and returns the right fold of `k` and `z` over those elements. What does our formula

      fun k z -> u (fun a b -> f a k b) z
        
do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists:

        []
        [2]
        [2; 4]
        [2; 4; 8]

(or rather, their list' versions). Then it takes the accumulated result `b` of previous steps in the fold, and it folds `k` and `b` over the list generated by `f a`. The result of doing so is passed on to the next step as the accumulated result so far.

So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed:

        0 ==>
        right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==>
        right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==>
        right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==>
        right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0

which indeed is the result of right-folding + and 0 over `[2; 2; 4; 2; 4; 8]`. If you trace through how this works, you should be able to persuade yourself that our formula:

      fun k z -> u (fun a b -> f a k b) z

will deliver just the same folds, for arbitrary choices of `k` and `z` (with the right types), and arbitrary list's `u` and appropriately-typed `f`s, as

        fun k z -> List.fold_right k (concat (map f u)) z

would.

For future reference, we might make two eta-reductions to our formula, so that we have instead:

      let l'_bind = fun k -> u (fun a -> f a k);;

Let's make some more tests:


    l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3]

    l'_bind (fun f z -> f 1 (f 2 z)) 
            (fun i -> fun f z -> f i (f (i+1) z)) ~~> <fun>

Sigh.  OCaml won't show us our own list.  So we have to choose an `f`
and a `z` that will turn our hand-crafted lists into standard OCaml
lists, so that they will print out.

        # let cons h t = h :: t;;  (* OCaml is stupid about :: *)
        # l'_bind (fun f z -> f 1 (f 2 z)) 
                          (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
        - : int list = [1; 2; 2; 3]

Ta da!

To bad this digression, though it ties together various
elements of the course, has *no relevance whatsoever* to the topic of
continuations...

Montague's PTQ treatment of DPs as generalized quantifiers
----------------------------------------------------------

We've hinted that Montague's treatment of DPs as generalized
quantifiers embodies the spirit of continuations (see de Groote 2001,
Barker 2002 for lengthy discussion).  Let's see why.  

First, we'll need a type constructor.  As you probably know, 
Montague replaced individual-denoting determiner phrases (with type `e`)
with generalized quantifiers (with [extensional] type `(e -> t) -> t`.
In particular, the denotation of a proper name like *John*, which
might originally denote a object `j` of type `e`, came to denote a
generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`.
Let's write a general function that will map individuals into their
corresponding generalized quantifier:

   gqize (a : e) = fun (p : e -> t) -> p a

This function wraps up an individual in a fancy box.  That is to say,
we are in the presence of a monad.  The type constructor, the unit and
the bind follow naturally.  We've done this enough times that we won't
belabor the construction of the bind function, the derivation is
similar to the List monad just given:

        type 'a continuation = ('a -> 'b) -> 'b
        c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
        c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
          fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)

How similar is it to the List monad?  Let's examine the type
constructor and the terms from the list monad derived above:

    type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
    l'_unit a = fun f -> f a                 
    l'_bind u f = fun k -> u (fun a -> f a k)

(We performed a sneaky but valid eta reduction in the unit term.)

The unit and the bind for the Montague continuation monad and the
homemade List monad are the same terms!  In other words, the behavior
of the List monad and the behavior of the continuations monad are
parallel in a deep sense.  To emphasize the parallel, we can
instantiate the type of the list' monad using the OCaml list type:

    type 'a c_list = ('a -> 'a list) -> 'a list

Have we really discovered that lists are secretly continuations?  Or
have we merely found a way of simulating lists using list
continuations?  Both perspectives are valid, and we can use our
intuitions about the list monad to understand continuations, and vice
versa (not to mention our intuitions about primitive recursion in
Church numerals too).  The connections will be expecially relevant
when we consider indefinites and Hamblin semantics on the linguistic
side, and non-determinism on the list monad side.

Refunctionalizing zippers
-------------------------

Manipulating trees with monads
------------------------------

This thread develops an idea based on a detailed suggestion of Ken
Shan's.  We'll build a series of functions that operate on trees,
doing various things, including replacing leaves, counting nodes, and
converting a tree to a list of leaves.  The end result will be an
application for continuations.

From an engineering standpoint, we'll build a tree transformer 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 during the remainder of the
course.  Here's a type constructor for binary trees:

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

These are trees in which the internal nodes do not have labels.  [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.,


<pre>
let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
               (Node ((Leaf 5),(Node ((Leaf 7),
                                      (Leaf 11))))))

    .
 ___|___
 |     |
 .     .
_|__  _|__
|  |  |  |
2  3  5  .
        _|__
        |  |
        7  11
</pre>

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

<pre>
let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
  match t with Leaf x -> Leaf (newleaf x)
             | Node (l, r) -> Node ((treemap newleaf l),
                                    (treemap newleaf r));;
</pre>
`treemap` takes 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:

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

    .
 ___|____
 |      |
 .      .
_|__  __|__
|  |  |   |
4  6  10  .
        __|___
        |    |
        14   22
</pre>

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

<pre>
let square x = x * x;;
treemap square t1;;
- : int tree =ppp
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
</pre>

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

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

\tree (. (. (f2) (f3))(. (f5) (.(f7)(f11))))
<pre>
\f    .
  ____|____
  |       |
  .       .
__|__   __|__
|   |   |   |
f2  f3  f5  .
          __|___
          |    |
          f7  f11
</pre>

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

With previous readers, we always knew which kind of environment to
expect: either an assignment function (the original calculator
simulation), a world (the intensionality monad), an integer (the
Jacobson-inspired link monad), etc.  In this situation, it will be
enough for now to expect that our reader will expect a function of
type `int->int`.

<pre>
type 'a reader = (int->int) -> 'a;;  (* mnemonic: e for environment *)
let reader_unit (x:'a): 'a reader = fun _ -> x;;
let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;
</pre>

It's easy to figure out how to turn an `int` into an `int reader`:

<pre>
let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
int2int_reader 2 (fun i -> i + i);;
- : int = 4
</pre>

But what do we do when the integers are scattered over the leaves of a
tree?  A binary tree is not the kind of thing that we can apply a
function of type `int->int` to.

<pre>
let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader =
  match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
             | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
                                reader_bind (treemonadizer f r) (fun y ->
                                  reader_unit (Node (x, y))));;
</pre>

This function says: give me a function `f` that knows how to turn
something of type `'a` into an `'b reader`, and I'll show you how to 
turn an `'a tree` into an `'a tree reader`.  In more fanciful terms, 
the `treemonadizer` function builds plumbing that connects all of the
leaves of a tree into one connected monadic network; it threads the
monad through the leaves.

<pre>
# treemonadizer int2int_reader t1 (fun i -> i + i);;
- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
</pre>

Here, our environment is the doubling function (`fun i -> i + i`).  If
we apply the very same `int tree reader` (namely, `treemonadizer
int2int_reader t1`) to a different `int->int` function---say, the
squaring function, `fun i -> i * i`---we get an entirely different
result:

<pre>
# treemonadizer int2int_reader t1 (fun i -> i * i);;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
</pre>

Now that we have a tree transducer that accepts a 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 nodes in
the tree.

<pre>
type 'a state = int -> 'a * int;;
let state_unit x i = (x, i+.5);;
let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
</pre>

Gratifyingly, we can use the `treemonadizer` function without any
modification whatsoever, except for replacing the (parametric) type
`reader` with `state`:

<pre>
let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
  match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
             | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
                                state_bind (treemonadizer f r) (fun y ->
                                  state_unit (Node (x, y))));;
</pre>

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

<pre>
# treemonadizer state_unit t1 0;;
- : int tree * int =
(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)

    .
 ___|___
 |     |
 .     .
_|__  _|__
|  |  |  |
2  3  5  .
        _|__
        |  |
        7  11
</pre>

Notice that we've counted each internal node twice---it's a good
exercise to adjust the code to count each node once.

One more revealing example before getting down to business: replacing
`state` everywhere in `treemonadizer` with `list` gives us

<pre>
# treemonadizer (fun x -> [ [x; square x] ]) t1;;
- : int list tree list =
[Node
  (Node (Leaf [2; 4], Leaf [3; 9]),
   Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
</pre>

Unlike the previous cases, instead of turning a tree into a function
from some input to a result, this transformer replaces each `int` with
a list of `int`'s.

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

<pre>
type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
let continuation_unit x c = c x;;
let continuation_bind u f c = u (fun a -> f a c);;

let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
  match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
             | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
                                continuation_bind (treemonadizer f r) (fun y ->
                                  continuation_unit (Node (x, y))));;
</pre>

We use the continuation monad described above, and insert the
`continuation` type in the appropriate place in the `treemonadizer` code.
We then compute:

<pre>
# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
- : int list = [2; 3; 5; 7; 11]
</pre>

We have found a way of collapsing a tree into a list of its leaves.

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 the
continuation unit as our first argument to `treemonadizer`, and then
apply the result to the identity function:

<pre>
# treemonadizer continuation_unit t1 (fun x -> x);;
- : int tree =
Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
</pre>

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

<pre>
(* Simulating the tree reader: distributing a operation over the leaves *)
# treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))

(* Simulating the int list tree list *)
# treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
- : int list tree =
Node
 (Node (Leaf [2; 4], Leaf [3; 9]),
  Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))

(* Counting leaves *)
# treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
- : int = 5
</pre>

We could simulate the tree state example too, but it would require
generalizing the type of the continuation monad to 

    type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;;

The tree monad
--------------

Of course, by now you may have realized that we have discovered a new
monad, the tree monad:

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

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

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

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 `fa`:

\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
<pre>
                .                         .            
              __|__                     __|__   
              |   |                     |   |   
              a1  .                    fa1  .   
                 _|__                     __|__ 
                 |  |                     |   | 
                 .  a5                    .  fa5
   bind         _|__       f   =        __|__   
                |  |                    |   |   
                .  a4                   .  fa4  
              __|__                   __|___       
              |   |                   |    |       
              a2  a3                 fa2  fa3         
</pre>   

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 (fa) g)

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

\tree (. (. (. (. (a1)(a2)))))
\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))  ))
<pre>
                                           .            
                                       ____|____        
          .               .            |       |        
bind    __|__   f  =    __|_    =      .       .         
        |   |           |   |        __|__   __|__      
        a1  a2         fa1 fa2       |   |   |   |      
                                     a1  a1  a1  a1  
</pre>

Now when we bind this tree to `g`, we get

<pre>
           .            
       ____|____        
       |       |        
       .       .        
     __|__   __|__      
     |   |   |   |      
    ga1 ga1 ga1 ga1  
</pre>

At this point, it should be easy to convince yourself that
using the recipe on the right hand side of the associative law will
built 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](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
that is intended to 
represent non-deterministic computations as a tree.