[[!toc]]
Manipulating trees with monads
------------------------------
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, replacing leaves 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 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 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 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.
\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))
\f .
_____|____
| |
. .
__|___ __|___
| | | |
f 2 f 3 f 5 .
__|___
| |
f 7 f 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
`f` returns an `int tree` in which each leaf `i` has been replaced
with `f 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 `λ f`) 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;; (* mnemonic: e for environment *)
let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;;
It would be a simple matter to turn an *integer* into an `int reader`:
let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;;
int_readerize 2 (fun i -> i + i);;
- : int = 4
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?
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' ->
reader_bind (tree_monadize f r) (fun r' ->
reader_unit (Node (l', r'))));;
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---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.
# tree_monadize int_readerize t1 double;;
- : 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
int_readerize t1`) to a different `int -> int` function---say, the
squaring function, `fun i -> i * i`---we get an entirely different
result:
# tree_monadize int_readerize t1 square;;
- : 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:
# tree_monadize (fun a -> fun s -> (a, s+1)) t1 0;;
- : 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 `fun a -> fun s -> (a, s+1)`
takes an `int` and wraps it in an `int state` monadic box that
increments the state. 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 state-incrementing for each of its leaves.
We can use the state monad to replace 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 -> (s+1, s+1)) t1 0;;
- : int tree * int =
(Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5)
The key thing to notice is that instead of copying `a` into the
monadic box, we throw away the `a` and put a copy of the state in
instead.
Reversing the 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' ->
state_bind (tree_monadize f l) (fun l' ->
state_unit (Node (l', r'))));;
# tree_monadize_rev (fun a -> fun s -> (s+1, s+1)) t1 0;;
- : int tree * int =
(Node (Node (Leaf 5, Leaf 4), Node (Leaf 3, Node (Leaf 2, Leaf 1))), 5)
We will need below to depend on controlling the order in which nodes
are visited when we use the continuation monad to solve the
same-fringe problem.
One more revealing example before getting down to business: replacing
`state` everywhere in `tree_monadize` with `list` gives us
# tree_monadize (fun i -> [ [i; square i] ]) t1;;
- : int list tree list =
[Node
(Node (Leaf [2; 4], Leaf [3; 9]),
Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
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. We might also have done this with a Reader monad, though then our environments would need to be of type `int -> int list`. Experiment with what happens if you supply the `tree_monadize` based on the List monad an operation like `fun -> [ i; [2*i; 3*i] ]`. Use small trees for your experiment.
[Why is the argument to `tree_monadize` `int -> int list list` instead
of `int -> int list`? Well, as usual, the List monad bind operation
will erase the outer list box, so if we want to replace the leaves
with lists, we have to nest the replacement lists inside a disposable
box.]
Now for the main point. What if we wanted to convert a tree to a list
of leaves?
type ('a, 'r) 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 -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) 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 `int`s into `'b`-wrapping Continuation monads, it will give us back a way to turn `int tree`s into corresponding `'b tree`-wrapping Continuation monads.
So for example, we compute:
# tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;
- : 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 -> fun 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?
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)))
(* Simulating the int list tree list *)
# tree_monadize (fun a -> fun k -> k [a; square a]) t1 (fun t -> t);;
- : int list tree =
Node
(Node (Leaf [2; 4], Leaf [3; 9]),
Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
(* Counting leaves *)
# tree_monadize (fun a -> fun k -> 1 + k a) t1 (fun t -> 0);;
- : int = 5
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;;
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](/code/tree_monadize.ml).
Using continuations to solve the same fringe problem
----------------------------------------------------
We've seen two solutions to the same fringe problem so far.
The simplest is to map each tree to a list of its leaves, then compare
the lists. But if the fringes differ in an early position, we've
wasted our time visiting the rest of the tree.
The second solution was to use tree zippers and mutable state to
simulate coroutines. We would unzip the first tree until we found the
next leaf, then store the zipper structure in the mutable variable
while we turned our attention to the other tree. Because we stop as
soon as we find the first mismatched leaf, this solution does not have
the flaw just mentioned of the solution that maps both trees to a list
of leaves before beginning comparison.
Since zippers are just continuations reified, we expect that the
solution in terms of zippers can be reworked using continuations, and
this is indeed the case. To make this work in the most convenient
way, we need to use the fully general type for continuations just mentioned.
tree_monadize (fun a k -> a, k a) t1 (fun t -> 0);;
The Binary Tree monad
---------------------
Of course, by now you may have realized that we have discovered a new
monad, the Binary 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 `f a`:
\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
. .
__|__ __|__
| | | |
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
\tree (. (. (. (. (a1) (a2)))))
\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))
.
____|____
. . | |
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
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.
What's this have to do with tree\_mondadize?
--------------------------------------------
So we've defined a Tree monad:
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);;
What's this have to do with the `tree_monadize` functions we defined earlier?
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' ->
reader_bind (tree_monadize f r) (fun r' ->
reader_unit (Node (l', r'))));;
... and so on for different monads?
The answer is that each of those `tree_monadize` functions is adding a Tree monad *layer* to a pre-existing Reader (and so on) monad. We discuss that further here: [[Monad Transformers]].