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
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
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
fun e -> . _____|____ | | . . __|___ __|___ | | | | e 2 e 3 e 5 . __|___ | | e 7 e 11
That is, we want to transform the ordinary tree
t1 (of type
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
[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
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;; 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
let asker : int -> 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
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' -> 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, such as
reader_unit---and I'll show you how to
'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
------------ 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 asker 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,
asker t1) to a different
int -> int function---say, the
fun i -> i * i---we get an entirely different
# tree_monadize asker 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:
# let incrementer = fun a -> fun s -> (a, s+1);; # tree_monadize incrementer 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
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
operations. It's not obvious that this will type correctly, so think
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
list lets us do:
# let decider i = if i = 2 then [20; 21] else [i];; # tree_monadize decider t1;; - : 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
'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
fun a k -> a :: k a does when you apply it to a
int, and the continuation
fun _ -> . Then given what we've
tree_monadize, what should we expect
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
(* 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
(store -> 'result, 'a) continuation.
Andre Filinsky has proposed that the continuation monad is able to simulate any other monad (Google for "mother of all monads").
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
bind u f is a tree with the same strucure as
except that each leaf
a has been replaced with the tree returned by
. . __|__ __|__ | | /\ | 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
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.
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: .