let rec treemap (newleaf : 'a -> 'b) (t : 'a tree) : 'b tree =
match t with
- | Leaf x -> Leaf (newleaf x)
+ | Leaf i -> Leaf (newleaf i)
| Node (l, r) -> Node (treemap newleaf l,
treemap newleaf r);;
`treemap`. For instance, we can easily square each leaf instead by
supplying the appropriate `int -> int` operation in place of `double`:
- let square x = x * x;;
+ let square i = i * i;;
treemap square t1;;
- : int tree =ppp
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
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 .
_____|____
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 `f` returns an `int
-tree` in which each leaf `x` has been replaced with `f x`.
+tree` in which each leaf `i` has been replaced with `f i`.
With previous readers, we always knew which kind of environment to
expect: either an assignment function (the original calculator
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'))
+ | Leaf i -> reader_bind (f i) (fun i' -> reader_unit (Leaf i'))
| Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
reader_bind (treemonadizer f r) (fun y ->
reader_unit (Node (x, y))));;
the tree.
type 'a state = int -> 'a * int;;
- let state_unit a = fun i -> (a, i);;
- let state_bind u f = fun i -> let (a, i') = u i in f a (i' + 1);;
+ let state_unit a = fun s -> (a, s);;
+ let state_bind_and_count u f = fun s -> let (a, s') = u s in f a (s' + 1);;
Gratifyingly, we can use the `treemonadizer` function without any
modification whatsoever, except for replacing the (parametric) type
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 ->
+ | Leaf i -> state_bind_and_count (f i) (fun i' -> state_unit (Leaf i'))
+ | Node (l, r) -> state_bind_and_count (treemonadizer f l) (fun x ->
+ state_bind_and_count (treemonadizer f r) (fun y ->
state_unit (Node (x, y))));;
Then we can count the number of nodes in the tree:
One more revealing example before getting down to business: replacing
`state` everywhere in `treemonadizer` with `list` gives us
- # treemonadizer (fun x -> [ [x; square x] ]) t1;;
+ # treemonadizer (fun i -> [ [i; square i] ]) t1;;
- : int list tree list =
[Node
(Node (Leaf [2; 4], Leaf [3; 9]),
from some input to a result, this transformer replaces each `int` with
a list of `int`'s.
+<!--
+We don't make it clear why the fun has to be int -> int list list, instead of int -> int list
+-->
+
+
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 x c = c x;;
- let continuation_bind u f c = u (fun a -> f a c);;
+ let continuation_unit a = fun k -> k a;;
+ let continuation_bind u f = fun k -> u (fun a -> f a k);;
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'))
+ | Leaf i -> continuation_bind (f i) (fun i' -> continuation_unit (Leaf i'))
| Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
continuation_bind (treemonadizer f r) (fun y ->
continuation_unit (Node (x, y))));;
`continuation` type in the appropriate place in the `treemonadizer` code.
We then compute:
- # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
+ # treemonadizer (fun a 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.
continuation unit as our first argument to `treemonadizer`, and then
apply the result to the identity function:
- # treemonadizer continuation_unit t1 (fun x -> x);;
+ # treemonadizer continuation_unit t1 (fun i -> i);;
- : int tree =
Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
interesting functions for the first argument of `treemonadizer`:
(* Simulating the tree reader: distributing a operation over the leaves *)
- # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
+ # treemonadizer (fun a k -> k (square a)) t1 (fun i -> i);;
- : 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);;
+ # treemonadizer (fun a k -> k [a; square a]) t1 (fun i -> i);;
- : 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);;
+ # treemonadizer (fun a k -> 1 + k a) t1 (fun i -> 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;;
+ type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
The binary tree monad
---------------------
monad, the binary tree monad:
type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
- let tree_unit (x: 'a) = Leaf x;;
+ let tree_unit (a: 'a) = Leaf a;;
let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree =
match u with
- | Leaf x -> f x
+ | 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 = fa
+ 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 `fa`:
+except that each leaf `a` has been replaced with `f a`:
-\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
+\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
. .
__|__ __|__
| | | |
- a1 . fa1 .
+ a1 . f a1 .
_|__ __|__
| | | |
- . a5 . fa5
+ . a5 . f a5
bind _|__ f = __|__
| | | |
- . a4 . fa4
+ . a4 . f a4
__|__ __|___
| | | |
- a2 a3 fa2 fa3
+ a2 a3 f a2 f a3
Given this equivalence, the right identity law
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))) ))
+\tree (. (. (. (. (a1) (a2)))))
+\tree (. (. (. (. (a1) (a1)) (. (a1) (a1)))))
.
____|____
. . | |
bind __|__ f = __|_ = . .
| | | | __|__ __|__
- a1 a2 fa1 fa2 | | | |
+ a1 a2 f a1 f a2 | | | |
a1 a1 a1 a1
Now when we bind this tree to `g`, we get
- .
- ____|____
- | |
- . .
- __|__ __|__
- | | | |
- ga1 ga1 ga1 ga1
+ .
+ _____|______
+ | |
+ . .
+ __|__ __|__
+ | | | |
+ 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