Our first task will be to replace each leaf with its double:
- let rec treemap (newleaf : 'a -> 'b) (t : 'a tree) : 'b tree =
+ let rec tree_map (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree =
match t with
- | Leaf i -> Leaf (newleaf i)
- | Node (l, r) -> Node (treemap newleaf l,
- treemap newleaf r);;
+ | Leaf i -> Leaf (leaf_modifier i)
+ | Node (l, r) -> Node (tree_map leaf_modifier l,
+ tree_map leaf_modifier r);;
-`treemap` takes a function that transforms old leaves into new leaves,
+`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;;
- treemap double t1;;
+ tree_map double t1;;
- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
| |
14 22
-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
+We could have built the doubling operation right into the `tree_map`
+code. However, because we've left what to do to each leaf as 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
+`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;;
- treemap square t1;;
+ tree_map square t1;;
- : int tree =ppp
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
-Note that what `treemap` does is take some global, contextual
+Note that what `tree_map` 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
+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 treemap function more and
+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 treemap function return a (monadized)
+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.
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`.
+Jacobson-inspired link monad), etc. In the present case, it will be
+enough to expect that our "environment" will be some function of type
+`int -> int`.
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's easy to figure out how to turn an `int` into an `int reader`:
+It would be a simple matter to turn an *integer* into an `int reader`:
- let int2int_reader : 'a -> 'b reader = fun (a : 'a) -> fun (op : 'a -> 'b) -> op a;;
- int2int_reader 2 (fun i -> i + i);;
+ 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 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
+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.
- let rec treemonadizer (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
+But we can do this:
+
+ let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
match t with
| 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 ->
+ | Node (l, r) -> reader_bind (tree_monadize f l) (fun x ->
+ reader_bind (tree_monadize f r) (fun y ->
reader_unit (Node (x, y))));;
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
+turn an `'a tree` into an `'b tree reader`. 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 leaves.
+`'b reader` monad through the original tree's leaves.
- # treemonadizer int2int_reader t1 (fun i -> i + i);;
+ # 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, `treemonadizer
-int2int_reader t1`) to a different `int -> int` function---say, the
+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:
- # treemonadizer int2int_reader t1 (fun i -> i * i);;
+ # 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.
+
+<!-- FIXME -->
+
For instance, we can use a state monad to count the number of nodes in
the tree.
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
+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 treemonadizer (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
+ let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
match t with
| 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 ->
+ | Node (l, r) -> state_bind_and_count (tree_monadize f l) (fun x ->
+ state_bind_and_count (tree_monadize f r) (fun y ->
state_unit (Node (x, y))));;
Then we can count the number of nodes in the tree:
- # treemonadizer state_unit t1 0;;
+ # tree_monadize state_unit t1 0;;
- : int tree * int =
(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
One more revealing example before getting down to business: replacing
-`state` everywhere in `treemonadizer` with `list` gives us
+`state` everywhere in `tree_monadize` with `list` gives us
- # treemonadizer (fun i -> [ [i; square i] ]) t1;;
+ # tree_monadize (fun i -> [ [i; square i] ]) t1;;
- : int list tree list =
[Node
(Node (Leaf [2; 4], Leaf [3; 9]),
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 =
+ let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
match t with
| 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 ->
+ | Node (l, r) -> continuation_bind (tree_monadize f l) (fun x ->
+ continuation_bind (tree_monadize f r) (fun y ->
continuation_unit (Node (x, y))));;
We use the continuation monad described above, and insert the
-`continuation` type in the appropriate place in the `treemonadizer` code.
+`continuation` type in the appropriate place in the `tree_monadize` code.
We then compute:
- # treemonadizer (fun a k -> a :: (k a)) t1 (fun t -> []);;
+ # tree_monadize (fun a k -> a :: (k a)) t1 (fun t -> []);;
- : int list = [2; 3; 5; 7; 11]
+<!-- FIXME: what if we had fun t -> [-t]? why `t`? -->
+
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
+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:
- # treemonadizer continuation_unit t1 (fun i -> i);;
+ # tree_monadize continuation_unit t1 (fun i -> i);;
- : 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 `treemonadizer`:
+interesting functions for the first argument of `tree_monadize`:
(* Simulating the tree reader: distributing a operation over the leaves *)
- # treemonadizer (fun a k -> k (square a)) t1 (fun i -> i);;
+ # tree_monadize (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 k -> k [a; square a]) t1 (fun i -> i);;
+ # tree_monadize (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 k -> 1 + k a) t1 (fun i -> 0);;
+ # tree_monadize (fun a k -> 1 + k a) t1 (fun i -> 0);;
- : int = 5
We could simulate the tree state example too, but it would require
As for the associative law,
- Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
+ 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