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 =
+ let rec tree_map (t : 'a tree) (leaf_modifier : 'a -> 'b): '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);;
+ | Node (l, r) -> Node (tree_map l leaf_modifier,
+ tree_map r leaf_modifier);;
-`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:
+`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;;
+ tree_map t1 double;;
- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
in place of `double`:
let square i = i * i;;
- tree_map square t1;;
+ tree_map t1 square;;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
[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
+expects contextual information (here, the `\f`) that can be
used to compute the content of indexicals embedded arbitrarily deeply
in the tree.]
But we can do this:
- let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> 'b reader) : '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' ->
+ | Node (l, r) -> reader_bind (tree_monadize l f) (fun l' ->
+ reader_bind (tree_monadize r f) (fun r' ->
reader_unit (Node (l', r'))));;
This function says: give me a function `f` that knows how to turn
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;;
+ # tree_monadize t1 int_readerize 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
+t1 int_readerize`) 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;;
+ # tree_monadize t1 int_readerize square;;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
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 =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> 'b state) : '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' ->
+ | Node (l, r) -> state_bind (tree_monadize l f) (fun l' ->
+ state_bind (tree_monadize r f) (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;;
+ # tree_monadize t1 (fun a -> fun s -> (a, s+1)) 0;;
- : int tree * int =
(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
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;;
+ # tree_monadize t1 (fun a -> fun s -> (s+1, s+1)) 0;;
- : int tree * int =
(Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5)
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 =
+ let rec tree_monadize_rev (t : 'a tree) (f : 'a -> 'b state) : '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 *)
+ | Node (l, r) -> state_bind (tree_monadize r f) (fun r' -> (* R first *)
+ state_bind (tree_monadize l f) (fun l'-> (* Then L *)
state_unit (Node (l', r'))));;
- # tree_monadize_rev (fun a -> fun s -> (s+1, s+1)) t1 0;;
+ # tree_monadize_rev t1 (fun a -> fun s -> (s+1, s+1)) 0;;
- : int tree * int =
(Node (Node (Leaf 5, Leaf 4), Node (Leaf 3, Node (Leaf 2, Leaf 1))), 5)
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;;
+ # tree_monadize t1 (fun i -> [ [i; square i] ]);;
- : 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 tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> ('b, 'r) continuation) : ('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' ->
+ | Node (l, r) -> continuation_bind (tree_monadize l f) (fun l' ->
+ continuation_bind (tree_monadize r f) (fun r' ->
continuation_unit (Node (l', r'))));;
We use the Continuation monad described above, and insert the
So for example, we compute:
- # tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;
+ # tree_monadize t1 (fun a k -> a :: k ()) (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
-operation `fun a -> fun k -> a :: k a` does when you apply it to a
+operation `fun a 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?
-In a moment, we'll return to the same-fringe problem. Since the
+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 its needed...
+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
`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);;
+ # tree_monadize t1 continuation_unit (fun t -> t);;
- : int tree =
Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
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);;
+ # tree_monadize t1 (fun a -> fun k -> k (square a)) (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);;
+ # tree_monadize t1 (fun a -> fun k -> k [a; square a]) (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);;
+ # tree_monadize t1 (fun a -> fun k -> 1 + k a) (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
+[To be fixed: exactly which kind of monad each of these computations simulates.]
- 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 problem, recall, is to take two trees and decide whether they have
-the same leaves in the same order.
-
-<pre>
- ta tb tc
- . . .
-_|__ _|__ _|__
-| | | | | |
-1 . . 3 1 .
- _|__ _|__ _|__
- | | | | | |
- 2 3 1 2 3 2
-
-let ta = Node (Leaf 1, Node (Leaf 2, Leaf 3));;
-let tb = Node (Node (Leaf 1, Leaf 2), Leaf 3);;
-let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));;
-</pre>
-
-So `ta` and `tb` are different trees that have the same fringe, but
-`ta` and `tc` are not.
-
-The simplest solution is to map each tree to a list of its leaves,
-then compare the lists. But because we will have computed the entire
-fringe before starting the comparison, if the fringes differ in an
-early position, we've wasted our time examining the rest of the trees.
-
-The second solution was to use tree zippers and mutable state to
-simulate coroutines (see [[coroutines and aborts]]). In that
-solution, we pulled the zipper on the first tree until we found the
-next leaf, then stored the zipper structure in the mutable variable
-while we turned our attention to the other tree. Because we stopped
-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. Before we can arrive at a solution, however,
-we must define a data structure called a stream:
-
- type 'a stream = End | Next of 'a * (unit -> 'a stream);;
-
-A stream is like a list in that it contains a series of objects (all
-of the same type, here, type `'a`). The first object in the stream
-corresponds to the head of a list, which we pair with a stream
-representing the rest of a the list. There is a special stream called
-`End` that represents a stream that contains no (more) elements,
-analogous to the empty list `[]`.
-
-Actually, we pair each element not with a stream, but with a thunked
-stream, that is, a function from the unit type to streams. The idea
-is that the next element in the stream is not computed until we forced
-the thunk by applying it to the unit:
-
-<pre>
-# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));;
-val make_int_stream : int -> int stream = <fun>
-# let int_stream = make_int_stream 1;;
-val int_stream : int stream = Next (1, <fun>) (* First element: 1 *)
-# match int_stream with Next (i, rest) -> rest;;
-- : unit -> int stream = <fun> (* Rest: a thunk *)
-
-(* Force the thunk to compute the second element *)
-# (match int_stream with Next (i, rest) -> rest) ();;
-- : int stream = Next (2, <fun>)
-</pre>
-
-You can think of `int_stream` as a functional object that provides
-access to an infinite sequence of integers, one at a time. It's as if
-we had written `[1;2;...]` where `...` meant "continue indefinitely".
-
-So, with streams in hand, we need only rewrite our continuation tree
-monadizer so that instead of mapping trees to lists, it maps them to
-streams. Instead of
-
- # tree_monadize (fun a k -> a :: k a) t1 (fun t -> []);;
- - : int list = [2; 3; 5; 7; 11]
-
-as above, we have
-
- # tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End);;
- - : int stream = Next (2, <fun>)
-
-We can see the first element in the stream, the first leaf (namely,
-2), but in order to see the next, we'll have to force a thunk.
-
-Then to complete the same-fringe function, we simply convert both
-trees into leaf-streams, then compare the streams element by element.
-The code is enitrely routine, but for the sake of completeness, here it is:
-
-<pre>
-let rec compare_streams stream1 stream2 =
- match stream1, stream2 with
- | End, End -> true (* Done! Fringes match. *)
- | Next (next1, rest1), Next (next2, rest2) when next1 = next2 -> compare_streams (rest1 ()) (rest2 ())
- | _ -> false;;
-
-let same_fringe t1 t2 =
- let stream1 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End) in
- let stream2 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t2 (fun _ -> End) in
- compare_streams stream1 stream2;;
-</pre>
+We could simulate the tree state example too by setting the relevant
+type to `('a, 'state -> 'result) continuation`.
+In fact, Andre Filinsky has suggested that the continuation monad is
+able to simulate any other monad (Google for "mother of all monads").
-Notice the forcing of the thunks in the recursive call to
-`compare_streams`. So indeed:
+We would eventually want to generalize the continuation type to
-<pre>
-# same_fringe ta tb;;
-- : bool = true
-# same_fringe ta tc;;
-- : bool = false
-</pre>
-
-Now, this implementation is a bit silly, since in order to convert the
-trees to leaf streams, our tree_monadizer function has to visit every
-node in the tree. But if we needed to compare each tree to a large
-set of other trees, we could arrange to monadize each tree only once,
-and then run compare_streams on the monadized trees.
+ type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
-By the way, what if you have reason to believe that the fringes of
-your trees are more likely to differ near the right edge than the left
-edge? If we reverse evaluation order in the tree_monadizer function,
-as shown above when we replaced leaves with their ordinal position,
-then the resulting streams would produce leaves from the right to the
-left.
+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).
The idea of using continuations to characterize natural language meaning
------------------------------------------------------------------------
This sentence means (roughly)
- ∀ x . yesterday(saw x) john
+ forall x . yesterday(saw x) john
That is, the quantifier *everyone* contributes a variable in the
direct object position, and a universal quantifier that takes scope
Then we can crudely approximate quantification as follows:
<pre>
-# tree_monadize lex sentence1 (fun x -> x);;
+# tree_monadize sentence1 lex (fun x -> x);;
- : string tree =
Node
(Leaf "forall x",
<pre>
# let sentence2 = Node (Leaf "everyone", Node (Leaf "saw", Leaf "someone"));;
-# tree_monadize lex sentence2 (fun x -> x);;
+# tree_monadize sentence2 lex (fun x -> x);;
- : string tree =
Node
(Leaf "forall x",
inverse scope:
<pre>
-# tree_monadize_rev lex sentence2 (fun x -> x);;
+# tree_monadize_rev sentence2 lex (fun x -> x);;
- : string tree =
Node
(Leaf "exists y",
that is intended to represent non-deterministic computations as a tree.
-What's this have to do with tree\_mondadize?
+What's this have to do with tree\_monadize?
--------------------------------------------
So we've defined a Tree monad:
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 =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> 'b reader) : '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' ->
+ | Node (l, r) -> reader_bind (tree_monadize l f) (fun l' ->
+ reader_bind (tree_monadize r f) (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]].
+Well, notice that `tree\_monadizer` takes arguments whose types
+resemble that of a monadic `bind` function. Here's a schematic bind
+function compared with `tree\_monadizer`:
+
+ bind (u:'a Monad) (f: 'a -> 'b Monad): 'b Monad
+ tree\_monadizer (u:'a Tree) (f: 'a -> 'b Monad): 'b Tree Monad
+
+Comparing these types makes it clear that `tree\_monadizer` provides a
+way to distribute an arbitrary monad M across the leaves of any tree to
+form a new tree inside an M box.
+The more general 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]].