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
+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 transformer that
+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
+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
| 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:
+`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;;
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
+each leaf instead, by supplying the appropriate `int -> int` operation
in place of `double`:
let square i = i * i;;
tree that is ready to accept any `int -> int` function and produce the
updated tree.
- \f .
- _____|____
- | |
- . .
- __|___ __|___
- | | | |
- f 2 f 3 f 5 .
- __|___
- | |
- f 7 f 11
+ fun e -> .
+ _____|____
+ | |
+ . .
+ __|___ __|___
+ | | | |
+ e 2 e 3 e 5 .
+ __|___
+ | |
+ e 7 e 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`.
+`e` returns an `int tree` in which each leaf `i` has been replaced
+with `e 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
+expects contextual information (here, the `e`) that can be
used to compute the content of indexicals embedded arbitrarily deeply
in the tree.]
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 *)
+ 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;;
+ 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);;
+ let asker : int -> int reader =
+ fun (a : int) ->
+ fun (modifier : int -> int) -> modifier a;;
+ asker 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?
+`asker a` is a monadic box that waits for an an environment (here, the argument `modifier`) and returns what that environment maps `a` to.
+
+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.
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
+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 `asker` or `reader_unit`---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:
------------
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 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, `tree_monadize
-int_readerize t1`) to a different `int -> int` function---say, the
+asker 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;;
+ # tree_monadize asker t1 square;;
- : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
Then we can count the number of leaves in the tree:
- # tree_monadize (fun a -> fun s -> (a, s+1)) t1 0;;
+ # 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
+ .
+ ___|___
+ | |
+ . .
+ ( _|__ _|__ , 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
+Why does this work? Because the operation `incrementer`
+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 state-incrementing for each of its leaves.
+that does the same store-incrementing for each of its leaves.
-We can use the state monad to replace leaves with a number
+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 -> (s+1, s+1)) t1 0;;
- - : int tree * int =
- (Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5)
+ # 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 copying `a` into the
-monadic box, we throw away the `a` and put a copy of the state in
-instead.
+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 order requires reversing the order of the state_bind
+Reversing the annotation 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
+ 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)
- # 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.
+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 `tree_monadize` with `list` gives us
+`state` everywhere in `tree_monadize` with `list` lets us do:
- # 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])))]
+ # 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 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.]
+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 `int`s for its leaves.
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;;
+ 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 -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
+ 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' ->
So for example, we compute:
- # tree_monadize (fun a -> fun k -> a :: k a) t1 (fun t -> []);;
+ # 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
-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?
+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
- : 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
+It's not immediately obvious to us how to simulate the List monadization of the tree using this technique.
- type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
+We could simulate the tree annotating example by setting the relevant
+type to `(store -> 'result, 'a) continuation`.
-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]
+Andre Filinsky has proposed that the continuation monad is
+able to simulate any other monad (Google for "mother of all monads").
-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>
-
-Notice the forcing of the thunks in the recursive call to
-`compare_streams`. So indeed:
-
-<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.
-
-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
------------------------------------------------------------------------
**Quantification and default quantifier scope construal**.
-We saw in the copy-string example and in the same-fringe example that
-local properties of a tree (whether a character is `S` or not, which
+We saw in the copy-string example ("abSd") and in the same-fringe example that
+local properties of a structure (whether a character is `'S'` or not, which
integer occurs at some leaf position) can control global properties of
the computation (whether the preceeding string is copied or not,
whether the computation halts or proceeds). Local control of
This sentence means (roughly)
- &Forall; 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
over the whole sentence. If we have a lexical meaning function like
the following:
-<pre>
-let lex (s:string) k = match s with
- | "everyone" -> Node (Leaf "forall x", k "x")
- | "someone" -> Node (Leaf "exists y", k "y")
- | _ -> k s;;
-
-let sentence1 = Node (Leaf "John",
- Node (Node (Leaf "saw",
- Leaf "everyone"),
- Leaf "yesterday"));;
-</pre>
+ let lex (s:string) k = match s with
+ | "everyone" -> Node (Leaf "forall x", k "x")
+ | "someone" -> Node (Leaf "exists y", k "y")
+ | _ -> k s;;
Then we can crudely approximate quantification as follows:
-<pre>
-# tree_monadize lex sentence1 (fun x -> x);;
-- : string tree =
-Node
- (Leaf "forall x",
- Node (Leaf "John", Node (Node (Leaf "saw", Leaf "x"), Leaf "yesterday")))
-</pre>
+ # let sentence1 = Node (Leaf "John",
+ Node (Node (Leaf "saw",
+ Leaf "everyone"),
+ Leaf "yesterday"));;
+
+ # tree_monadize lex sentence1 (fun x -> x);;
+ - : string tree =
+ Node
+ (Leaf "forall x",
+ Node (Leaf "John", Node (Node (Leaf "saw", Leaf "x"), Leaf "yesterday")))
In order to see the effects of evaluation order,
observe what happens when we combine two quantifiers in the same
sentence:
-<pre>
-# let sentence2 = Node (Leaf "everyone", Node (Leaf "saw", Leaf "someone"));;
-# tree_monadize lex sentence2 (fun x -> x);;
-- : string tree =
-Node
- (Leaf "forall x",
- Node (Leaf "exists y", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
-</pre>
+ # let sentence2 = Node (Leaf "everyone", Node (Leaf "saw", Leaf "someone"));;
+ # tree_monadize lex sentence2 (fun x -> x);;
+ - : string tree =
+ Node
+ (Leaf "forall x",
+ Node (Leaf "exists y", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
The universal takes scope over the existential. If, however, we
-replace the usual tree_monadizer with tree_monadizer_rev, we get
+replace the usual `tree_monadizer` with `tree_monadizer_rev`, we get
inverse scope:
-<pre>
-# tree_monadize_rev lex sentence2 (fun x -> x);;
-- : string tree =
-Node
- (Leaf "exists y",
- Node (Leaf "forall x", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
-</pre>
+ # tree_monadize_rev lex sentence2 (fun x -> x);;
+ - : string tree =
+ Node
+ (Leaf "exists y",
+ Node (Leaf "forall x", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
There are many crucially important details about quantification that
-are being simplified here, and the continuation treatment here is not
+are being simplified here, and the continuation treatment used here is not
scalable for a number of reasons. Nevertheless, it will serve to give
an idea of how continuations can provide insight into the behavior of
-quantifiers.
+quantifiers.
-The Binary Tree monad
----------------------
+The 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,
+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);;
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`:
+except that each leaf `a` has been replaced with the tree returned by `f a`:
. .
__|__ __|__
- | | | |
+ | | /\ |
a1 . f a1 .
_|__ __|__
- | | | |
+ | | | /\
. a5 . f a5
bind _|__ f = __|__
- | | | |
+ | | | /\
. a4 . f a4
__|__ __|___
- | | | |
+ | | /\ /\
a2 a3 f a2 f a3
Given this equivalence, the right identity law
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.
+build the exact same final tree.
So binary trees are a monad.
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:
-
- 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]].
+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: [[Monad Transformers]].