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.
-\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))
-
- \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' ->
- state_bind (tree_monadize f l) (fun l' ->
+ | 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 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?
+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 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?
+
+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
- : 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.
+
+We could simulate the tree annotating example by setting the relevant
+type to `(store -> 'result, 'a) continuation`.
- type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;;
+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](/code/tree_monadize.ml).
-Using continuations to solve the same fringe problem
-----------------------------------------------------
+The idea of using continuations to characterize natural language meaning
+------------------------------------------------------------------------
+
+We might a philosopher or a linguist be interested in continuations,
+especially if efficiency of computation is usually not an issue?
+Well, the application of continuations to the same-fringe problem
+shows that continuations can manage order of evaluation in a
+well-controlled manner. In a series of papers, one of us (Barker) and
+Ken Shan have argued that a number of phenomena in natural langauge
+semantics are sensitive to the order of evaluation. We can't
+reproduce all of the intricate arguments here, but we can give a sense
+of how the analyses use continuations to achieve an analysis of
+natural language meaning.
+
+**Quantification and default quantifier scope construal**.
-We've seen two solutions to the same fringe problem so far.
-The simplest is to map each tree to a list of its leaves, then compare
-the lists. But if the fringes differ in an early position, we've
-wasted our time visiting the rest of the tree.
+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
+surrounding context is a reasonable description of in-situ
+quantification.
-The second solution was to use tree zippers and mutable state to
-simulate coroutines. We would unzip the first tree until we found the
-next leaf, then store the zipper structure in the mutable variable
-while we turned our attention to the other tree. Because we stop 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.
+ (1) John saw everyone yesterday.
-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. To make this work in the most convenient
-way, we need to use the fully general type for continuations just mentioned.
+This sentence means (roughly)
-tree_monadize (fun a k -> a, k a) t1 (fun t -> 0);;
+ 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:
+ let lex (s:string) k = match s with
+ | "everyone" -> Node (Leaf "forall x", k "x")
+ | "someone" -> Node (Leaf "exists y", k "y")
+ | _ -> k s;;
-The Binary Tree monad
----------------------
+Then we can crudely approximate quantification as follows:
+
+ # 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")))
-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,
+In order to see the effects of evaluation order,
+observe what happens when we combine two quantifiers in the same
+sentence:
+
+ # 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
+inverse scope:
+
+ # 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 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.
+
+
+The Tree 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`:
-
-\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5)))
+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
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)))))
-
.
____|____
. . | |
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]].