Manipulating trees with monads
------------------------------
-This topic develops an idea based on a detailed suggestion of Ken
-Shan's. We'll build a series of functions that operate on trees,
-doing various things, including replacing leaves, counting nodes, and
-converting a tree to a list of leaves. The end result will be an
-application for continuations.
+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
+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
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
+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
the utility of replacing one monad with other.
First, we'll be needing a lot of trees for the remainder of the
course. Here again is a type constructor for leaf-labeled, binary trees:
- type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
+ type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree);;
[How would you adjust the type constructor to allow for labels on the
internal nodes?]
let t1 = Node (Node (Leaf 2, Leaf 3),
Node (Leaf 5, Node (Leaf 7,
- Leaf 11)))
+ Leaf 11)))
.
___|___
| |
14 22
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
-`tree_map`. For instance, we can easily square each leaf instead by
-supplying the appropriate `int -> int` operation in place of `double`:
+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
+in place of `double`:
let square i = i * i;;
tree_map square t1;;
- - : int tree =ppp
+ - : int tree =
Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
Note that what `tree_map` does is take some unchanging contextual
information---what to do to each leaf---and supplies that information
to each subpart of the computation. In other words, `tree_map` has the
-behavior of a reader monad. Let's make that explicit.
+behavior of a Reader monad. Let's make that explicit.
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 `tree_map` 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.
-\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11))))
-
\f .
_____|____
| |
f 7 f 11
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 `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
-simulation), a world (the intensionality monad), an integer (the
-Jacobson-inspired link monad), etc. In the present case, we expect that our "environment" will be some function of type `int -> int`. "Looking up" some `int` in the environment will return us the `int` that comes out the other side of that function.
+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`.
+
+[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
+used to compute the content of indexicals embedded arbitrarily deeply
+in the tree.]
+
+With our previous applications of the Reader monad, we always knew
+which kind of environment to expect: either an assignment function, as
+in the original calculator simulation; a world, as in the
+intensionality monad; an individual, as in the Jacobson-inspired link
+monad; etc. In the present case, we expect that our "environment"
+will be some function of type `int -> int`. "Looking up" some `int` in
+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 *)
let reader_unit (a : 'a) : 'a reader = fun _ -> a;;
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.
-For instance, we can use a state monad to count the number of leaves in
+For instance, we can use a State monad to count the number of leaves in
the tree.
type 'a state = int -> 'a * int;;
___|___
| |
. .
- _|__ _|__
+ _|__ _|__ , 5
| | | |
2 3 5 .
_|__
| |
7 11
-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 `tree_monadize` function, it then wraps an `int tree` in a box, one that does the same state-incrementing for each of its leaves.
+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
+`tree_monadize` function, it then wraps an `int tree` in a box, one
+that does the same state-incrementing for each of its leaves.
+
+We can use the state monad to replace 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)
+
+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.
+
+Reversing the 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
+ | 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 -> (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.
One more revealing example before getting down to business: replacing
`state` everywhere in `tree_monadize` with `list` gives us
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.
-
-
-<!--
-FIXME: We don't make it clear why the fun has to be int -> int list list, instead of int -> int list
--->
+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.]
Now for the main point. What if we wanted to convert a tree to a list
of leaves?
continuation_bind (tree_monadize f r) (fun r' ->
continuation_unit (Node (l', r'))));;
-We use the continuation monad described above, and insert the
-`continuation` type in the appropriate place in the `tree_monadize` code. Then if we give the `tree_monadize` function an operation that converts `int`s into `'b`-wrapping continuation monads, it will give us back a way to turn `int tree`s into corresponding `'b tree`-wrapping continuation monads.
+We use the Continuation monad described above, and insert the
+`continuation` type in the appropriate place in the `tree_monadize` code. Then if we give the `tree_monadize` function an operation that converts `int`s into `'b`-wrapping Continuation monads, it will give us back a way to turn `int tree`s into corresponding `'b tree`-wrapping Continuation monads.
So for example, we compute:
- # tree_monadize (fun a -> fun k -> a :: (k a)) t1 (fun t -> []);;
+ # tree_monadize (fun a -> fun 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. 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 _ -> []`.
+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?
+
+In a moment, 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...
-The continuation monad is amazingly flexible; we can use it to
+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
`continuation_unit` as our first argument to `tree_monadize`, and then
# 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
+[To be fixed: exactly which kind of monad each of these computations simulates.]
+
+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").
+
+We would eventually want to generalize the continuation type to
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]
-The binary tree monad
+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.
+
+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 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
+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.
+
+ (1) John saw everyone yesterday.
+
+This sentence means (roughly)
+
+ 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>
+
+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>
+
+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>
+
+The universal takes scope over the existential. If, however, we
+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>
+
+There are many crucially important details about quantification that
+are being simplified here, and the continuation treatment 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 Binary Tree monad
---------------------
Of course, by now you may have realized that we have discovered a new
-monad, the binary tree monad:
+monad, the Binary 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);;
- let tree_unit (a: 'a) = Leaf a;;
+ 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));;
+ | 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:
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)))
-
. .
__|__ __|__
| | | |
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)))))
-
.
____|____
. . | |
[SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree)
that is intended to represent non-deterministic computations as a tree.
+
+What's this have to do with tree\_mondadize?
+--------------------------------------------
+
+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]].
+