X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=manipulating_trees_with_monads.mdwn;h=3247ce6232d7e3ee8d84cbb438520c723806bfac;hp=038cdc3697a4ab501734dac5209ffe2402d0ecd3;hb=c3395c03cdde647d084ca1ca2917adf7c3ec3838;hpb=8f067600295d47935d4ec86e612c189d0e39b0d7
diff --git a/manipulating_trees_with_monads.mdwn b/manipulating_trees_with_monads.mdwn
index 038cdc36..3247ce62 100644
--- a/manipulating_trees_with_monads.mdwn
+++ b/manipulating_trees_with_monads.mdwn
@@ -325,11 +325,11 @@ 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
@@ -360,142 +360,18 @@ interesting functions for the first argument of `tree_monadize`:
# 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.]
- 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.
-
-
- 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));;
-
-
-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:
-
-
-# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));;
-val make_int_stream : int -> int stream =
-# let int_stream = make_int_stream 1;;
-val int_stream : int stream = Next (1, ) (* First element: 1 *)
-# match int_stream with Next (i, rest) -> rest;;
-- : unit -> int stream = (* Rest: a thunk *)
-
-(* Force the thunk to compute the second element *)
-# (match int_stream with Next (i, rest) -> rest) ();;
-- : int stream = Next (2, )
-
-
-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
+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").
- # 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, )
-
-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:
-
-
-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;;
-
-
-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
-
-# same_fringe ta tb;;
-- : bool = true
-# same_fringe ta tc;;
-- : bool = false
-
-
-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
------------------------------------------------------------------------
@@ -525,7 +401,7 @@ quantification.
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
@@ -669,7 +545,7 @@ called a
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: