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
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>
-
-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
------------------------------------------------------------------------
--- /dev/null
+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>
+
+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.
+