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diff git a/assignment9.mdwn b/assignment9.mdwn
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 a/assignment9.mdwn
+++ b/assignment9.mdwn
@@ 4,7 +4,6 @@ Using continuations to solve the samefringe problem
The 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.
@@ 26,106 +25,157 @@ 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.
+We've seen two solutions to the same fringe problem so far.
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.
+simulate coroutines (see [[coroutines and aborts]], and
+[[assignment8]]). In that solution, we pulled the zipper on the first
+tree until we found the next leaf, then stored the zipper structure in
+a mutable variable while we turned our attention to the other tree.
+This solution is efficient: the zipper doesn't visit any leaves beyond
+the first mismatch.
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. Your assignment is to show how.
TODO LIST for solving the problem

+The first step is to review your answer to [[assignment8]], and make
+sure you understand what is going on.
+
1. Review the simple but inefficient solution (easy).
2. Understand the zipper/mutable state solution in [[coroutines and aborts]] (harder).
+Two strategies for solving the problem
+
3. Two obvious approaches:
a. Review the listzipper/listcontinuation example given in
+1. Review the listzipper/listcontinuation example given in
class in [[from list zippers to continuations]]; then
figure out how to refunctionalize the zippers used in the zipper
solution.
b. Review the tree_monadizer application of continuations that maps a
tree to a list of leaves in [[manipulating trees with monads]]. Spend
some time trying to understand exactly what it does. Suggestion:
compute the transformation for a tree with two leaves, performing all
beta reduction by hand using the definitions for bind_continuation and
so on. If you take this route, study the description of streams (a
particular kind of data structure) below. The goal will be to arrange
for the continuationflavored tree_monadizer to transform a tree into
a stream instead of into a list. Once you've done that, completing
the samefringe problem will be easy.
+2. Review how the continuationflavored `tree_monadizer` managed to
+ map a tree to a list of its leaves, in [[manipulating trees with monads]].
+ Spend some time trying to understand exactly what it
+ does: compute the treetolist transformation for a tree with two
+ leaves, performing all beta reduction by hand using the
+ definitions for `continuation_bind`, `continuation_unit` and so on.
+ If you take this route, study the description of **streams** (a
+ particular kind of data structure) below. The goal will be to
+ arrange for the continuationflavored `tree_monadizer` to transform
+ a tree into a stream instead of into a list. Once you've done
+ that, completing the samefringe problem will be easy.

Whichever method you choose, here are some goals to consider.
1. Make sure that your solution gives the right results on the trees
given above.
+given above (`ta`, `tb`, and `tc`).
2. Make sure your function works on trees that contain only a single
leaf, and when the two trees have different numbers of leaves.
+leaf, as well as when the two trees have different numbers of leaves.
3. Figure out a way to prove that your solution satisfies the
requirements of the problem, in particular, that when the trees differ
+3. Figure out a way to prove that your solution satisfies the main
+requirement of the problem; in particular, that when the trees differ
in an early position, your code does not waste time visiting the rest
of the tree. One way to do this is to add print statements to your
functions so that every time you visit a leaf (say), a message is
printed on the output.
+printed on the output. (In OCaml: `print_int 1` prints an `int`, `print_string "foo"` prints a `string`, `print_newline ()` prints a line break, and `print_endline "foo"` prints a string followed by a line break.) If two trees differ in the middle of their fringe, you should show that your solution prints debugging information for the first half of the fringe, but then stops.
4. What if you had some reason to believe that the trees you were
going to compare were more likely to differ in the rightmost region?
What would you have to change in your solution so that it compared the
fringe from right to left?
+What would you have to change in your solution so that it worked from
+right to left?
Streams

A stream is like a list in that it contains a series of objects (all
of the same type, here, type `'a`). It differs from a list in that
the tail of the list is left uncomputed until needed. We will turn
the stream on and off by thunking it (see class notes for [[week6]]
on thunks, as well as [[assignment5]]).
+A stream is like a list in that it wraps a series of elements of a single type.
+It differs from a list in that the tail of the series is left uncomputed
+until needed. We will turn the stream on and off by thunking it (see
+class notes for [[week6]] on thunks, as well as [[assignment5]]).
type 'a stream = End  Next of 'a * (unit > 'a stream);;
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:
+Streams that are not empty contain a first object, paired with a
+thunked stream representing the rest of the series. In order to get
+access to the next element in the stream, we must *force* the thunk by
+applying it to the unit. Watch the behavior of this stream in detail.
+This stream delivers the natural numbers, in order: 1, 2, 3, ...
# 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 *)
+
+# let tail = match int_stream with Next (i, rest) > rest;;
+val tail : unit > int stream = [fun] (* Tail: a thunk *)
(* Force the thunk to compute the second element *)
# (match int_stream with Next (i, rest) > rest) ();;
 : int stream = Next (2, [fun])
+# tail ();;
+ : int stream = Next (2, [fun]) (* Second element: 2 *)
+
+# match tail () with Next (_, rest) > rest ();;
+ : int stream = Next (3, [fun]) (* Third element: 3 *)
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".

+we had written `[1;2;...]` where `...` meant "continue for as long as
+some other process needs new integers".
+
+
+