figure out how to re-functionalize the zippers used in the zipper
solution.
-2. Review how the continuation-flavored tree\_monadizer managed to
+2. Review how the continuation-flavored `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 tree-to-list transformation for a tree with two
leaves, performing all beta reduction by hand using the
- definitions for bind\_continuation, unit\_continuation and so on.
+ 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 continuation-flavored tree_monadizer to transform
+ arrange for the continuation-flavored `tree_monadizer` to transform
a tree into a stream instead of into a list. Once you've done
that, completing the same-fringe problem will be easy.
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. 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.
+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?
Streams
-------
-A stream is like a list in that it contains a series of objects. It
-differs from a list in that the tail of the list is left uncomputed
+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]]).
access to an infinite sequence of integers, one at a time. It's as if
we had written `[1;2;...]` where `...` meant "continue for as long as
some other process needs new integers".
+
+
+<!--
+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 entirely 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:
+
+ # same_fringe ta tb;;
+ - : bool = true
+ # same_fringe ta tc;;
+ - : bool = false
+
+Now, you might think that 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, so we'd have to traverse
+the entire tree at some point. But you'd be wrong: part of what gets
+suspended in the thunking of the stream is the computation of the rest
+of the monadized tree.
+
+-->