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 = [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])
```
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, ) 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:
```# 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. Proving this claim requires adding print statements (or other tracing technology) within the tree monadizer function. 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.