## Refunctionalizing zippers: from lists to continuations

If zippers are continuations reified (defuntionalized), then one route to continuations is to re-functionalize a zipper. Then the concreteness and understandability of the zipper provides a way of understanding an equivalent treatment using continuations.

Let's work with lists of chars for a change. We'll sometimes write "abSd" as an abbreviation for
['a'; 'b'; 'S'; 'd'].

We will set out to compute a deceptively simple-seeming task: given a string, replace each occurrence of 'S' in that string with a copy of the string up to that point.

We'll define a function t (for "task") that maps strings to their updated version.

Expected behavior:

t "abSd" ~~> "ababd"


In linguistic terms, this is a kind of anaphora resolution, where 'S' is functioning like an anaphoric element, and the preceding string portion is the antecedent.

This task can give rise to considerable complexity. Note that it matters which 'S' you target first (the position of the * indicates the targeted 'S'):

    t "aSbS"
*
~~> t "aabS"
*
~~> "aabaab"


versus

    t "aSbS"
*
~~> t "aSbaSb"
*
~~> t "aabaSb"
*
~~> "aabaaabab"


versus

    t "aSbS"
*
~~> t "aSbaSb"
*
~~> t "aSbaaSbab"
*
~~> t "aSbaaaSbaabab"
*
~~> ...


Apparently, this task, as simple as it is, is a form of computation, and the order in which the 'S's get evaluated can lead to divergent behavior.

For now, we'll agree to always evaluate the leftmost 'S', which guarantees termination, and a final string without any 'S' in it.

This is a task well-suited to using a zipper. We'll define a function tz (for task with zippers), which accomplishes the task by mapping a char list zipper to a char list. We'll call the two parts of the zipper unzipped and zipped; we start with a fully zipped list, and move elements to the unzipped part by pulling the zipper down until the entire list has been unzipped, at which point the zipped half of the zipper will be empty.

type 'a list_zipper = ('a list) * ('a list);;

let rec tz (z : char list_zipper) =
match z with
| (unzipped, []) -> List.rev(unzipped) (* Done! *)
| (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped)
| (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *)

# tz ([], ['a'; 'b'; 'S'; 'd']);;
- : char list = ['a'; 'b'; 'a'; 'b'; 'd']

# tz ([], ['a'; 'S'; 'b'; 'S']);;
- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']


Note that the direction in which the zipper unzips enforces the evaluate-leftmost rule. Task completed.

One way to see exactly what is going on is to watch the zipper in action by tracing the execution of tz. By using the #trace directive in the OCaml interpreter, the system will print out the arguments to tz each time it is called, including when it is called recursively within one of the match clauses. Note that the lines with left-facing arrows (<--) show (both initial and recursive) calls to tz, giving the value of its argument (a zipper), and the lines with right-facing arrows (-->) show the output of each recursive call, a simple list.

# #trace tz;;
t1 is now traced.
# tz ([], ['a'; 'b'; 'S'; 'd']);;
tz <-- ([], ['a'; 'b'; 'S'; 'd'])       (* Initial call *)
tz <-- (['a'], ['b'; 'S'; 'd'])         (* Pull zipper *)
tz <-- (['b'; 'a'], ['S'; 'd'])         (* Pull zipper *)
tz <-- (['b'; 'a'; 'b'; 'a'], ['d'])    (* Special 'S' step *)
tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], [])  (* Pull zipper *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd']        (* Output reversed *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
- : char list = ['a'; 'b'; 'a'; 'b'; 'd']


The nice thing about computations involving lists is that it's so easy to visualize them as a data structure. Eventually, we want to get to a place where we can talk about more abstract computations. In order to get there, we'll first do the exact same thing we just did with concrete zipper using procedures instead.

Think of a list as a procedural recipe: ['a'; 'b'; 'c'; 'd'] is the result of the computation 'a'::('b'::('c'::('d'::[]))) (or, in our old style, make_list 'a' (make_list 'b' (make_list 'c' (make_list 'd' empty)))). The recipe for constructing the list goes like this:

(1) make a new list whose first element is 'd' and whose tail is the list constructed in step (0)

## (2) make a new list whose first element is 'c' and whose tail is the list constructed in step (1)

(3) make a new list whose first element is 'b' and whose tail is the list constructed in step (2)
(4) make a new list whose first element is 'a' and whose tail is the list constructed in step (3)

What is the type of each of these steps? Well, it will be a function from the result of the previous step (a list) to a new list: it will be a function of type char list -> char list. We'll call each step (or group of steps) a continuation of the previous steps. So in this context, a continuation is a function of type char list -> char list. For instance, the continuation corresponding to the portion of the recipe below the horizontal line is the function fun (tail : char list) -> 'a'::('b'::tail). What is the continuation of the 4th step? That is, after we've built up 'a'::('b'::('c'::('d'::[]))), what more has to happen to that for it to become the list ['a'; 'b'; 'c'; 'd']? Nothing! Its continuation is the function that does nothing: fun tail -> tail.

In what follows, we'll be thinking about the result list that we're building up in this procedural way. We'll treat our input list just as a plain old static list data structure, that we recurse through in the normal way we're accustomed to. We won't need a zipper data structure, because the continuation-based representation of our result list will take over the same role.

So our new function tc (for task with continuations) takes an input list (not a zipper) and a also takes a continuation k (it's conventional to use k for continuation variables). k is a function that represents how the result list is going to continue being built up after this invocation of tc delivers up a value. When we invoke tc for the first time, we expect it to deliver as a value the very de-S'd list we're seeking, so the way for the list to continue being built up is for nothing to happen to it. That is, our initial invocation of tc will supply fun tail -> tail as the value for k. Here is the whole tc function. Its structure and behavior follows tz from above, which we've repeated here to facilitate detailed comparison:

let rec tz (z : char list_zipper) =
match z with
| (unzipped, []) -> List.rev(unzipped) (* Done! *)
| (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped)
| (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *)

let rec tc (l: char list) (k: (char list) -> (char list)) =
match l with
| [] -> List.rev (k [])
| 'S'::zipped -> tc zipped (fun tail -> k (k tail))
| target::zipped -> tc zipped (fun tail -> target::(k tail));;

# tc ['a'; 'b'; 'S'; 'd'] (fun tail -> tail);;
- : char list = ['a'; 'b'; 'a'; 'b']

# tc ['a'; 'S'; 'b'; 'S'] (fun tail -> tail);;
- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']


To emphasize the parallel, we've re-used the names zipped and target. The trace of the procedure will show that these variables take on the same values in the same series of steps as they did during the execution of tz above: there will once again be one initial and four recursive calls to tc, and zipped will take on the values "bSd", "Sd", "d", and "" (and, once again, on the final call, the first match clause will fire, so the the variable zipped will not be instantiated).

We have not named the continuation argument unzipped, although that is what the parallel would suggest. The reason is that unzipped (in tz) is a list, but k (in tc) is a function. That's the most crucial difference between the solutions---it's the point of the excercise, and it should be emphasized. For instance, you can see this difference in the fact that in tz, we have to glue together the two instances of unzipped with an explicit (and, computationally speaking, relatively inefficient) List.append. In the tc version of the task, we simply compose k with itself: k o k = fun tail -> k (k tail).

A call tc ['a'; 'b'; 'S'; 'd'] would yield a partially-applied function; it would still wait for another argument, a continuation of type char list -> char list. So we have to give it an "initial continuation" to get started. As mentioned above, we supply the identity function as the initial continuation. Why did we choose that? Again, if you have already constructed the result list "ababd", what's the desired continuation? What's the next step in the recipe to produce the desired result, i.e, the very same list, "ababd"? Clearly, the identity function.

A good way to test your understanding is to figure out what the continuation function k must be at the point in the computation when tc is applied to the argument "Sd". Two choices: is it fun tail -> 'a'::'b'::tail, or it is fun tail -> 'b'::'a'::tail? The way to see if you're right is to execute the following command and see what happens:

tc ['S'; 'd'] (fun tail -> 'a'::'b'::tail);;


There are a number of interesting directions we can go with this task. The reason this task was chosen is because the task itself (as opposed to the functions used to implement the task) can be viewed as a simplified picture of a computation using continuations, where 'S' plays the role of a continuation operator. (It works like the Scheme operators shift or control; the differences between them don't manifest themselves in this example. See Ken Shan's paper Shift to control, which inspired some of the discussion in this topic.) In the analogy, the input list portrays a sequence of functional applications, where [f1; f2; f3; x] represents f1(f2(f3 x)). The limitation of the analogy is that it is only possible to represent computations in which the applications are always right-branching, i.e., the computation ((f1 f2) f3) x cannot be directly represented.

One way to extend this exercise would be to add a special symbol '#', and then the task would be to copy from the target 'S' only back to the closest '#'. This would allow our task to simulate delimited continuations with embedded prompts (also called resets).

Here is some Scheme code implementing the tz and tc functions, first as presented above, and second with the variant just mentioned, using '#'. There's also a third kind of implementation, which is akin to the tc version, but doesn't explicitly pass a k argument, and instead uses these unfamiliar operations reset and shift. We'll be explaining what these do shortly. (The reason this code is in Scheme is because that's the language in which it's easiest to work with operations like reset and shift.)