resolution, where `'S'` is functioning like an anaphoric element, and
the preceding string portion is the antecedent.
resolution, where `'S'` is functioning like an anaphoric element, and
the preceding string portion is the antecedent.
-Aparently, this task, as simple as it is, is a form of computation,
+Apparently, this task, as simple as it is, is a form of computation,
`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
`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
- 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 *)
+ 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'; 'b'; 'S'; 'd']);;
- : char list = ['a'; 'b'; 'a'; 'b'; 'd']
# tz ([], ['a'; 'S'; 'b'; 'S']);;
- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
# tz ([], ['a'; 'S'; 'b'; 'S']);;
- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
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
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 (recurcively) called. Note that the
-lines with left-facing arrows (`<--`) show (recursive) calls to `tz`,
+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.
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.
tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *)
tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *)
tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *)
tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *)
tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], []) (* Pull zipper *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd'] (* Output reversed *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], []) (* Pull zipper *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd'] (* Output reversed *)
tz --> ['a'; 'b'; 'a'; 'b'; 'd']
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
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
Think of a list as a procedural recipe: `['a'; 'b'; 'S'; 'd']` is the result of
the computation `'a'::('b'::('S'::('d'::[])))` (or, in our old style,
Think of a list as a procedural recipe: `['a'; 'b'; 'S'; 'd']` is the result of
the computation `'a'::('b'::('S'::('d'::[])))` (or, in our old style,
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
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
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)`.
This means that we can now represent the unzipped part of our
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)`.
This means that we can now represent the unzipped part of our
describing how to finish building a list. We'll write a new
function, `tc` (for task with continuations), that will take an input
describing how to finish building a list. We'll write a new
function, `tc` (for task with continuations), that will take an input
The structure and the behavior will follow that of `tz` above, with
some small but interesting differences. We've included the orginal
`tz` to facilitate detailed comparison:
The structure and the behavior will follow that of `tz` above, with
some small but interesting differences. We've included the orginal
`tz` to facilitate detailed comparison:
| (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped)
| (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *)
| (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped)
| (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *)
- | [] -> List.rev (c [])
- | 'S'::zipped -> tc zipped (fun tail -> c (c tail))
- | target::zipped -> tc zipped (fun tail -> target::(c tail));;
+ | [] -> List.rev (k [])
+ | 'S'::zipped -> tc zipped (fun tail -> k (k tail))
+ | target::zipped -> tc zipped (fun tail -> target::(k tail));;
`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
`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
four recursive calls to `tc`, and `zipped` will take on the values
`"bSd"`, `"Sd"`, `"d"`, and `""` (and, once again, on the final call,
four recursive calls to `tc`, and `zipped` will take on the values
`"bSd"`, `"Sd"`, `"d"`, and `""` (and, once again, on the final call,
-I have not called the functional argument `unzipped`, although that is
-what the parallel would suggest. The reason is that `unzipped` is a
-list, but `c` is a function. That's the most crucial difference, the
+We have not named the functional 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
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
-relatively inefficient) `List.append`.
-In the `tc` version of the task, we simply compose `c` with itself:
-`c o c = fun tail -> c (c tail)`.
+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']` yields a partially-applied function; it still waits for another argument, a continuation of type `char list -> char list`. We have to give it an "initial continuation" to get started. Here we supply *the identity function* as the initial continuation. Why did we choose that? Well, if
you have already constructed the initial list `"abSd"`, what's the desired continuation? What's the next step in the recipe to produce the desired result, i.e, the very same list, `"abSd"`? Clearly, the identity function.
A good way to test your understanding is to figure out what the
A call `tc ['a'; 'b'; 'S'; 'd']` yields a partially-applied function; it still waits for another argument, a continuation of type `char list -> char list`. We have to give it an "initial continuation" to get started. Here we supply *the identity function* as the initial continuation. Why did we choose that? Well, if
you have already constructed the initial list `"abSd"`, what's the desired continuation? What's the next step in the recipe to produce the desired result, i.e, the very same list, `"abSd"`? Clearly, the identity function.
A good way to test your understanding is to figure out what the
-continuation function `c` must be at the point in the computation when
-`tc` is called with the first argument `"Sd"`. Two choices: is it
+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.
`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.
-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.) In the analogy, the input list portrays a
+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](http://www.cs.rutgers.edu/~ccshan/recur/recur.pdf),
+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
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
The reason the task is well-suited to the list zipper is in part
because the list monad has an intimate connection with continuations.
The reason the task is well-suited to the list zipper is in part
because the list monad has an intimate connection with continuations.