edits

[lambda.git] / from_lists_to_continuations.mdwn

diff --git a/from_lists_to_continuations.mdwn b/from_lists_to_continuations.mdwn
index 0b3f464..3dc6dde 100644 (file)
--- a/from_lists_to_continuations.mdwn
+++ b/from_lists_to_continuations.mdwn
@@ -1,4 +1,3 @@
-

 Refunctionalizing zippers: from lists to continuations
 ------------------------------------------------------
 
 Refunctionalizing zippers: from lists to continuations
 ------------------------------------------------------
 


@@ -29,7 +28,7 @@ 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.
 
 resolution, where `'S'` is functioning like an anaphoric element, and
 the preceding string portion is the antecedent.
 

-This deceptively simple task gives rise to some mind-bending complexity.
+This simple task gives rise to considerable complexity.

 Note that it matters which 'S' you target first (the position of the *
 indicates the targeted 'S'):
 
 Note that it matters which 'S' you target first (the position of the *
 indicates the targeted 'S'):
 


@@ -67,7 +66,7 @@ versus
 ~~> ...
 </pre>
 
 ~~> ...
 </pre>
 

-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,

 and the order in which the `'S'`s get evaluated can lead to divergent
 behavior.
 
 and the order in which the `'S'`s get evaluated can lead to divergent
 behavior.
 


@@ -78,16 +77,17 @@ 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
 `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 zipped part by pulling the zipped down until the
-entire list has been unzipped (and so the zipped half of the zipper is empty).
+move elements from the zipped part 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).

 
 <pre>
 type 'a list_zipper = ('a list) * ('a list);;
 
 let rec tz (z:char list_zipper) = 
 
 <pre>
 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 *)
+  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']


@@ -102,7 +102,7 @@ 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
 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
+arguments to `tz` each time it is (recursively) called.  Note that the

 lines with left-facing arrows (`<--`) show (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
 lines with left-facing arrows (`<--`) show (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


@@ -129,11 +129,11 @@ 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
 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.  
+concrete zippers using procedures instead.

 
 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
 
 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, `makelist a (makelist b (makelist S (makelist c empty)))`).
+style, `makelist 'a' (makelist 'b' (makelist 'S' (makelist 'c' empty)))`).

 The recipe for constructing the list goes like this:
 
 <pre>
 The recipe for constructing the list goes like this:
 
 <pre>


@@ -148,25 +148,25 @@ The recipe for constructing the list goes like this:
 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

-a **continuation** of the recipe.  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)`.
-
-This means that we can now represent the unzipped part of our
-zipper--the part we've already unzipped--as a continuation: a function
-describing how to finish building the list.  We'll write a new
-function, `tc` (for task with continuations), that will take an input
-list (not a zipper!) and a continuation and return a processed list.
-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:
+(or group of steps) a **continuation** of the recipe.  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)`.
+
+This means that we can now represent the unzipped part of our zipper
+as a continuation: a function describing how to finish building the
+list.  We'll write a new function, `tc` (for task with continuations),
+that will take an input list (not a zipper!) and a continuation and
+return a processed list.  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:

 
 <pre>
 let rec tz (z:char list_zipper) = 
 
 <pre>
 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 *)
+  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) (c: (char list) -> (char list)) =
   match l with [] -> List.rev (c [])
 
 let rec tc (l: char list) (c: (char list) -> (char list)) =
   match l with [] -> List.rev (c [])


@@ -174,13 +174,13 @@ let rec tc (l: char list) (c: (char list) -> (char list)) =
              | target::zipped -> tc zipped (fun x -> target::(c x));;
 
 # tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);;
              | target::zipped -> tc zipped (fun x -> target::(c x));;
 
 # tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);;

-- : char list = ['a'; 'b'; 'a'; 'b']
+- : char list = ['a'; 'b'; 'a'; 'b'; 'd']

 
 # tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);;
 - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
 </pre>
 
 
 # tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);;
 - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
 </pre>
 

-To emphasize the parallel, I've re-used the names `zipped` and
+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
 `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


@@ -190,46 +190,49 @@ the first `match` clause will fire, so the the variable `zipper` will
 not be instantiated).
 
 I have not called the functional argument `unzipped`, although that is
 not be instantiated).
 
 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
+what the parallel would suggest.  The reason is that `unzipped` (in
+`tz`) is a list, but `c` (in `tc`) is a function.  ('c' stands for
+'continuation', of course.)  That's the most crucial difference, 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 `List.append`.
-In the `tc` version of the task, we simply compose `c` with itself: 
-`c o c = fun x -> c (c x)`.
+together the two instances of `unzipped` with an explicit (and
+relatively computationally inefficient) `List.append`.  In the `tc`
+version of the task, we simply compose `c` with itself: `c o c = fun x
+-> c (c x)`.

 
 Why use the identity function as the initial continuation?  Well, if
 
 Why use the identity function as the initial continuation?  Well, if

-you have already constructed the list "abSd", what's the next step in
-the recipe to produce the desired result (which is the same list,
-"abSd")?  Clearly, the identity continuation.
+you have already constructed the initial list `"abSd"`, what's the next
+step in the recipe to produce the desired result, i.e, the very same
+list, `"abSd"`?  Clearly, the identity continuation.

 
 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
 
 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

-`fun x -> a::b::x`, or it is `fun x -> b::a::x`?  
-The way to see if you're right is to execute the following 
-command and see what happens:
+`fun x -> a::b::x`, or it is `fun x -> b::a::x`?  The way to see if
+you're right is to execute the following command and see what happens:

 
     tc ['S'; 'd'] (fun x -> 'a'::'b'::x);;
 
 There are a number of interesting directions we can go with this task.
 
     tc ['S'; 'd'] (fun x -> 'a'::'b'::x);;
 
 There are a number of interesting directions we can go with this task.

-The task was chosen because the computation can be viewed as a
+The reason this task was chosen is because it can be viewed as a

 simplified picture of a computation using continuations, where `'S'`
 plays the role of a control operator with some similarities to what is
 simplified picture of a computation using continuations, where `'S'`
 plays the role of a control operator with some similarities to what is

-often called `shift`.  In the analogy, the list portrays a string 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.
+often called `shift`.  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 possibile development is that we could add a special symbol `'#'`,
 and then the task would be to copy from the target `'S'` only back to
 the closest `'#'`.  This would allow the task to simulate delimited
 
 One possibile development is that we could add a special symbol `'#'`,
 and then the task would be to copy from the target `'S'` only back to
 the closest `'#'`.  This would allow the task to simulate delimited

-continuations (for right-branching computations).
+continuations with embedded prompts.
+
+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 following section explores this connection.  We'll return to the
+list task after talking about generalized quantifiers below.
+

 
 

-The task is well-suited to the list zipper because the list monad has
-an intimate connection with continuations.  The following section
-makes this connection.  We'll return to the list task after talking
-about generalized quantifiers below.