Expected behavior:
-<pre>
-t "abSd" ~~> "ababd"
-</pre>
+ t "abSd" ~~> "ababd"
In linguistic terms, this is a kind of anaphora
Note that it matters which 'S' you target first (the position of the *
indicates the targeted 'S'):
-<pre>
- t "aSbS"
- *
-~~> t "aabS"
- *
-~~> "aabaab"
-</pre>
+ t "aSbS"
+ *
+ ~~> t "aabS"
+ *
+ ~~> "aabaab"
versus
-<pre>
- t "aSbS"
- *
-~~> t "aSbaSb"
- *
-~~> t "aabaSb"
- *
-~~> "aabaaabab"
-</pre>
+ t "aSbS"
+ *
+ ~~> t "aSbaSb"
+ *
+ ~~> t "aabaSb"
+ *
+ ~~> "aabaaabab"
versus
-<pre>
- t "aSbS"
- *
-~~> t "aSbaSb"
- *
-~~> t "aSbaaSbab"
- *
-~~> t "aSbaaaSbaabab"
- *
-~~> ...
-</pre>
+ t "aSbS"
+ *
+ ~~> t "aSbaSb"
+ *
+ ~~> t "aSbaaSbab"
+ *
+ ~~> t "aSbaaaSbaabab"
+ *
+ ~~> ...
Aparently, 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
move elements to the zipped part by pulling the zipper down until the
entire list has been unzipped (and so the zipped half of the zipper is empty).
-<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 *)
-
-# 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']
-</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 *)
+
+ # 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 this implementation enforces the evaluate-leftmost rule.
Task completed.
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
-simple list.
-
-<pre>
-# #trace tz;;
-t1 is now traced.
-# tz ([], ['a'; 'b'; 'S'; 'd']);;
-tz <-- ([], ['a'; 'b'; 'S'; 'd'])
-tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *)
-tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *)
-tz <-- (['b'; 'a'; 'b'; 'a'], ['d']) (* Special 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']
-</pre>
+simple list.
+
+ # #trace tz;;
+ t1 is now traced.
+ # tz ([], ['a'; 'b'; 'S'; 'd']);;
+ tz <-- ([], ['a'; 'b'; 'S'; 'd'])
+ tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *)
+ tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *)
+ tz <-- (['b'; 'a'; 'b'; 'a'], ['d']) (* Special 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.
-
-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)))`).
-The recipe for constructing the list goes like this:
-
-<pre>
-(0) Start with the empty list []
-(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 'S' 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)
-</pre>
+concrete zipper using procedures.
+
+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,
+`make_list 'a' (make_list 'b' (make_list 'S' (make_list 'd' empty)))`). The
+recipe for constructing the list goes like this:
+
+> (0) Start with the empty list []
+> (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 'S' 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
(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)`.
+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
+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.
some small but interesting differences. We've included the orginal
`tz` to facilitate detailed comparison:
-<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 *)
-
-let rec tc (l: char list) (c: (char list) -> (char list)) =
- match l with [] -> List.rev (c [])
- | 'S'::zipped -> tc zipped (fun x -> c (c x))
- | target::zipped -> tc zipped (fun x -> target::(c x));;
-
-# tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);;
-- : char list = ['a'; 'b'; 'a'; 'b']
-
-# tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);;
-- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
-</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 *)
+
+ let rec tc (l: char list) (c: (char list) -> (char list)) =
+ match l with
+ | [] -> List.rev (c [])
+ | 'S'::zipped -> tc zipped (fun x -> c (c x))
+ | target::zipped -> tc zipped (fun x -> target::(c x));;
+
+ # tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);;
+ - : char list = ['a'; 'b'; 'a'; 'b']
+
+ # tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);;
+ - : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
To emphasize the parallel, I've re-used the names `zipped` and
`target`. The trace of the procedure will show that these variables
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:
+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
list task after talking about generalized quantifiers below.
-