-Let's work with lists of chars for a change. To maximize readability, we'll
-indulge in an abbreviatory convention that "abSd" abbreviates the
-list `['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:
-
-<pre>
-t "abSd" ~~> "ababd"
-</pre>
-
-
-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 deceptively simple task gives rise to some mind-bending complexity.
-Note that it matters which 'S' you target first (the position of the *
-indicates the targeted 'S'):
-
-<pre>
- t "aSbS"
- *
-~~> t "aabS"
- *
-~~> "aabaab"
-</pre>
-
-versus
-
-<pre>
- t "aSbS"
- *
-~~> t "aSbaSb"
- *
-~~> t "aabaSb"
- *
-~~> "aabaaabab"
-</pre>
-
-versus
-
-<pre>
- t "aSbS"
- *
-~~> t "aSbaSb"
- *
-~~> t "aSbaaSbab"
- *
-~~> t "aSbaaaSbaabab"
- *
-~~> ...
-</pre>
-
-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
-behavior.
-
-For now, we'll agree to always evaluate the leftmost `'S'`.
-
-This is a task well-suited to using a zipper. We'll define a function
-`tz`, which accomplished 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 zipped part 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>
-
-Note that this implementation 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 `t1`. By using the `#trace`
-directive in the Ocaml interpreter, the system will print out the
-arguments to `t1` each time it is (recurcively) 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
-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>
-
-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>
-
-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:
-
-<pre>
-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>
-
-To emphasize the parallel, I'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 `zipper` will
-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
-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)`.
-
-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.
-
-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
-
-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
-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`. &sset; &integral; 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.
-
-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).
-
-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.
-
-
-Rethinking the list monad
--------------------------
-
-To construct a monad, the key element is to settle on a type
-constructor, and the monad naturally follows from that. We'll remind
-you of some examples of how monads follow from the type constructor in
-a moment. This will involve some review of familair material, but
-it's worth doing for two reasons: it will set up a pattern for the new
-discussion further below, and it will tie together some previously
-unconnected elements of the course (more specifically, version 3 lists
-and monads).
-
-For instance, take the **Reader Monad**. Once we decide that the type
-constructor is
-
- type 'a reader = env -> 'a
-
-then the choice of unit and bind is natural:
-
- let r_unit (a : 'a) : 'a reader = fun (e : env) -> a
-
-Since the type of an `'a reader` is `env -> 'a` (by definition),
-the type of the `r_unit` function is `'a -> env -> 'a`, which is a
-specific case of the type of the *K* combinator. So it makes sense
-that *K* is the unit for the reader monad.
-
-Since the type of the `bind` operator is required to be
-
- r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader)
-
-We can reason our way to the correct `bind` function as follows. We
-start by declaring the types determined by the definition of a bind operation:
-
- let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ...
-
-Now we have to open up the `u` box and get out the `'a` object in order to
-feed it to `f`. Since `u` is a function from environments to
-objects of type `'a`, the way we open a box in this monad is
-by applying it to an environment:
-
- ... f (u e) ...
-
-This subexpression types to `'b reader`, which is good. The only
-problem is that we invented an environment `e` that we didn't already have ,
-so we have to abstract over that variable to balance the books:
-
- fun e -> f (u e) ...
-
-This types to `env -> 'b reader`, but we want to end up with `env ->
-'b`. Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment. So we end up as follows:
-
- r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) =
- f (u e) e
-
-And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does.
-
-[The bind we cite here is a condensed version of the careful `let a = u e in ...`
-constructions we provided in earlier lectures. We use the condensed
-version here in order to emphasize similarities of structure across
-monads.]
-
-The **State Monad** is similar. Once we've decided to use the following type constructor:
-
- type 'a state = store -> ('a, store)
-
-Then our unit is naturally:
-
- let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
-
-And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box:
-
- let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
- ... f (...) ...
-
-But unlocking the `u` box is a little more complicated. As before, we
-need to posit a state `s` that we can apply `u` to. Once we do so,
-however, we won't have an `'a`, we'll have a pair whose first element
-is an `'a`. So we have to unpack the pair:
-
- ... let (a, s') = u s in ... (f a) ...
-
-Abstracting over the `s` and adjusting the types gives the result:
-
- let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state =
- fun (s : store) -> let (a, s') = u s in f a s'
-
-The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we
-won't pause to explore it here, though conceptually its unit and bind
-follow just as naturally from its type constructor.
-
-Our other familiar monad is the **List Monad**, which we were told
-looks like this:
-
- type 'a list = ['a];;
- l_unit (a : 'a) = [a];;
- l_bind u f = List.concat (List.map f u);;
-
-Thinking through the list monad will take a little time, but doing so
-will provide a connection with continuations.