-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 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'): - -
-    t "aSbS"
-        *
-~~> t "aabS"
-          *
-~~> "aabaab"
-
- -versus - -
-    t "aSbS"
-          *
-~~> t "aSbaSb"
-        *
-~~> t "aabaSb"
-           *
-~~> "aabaaabab"
-
- -versus - -
-    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 -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 zipped part by pulling the zipped down until the -entire list has been unzipped (and so the zipped half of the zipper is 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 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 `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`, -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'])
-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: - -
-(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 -be a function of type `char list -> char list`. We'll call each step -(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--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: - -
-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']
-
-	... f (u e) ...
-
- -This subexpression types to `'b reader`, which is good. The only -problem is that we made use of an environment `e` that we didn't already have, -so we must abstract over that variable to balance the books: - - fun e -> f (u e) ... - -[To preview the discussion of the Curry-Howard correspondence, what -we're doing here is constructing an intuitionistic proof of the type, -and using the Curry-Howard labeling of the proof as our bind term.] - -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
-
-let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
-               (Node ((Leaf 5),(Node ((Leaf 7),
-                                      (Leaf 11))))))
-
-    .
- ___|___
- |     |
- .     .
-_|__  _|__
-|  |  |  |
-2  3  5  .
-        _|__
-        |  |
-        7  11
-
- -Our first task will be to replace each leaf with its double: - -
-let rec treemap (newleaf:'a -> 'b) (t:'a tree):('b tree) =
-  match t with Leaf x -> Leaf (newleaf x)
-             | Node (l, r) -> Node ((treemap newleaf l),
-                                    (treemap newleaf r));;
-
-`treemap` takes a function that transforms old leaves into new leaves, -and maps that function over all the leaves in the tree, leaving the -structure of the tree unchanged. For instance: - -
-let double i = i + i;;
-treemap double t1;;
-- : int tree =
-Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
-
-    .
- ___|____
- |      |
- .      .
-_|__  __|__
-|  |  |   |
-4  6  10  .
-        __|___
-        |    |
-        14   22
-
- -We could have built the doubling operation right into the `treemap` -code. However, because what to do to each leaf is a parameter, we can -decide to do something else to the leaves without needing to rewrite -`treemap`. For instance, we can easily square each leaf instead by -supplying the appropriate `int -> int` operation in place of `double`: - -
-let square x = x * x;;
-treemap square t1;;
-- : int tree =ppp
-Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
-
- -Note that what `treemap` does is take some global, contextual -information---what to do to each leaf---and supplies that information -to each subpart of the computation. In other words, `treemap` has the -behavior of a reader monad. Let's make that explicit. - -In general, we're on a journey of making our treemap function more and -more flexible. So the next step---combining the tree transducer with -a reader monad---is to have the treemap function return a (monadized) -tree that is ready to accept any `int->int` function and produce the -updated tree. - -\tree (. (. (f2) (f3))(. (f5) (.(f7)(f11)))) -
-\f    .
-  ____|____
-  |       |
-  .       .
-__|__   __|__
-|   |   |   |
-f2  f3  f5  .
-          __|___
-          |    |
-          f7  f11
-
- -That is, we want to transform the ordinary tree `t1` (of type `int -tree`) into a reader object of type `(int->int)-> int tree`: something -that, when you apply it to an `int->int` function returns an `int -tree` in which each leaf `x` has been replaced with `(f x)`. - -With previous readers, we always knew which kind of environment to -expect: either an assignment function (the original calculator -simulation), a world (the intensionality monad), an integer (the -Jacobson-inspired link monad), etc. In this situation, it will be -enough for now to expect that our reader will expect a function of -type `int->int`. - -
-type 'a reader = (int->int) -> 'a;;  (* mnemonic: e for environment *)
-
- -It's easy to figure out how to turn an `int` into an `int reader`: - -
-let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
-int2int_reader 2 (fun i -> i + i);;
-- : int = 4
-
- -But what do we do when the integers are scattered over the leaves of a -tree? A binary tree is not the kind of thing that we can apply a -function of type `int->int` to. - -
-  match t with Leaf x -> reader_bind (f x) (fun x' -> reader_unit (Leaf x'))
-             | Node (l, r) -> reader_bind (treemonadizer f l) (fun x ->
-
- -This function says: give me a function `f` that knows how to turn -something of type `'a` into an `'b reader`, and I'll show you how to -turn an `'a tree` into an `'a tree reader`. In more fanciful terms, -the `treemonadizer` function builds plumbing that connects all of the -leaves of a tree into one connected monadic network; it threads the -monad through the leaves. - -
-- : int tree =
-Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
-
- -Here, our environment is the doubling function (`fun i -> i + i`). If -we apply the very same `int tree reader` (namely, `treemonadizer -int2int_reader t1`) to a different `int->int` function---say, the -squaring function, `fun i -> i * i`---we get an entirely different -result: - -
-- : int tree =
-Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
-
- -Now that we have a tree transducer that accepts a monad as a -parameter, we can see what it would take to swap in a different monad. -For instance, we can use a state monad to count the number of nodes in -the tree. - -
-type 'a state = int -> 'a * int;;
-let state_unit x i = (x, i+.5);;
-let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
-
- -Gratifyingly, we can use the `treemonadizer` function without any -modification whatsoever, except for replacing the (parametric) type -`reader` with `state`: - -
-let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state =
-  match t with Leaf x -> state_bind (f x) (fun x' -> state_unit (Leaf x'))
-             | Node (l, r) -> state_bind (treemonadizer f l) (fun x ->
-                                state_bind (treemonadizer f r) (fun y ->
-                                  state_unit (Node (x, y))));;
-
- -Then we can count the number of nodes in the tree: - -
-- : int tree * int =
-(Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13)
-
-    .
- ___|___
- |     |
- .     .
-_|__  _|__
-|  |  |  |
-2  3  5  .
-        _|__
-        |  |
-        7  11
-
- -Notice that we've counted each internal node twice---it's a good -exercise to adjust the code to count each node once. - -One more revealing example before getting down to business: replacing -`state` everywhere in `treemonadizer` with `list` gives us - -
-# treemonadizer (fun x -> [ [x; square x] ]) t1;;
-- : int list tree list =
-[Node
-  (Node (Leaf [2; 4], Leaf [3; 9]),
-   Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))]
-
- -Unlike the previous cases, instead of turning a tree into a function -from some input to a result, this transformer replaces each `int` with -a list of `int`'s. - -Now for the main point. What if we wanted to convert a tree to a list -of leaves? - -
-type ('a, 'r) continuation = ('a -> 'r) -> 'r;;
-let continuation_unit x c = c x;;
-let continuation_bind u f c = u (fun a -> f a c);;
-
-let rec treemonadizer (f:'a -> ('b, 'r) continuation) (t:'a tree):(('b tree), 'r) continuation =
-  match t with Leaf x -> continuation_bind (f x) (fun x' -> continuation_unit (Leaf x'))
-             | Node (l, r) -> continuation_bind (treemonadizer f l) (fun x ->
-                                continuation_bind (treemonadizer f r) (fun y ->
-                                  continuation_unit (Node (x, y))));;
-
- -We use the continuation monad described above, and insert the -`continuation` type in the appropriate place in the `treemonadizer` code. -We then compute: - -
-# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
-- : int list = [2; 3; 5; 7; 11]
-
- -We have found a way of collapsing a tree into a list of its leaves. - -The continuation monad is amazingly flexible; we can use it to -simulate some of the computations performed above. To see how, first -note that an interestingly uninteresting thing happens if we use the -continuation unit as our first argument to `treemonadizer`, and then -apply the result to the identity function: - -
-# treemonadizer continuation_unit t1 (fun x -> x);;
-- : int tree =
-Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
-
- -That is, nothing happens. But we can begin to substitute more -interesting functions for the first argument of `treemonadizer`: - -
-(* Simulating the tree reader: distributing a operation over the leaves *)
-# treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
-- : int tree =
-Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
-
-(* Simulating the int list tree list *)
-# treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
-- : int list tree =
-Node
- (Node (Leaf [2; 4], Leaf [3; 9]),
-  Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121])))
-
-(* Counting leaves *)
-# treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
-- : int = 5
-
- -We could simulate the tree state example too, but it would require -generalizing the type of the continuation monad to - - type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;; - -The binary tree monad ---------------------- - -Of course, by now you may have realized that we have discovered a new -monad, the binary tree monad: - -
-type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);;
-let tree_unit (x:'a) = Leaf x;;
-let rec tree_bind (u:'a tree) (f:'a -> 'b tree):'b tree =
-  match u with Leaf x -> f x
-             | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
-
- -For once, let's check the Monad laws. The left identity law is easy: - - Left identity: bind (unit a) f = bind (Leaf a) f = fa - -To check the other two laws, we need to make the following -observation: it is easy to prove based on `tree_bind` by a simple -induction on the structure of the first argument that the tree -resulting from `bind u f` is a tree with the same strucure as `u`, -except that each leaf `a` has been replaced with `fa`: - -\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5))) -
-                .                         .
-              __|__                     __|__
-              |   |                     |   |
-              a1  .                    fa1  .
-                 _|__                     __|__
-                 |  |                     |   |
-                 .  a5                    .  fa5
-   bind         _|__       f   =        __|__
-                |  |                    |   |
-                .  a4                   .  fa4
-              __|__                   __|___
-              |   |                   |    |
-              a2  a3                 fa2  fa3
-
- -Given this equivalence, the right identity law - - Right identity: bind u unit = u - -falls out once we realize that - - bind (Leaf a) unit = unit a = Leaf a - -As for the associative law, - - Associativity: bind (bind u f) g = bind u (\a. bind (fa) g) - -we'll give an example that will show how an inductive proof would -proceed. Let `f a = Node (Leaf a, Leaf a)`. Then - -\tree (. (. (. (. (a1)(a2))))) -\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) )) -
-                                           .
-                                       ____|____
-          .               .            |       |
-bind    __|__   f  =    __|_    =      .       .
-        |   |           |   |        __|__   __|__
-        a1  a2         fa1 fa2       |   |   |   |
-                                     a1  a1  a1  a1
-
- -Now when we bind this tree to `g`, we get - -
-           .
-       ____|____
-       |       |
-       .       .
-     __|__   __|__
-     |   |   |   |
-    ga1 ga1 ga1 ga1
-