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:

(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 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']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 (and relatively 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 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 `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. 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 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 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. Rethinking the list monad ------------------------- To construct a monad, the key element is to settle on a type constructor, and the monad more or less 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 The reason this is a fairly natural choice is that because the type of an `'a reader` is `env -> 'a` (by definition), the type of the `r_unit` function is `'a -> env -> 'a`, which is an instance 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 traditional reader `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 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) eAnd 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. Recall that `List.map` takes a function and a list and returns the result to applying the function to the elements of the list: List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]] and List.concat takes a list of lists and erases the embdded list boundaries: List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3] And sure enough, l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] Now, why this unit, and why this bind? Well, ideally a unit should not throw away information, so we can rule out `fun x -> []` as an ideal unit. And units should not add more information than required, so there's no obvious reason to prefer `fun x -> [x,x]`. In other words, `fun x -> [x]` is a reasonable choice for a unit. As for bind, an `'a list` monadic object contains a lot of objects of type `'a`, and we want to make use of each of them (rather than arbitrarily throwing some of them away). The only thing we know for sure we can do with an object of type `'a` is apply the function of type `'a -> 'a list` to them. Once we've done so, we have a collection of lists, one for each of the `'a`'s. One possibility is that we could gather them all up in a list, so that `bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts the object returned by the second argument of `bind` to always be of type `'b list list`. We can elimiate that restriction by flattening the list of lists into a single list: this is just List.concat applied to the output of List.map. So there is some logic to the choice of unit and bind for the list monad. Yet we can still desire to go deeper, and see if the appropriate bind behavior emerges from the types, as it did for the previously considered monads. But we can't do that if we leave the list type as a primitive Ocaml type. However, we know several ways of implementing lists using just functions. In what follows, we're going to use type 3 lists, the right fold implementation (though it's important and intriguing to wonder how things would change if we used some other strategy for implementating lists). These were the lists that made lists look like Church numerals with extra bits embdded in them: empty list: fun f z -> z list with one element: fun f z -> f 1 z list with two elements: fun f z -> f 2 (f 1 z) list with three elements: fun f z -> f 3 (f 2 (f 1 z)) and so on. To save time, we'll let the OCaml interpreter infer the principle types of these functions (rather than inferring what the types should be ourselves): # fun f z -> z;; - : 'a -> 'b -> 'b =

let t1 = Node ((Node ((Leaf 2), (Leaf 3))), (Node ((Leaf 5),(Node ((Leaf 7), (Leaf 11)))))) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Our 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 22We 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 f11That 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 *) let reader_unit (x:'a): 'a reader = fun _ -> x;; let reader_bind (u: 'a reader) (f:'a -> 'c reader):'c reader = fun e -> f (u e) e;;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 = 4But 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.

let rec treemonadizer (f:'a -> 'b reader) (t:'a tree):('b tree) reader = 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 -> reader_bind (treemonadizer f r) (fun y -> reader_unit (Node (x, y))));;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.

# treemonadizer int2int_reader t1 (fun i -> i + i);; - : 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:

# treemonadizer int2int_reader t1 (fun i -> i * i);; - : 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:

# treemonadizer state_unit t1 0;; - : int tree * int = (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 13) . ___|___ | | . . _|__ _|__ | | | | 2 3 5 . _|__ | | 7 11Notice 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 = 5We 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 fa3Given 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 a1Now when we bind this tree to `g`, we get

. ____|____ | | . . __|__ __|__ | | | | ga1 ga1 ga1 ga1At this point, it should be easy to convince yourself that using the recipe on the right hand side of the associative law will built the exact same final tree. So binary trees are a monad. Haskell combines this monad with the Option monad to provide a monad called a [SearchTree](http://hackage.haskell.org/packages/archive/tree-monad/0.2.1/doc/html/src/Control-Monad-SearchTree.html#SearchTree) that is intended to represent non-deterministic computations as a tree.