... 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. ##[[List Monad as Continuation Monad]]## ##[[Manipulating Trees with Monads]]##