-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 *) -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 = 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. - -

-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 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 -- -At 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. - - -Refunctionalizing zippers: from lists to continuations ------------------------------------------------------- - -Let's work with lists of chars for a change. To maximize readability, we'll -indulge in an abbreviatory convention that "abc" abbreviates the -list `['a'; 'b'; 'c']`. - -Task 1: replace each occurrence of 'S' with a copy of the string up to -that point. - -Expected behavior: - -

-t1 "abSe" ~~> "ababe" -- - -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'): - -

- t1 "aSbS" - * -~~> t1 "aabS" - * -~~> "aabaab" -- -versus - -

- t1 "aSbS" - * -~~> t1 "aSbaSb" - * -~~> t1 "aabaSb" - * -~~> "aabaaabab" -- -versus - -

- t1 "aSbS" - * -~~> t1 "aSbaSb" - * -~~> t1 "aSbaaSbab" - * -~~> t1 "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, as usual, we'll agree to always evaluate the leftmost `'S'`. - -This is a task well-suited to using a zipper. - -

-type 'a list_zipper = ('a list) * ('a list);; - -let rec t1 (z:char list_zipper) = - match z with (sofar, []) -> List.rev(sofar) (* Done! *) - | (sofar, 'S'::rest) -> t1 ((List.append sofar sofar), rest) - | (sofar, fst::rest) -> t1 (fst::sofar, rest);; (* Move zipper *) - -# t1 ([], ['a'; 'b'; 'S'; 'e']);; -- : char list = ['a'; 'b'; 'a'; 'b'; 'e'] - -# t1 ([], ['a'; 'S'; 'b'; 'S']);; -- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -- -Note that this implementation enforces the evaluate-leftmost rule. -Task 1 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: - -

-# #trace t1;; -t1 is now traced. -# t1 ([], ['a'; 'b'; 'S'; 'e']);; -t1 <-- ([], ['a'; 'b'; 'S'; 'e']) -t1 <-- (['a'], ['b'; 'S'; 'e']) -t1 <-- (['b'; 'a'], ['S'; 'e']) -t1 <-- (['b'; 'a'; 'b'; 'a'], ['e']) -t1 <-- (['e'; 'b'; 'a'; 'b'; 'a'], []) -t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] -t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] -t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] -t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] -t1 --> ['a'; 'b'; 'a'; 'b'; 'e'] -- : char list = ['a'; 'b'; 'a'; 'b'; 'e'] -- -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'; 'c']` means (1) -start with the empty list `[]`; (2) make a new list whose first -element is 'c' and whose tail is the list construted in the previous -step; (3) make a new list whose first element is 'b' and whose tail is -the list constructed in the previous step; and (4) make a new list -whose first element is 'a' and whose tail is the list constructed in -the previous step. - -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`. - -This means that we can now represent the sofar part of our zipper--the -part we've already unzipped--as a continuation, a function describing -how to finish building the list: - -

-let rec t1c (l: char list) (c: (char list) -> (char list)) = - match l with [] -> c [] - | 'S'::rest -> t1c rest (fun x -> c (c x)) - | a::rest -> t1c rest (fun x -> List.append (c x) [a]);; - -# t1c ['a'; 'b'; 'S'] (fun x -> x);; -- : char list = ['a'; 'b'; 'a'; 'b'] - -# t1c ['a'; 'S'; 'b'; 'S'] (fun x -> x);; -- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -- -Note that we don't need to do any reversing. -