+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.