[[!toc]] Manipulating trees with monads ------------------------------ This topic develops an idea based on a suggestion of Ken Shan's. We'll build a series of functions that operate on trees, doing various things, including updating leaves with a Reader monad, counting nodes with a State monad, replacing leaves with a List monad, and converting a tree into a list of leaves with a Continuation monad. It will turn out that the continuation monad can simulate the behavior of each of the other monads. From an engineering standpoint, we'll build a tree transformer that deals in monads. We can modify the behavior of the system by swapping one monad for another. We've already seen how adding a monad can add a layer of funtionality without disturbing the underlying system, for instance, in the way that the Reader monad allowed us to add a layer of intensionality to an extensional grammar, but we have not yet seen the utility of replacing one monad with other. First, we'll be needing a lot of trees for the remainder of the course. Here again is a type constructor for leaf-labeled, binary trees: type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree);; [How would you adjust the type constructor to allow for labels on the internal nodes?] We'll be using trees where the nodes are integers, e.g., 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 tree_map (t : 'a tree) (leaf_modifier : 'a -> 'b): 'b tree = match t with | Leaf i -> Leaf (leaf_modifier i) | Node (l, r) -> Node (tree_map l leaf_modifier, tree_map r leaf_modifier);; `tree_map` takes a tree and 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;; tree_map t1 double;; - : 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 `tree_map` code. However, because we've made what to do to each leaf a parameter, we can decide to do something else to the leaves without needing to rewrite `tree_map`. For instance, we can easily square each leaf instead by supplying the appropriate `int -> int` operation in place of `double`: let square i = i * i;; tree_map t1 square;; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) Note that what `tree_map` does is take some unchanging contextual information---what to do to each leaf---and supplies that information to each subpart of the computation. In other words, `tree_map` has the behavior of a Reader monad. Let's make that explicit. In general, we're on a journey of making our `tree_map` function more and more flexible. So the next step---combining the tree transformer with a Reader monad---is to have the `tree_map` function return a (monadized) tree that is ready to accept any `int -> int` function and produce the updated tree. \f . _____|____ | | . . __|___ __|___ | | | | f 2 f 3 f 5 . __|___ | | f 7 f 11 That is, we want to transform the ordinary tree `t1` (of type `int tree`) into a reader monadic object of type `(int -> int) -> int tree`: something that, when you apply it to an `int -> int` function `f` returns an `int tree` in which each leaf `i` has been replaced with `f i`. [Application note: this kind of reader object could provide a model for Kaplan's characters. It turns an ordinary tree into one that expects contextual information (here, the `\f`) that can be used to compute the content of indexicals embedded arbitrarily deeply in the tree.] With our previous applications of the Reader monad, we always knew which kind of environment to expect: either an assignment function, as in the original calculator simulation; a world, as in the intensionality monad; an individual, as in the Jacobson-inspired link monad; etc. In the present case, we expect that our "environment" will be some function of type `int -> int`. "Looking up" some `int` in the environment will return us the `int` that comes out the other side of that function. type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *) let reader_unit (a : 'a) : 'a reader = fun _ -> a;; let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;; It would be a simple matter to turn an *integer* into an `int reader`: let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;; int_readerize 2 (fun i -> i + i);; - : int = 4 But how do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader? A tree is not the kind of thing that we can apply a function of type `int -> int` to. But we can do this: let rec tree_monadize (t : 'a tree) (f : 'a -> 'b reader) : 'b tree reader = match t with | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b)) | Node (l, r) -> reader_bind (tree_monadize l f) (fun l' -> reader_bind (tree_monadize r f) (fun r' -> reader_unit (Node (l', r'))));; This function says: give me a function `f` that knows how to turn something of type `'a` into an `'b reader`---this is a function of the same type that you could bind an `'a reader` to---and I'll show you how to turn an `'a tree` into an `'b tree reader`. That is, if you show me how to do this: ------------ 1 ---> | 1 | ------------ then I'll give you back the ability to do this: ____________ . | . | __|___ ---> | __|___ | | | | | | | 1 2 | 1 2 | ------------ And how will that boxed tree behave? Whatever actions you perform on it will be transmitted down to corresponding operations on its leaves. For instance, our `int reader` expects an `int -> int` environment. If supplying environment `e` to our `int reader` doubles the contained `int`: ------------ 1 ---> | 1 | applied to e ~~> 2 ------------ Then we can expect that supplying it to our `int tree reader` will double all the leaves: ____________ . | . | . __|___ ---> | __|___ | applied to e ~~> __|___ | | | | | | | | 1 2 | 1 2 | 2 4 ------------ In more fanciful terms, the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the `'b reader` monad through the original tree's leaves. # tree_monadize t1 int_readerize double;; - : 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, `tree_monadize t1 int_readerize`) to a different `int -> int` function---say, the squaring function, `fun i -> i * i`---we get an entirely different result: # tree_monadize t1 int_readerize square;; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) Now that we have a tree transformer that accepts a *reader* 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 leaves in the tree. type 'a state = int -> 'a * int;; let state_unit a = fun s -> (a, s);; let state_bind u f = fun s -> let (a, s') = u s in f a s';; Gratifyingly, we can use the `tree_monadize` function without any modification whatsoever, except for replacing the (parametric) type `'b reader` with `'b state`, and substituting in the appropriate unit and bind: let rec tree_monadize (t : 'a tree) (f : 'a -> 'b state) : 'b tree state = match t with | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b)) | Node (l, r) -> state_bind (tree_monadize l f) (fun l' -> state_bind (tree_monadize r f) (fun r' -> state_unit (Node (l', r'))));; Then we can count the number of leaves in the tree: # tree_monadize t1 (fun a -> fun s -> (a, s+1)) 0;; - : int tree * int = (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5) . ___|___ | | . . _|__ _|__ , 5 | | | | 2 3 5 . _|__ | | 7 11 Note that the value returned is a pair consisting of a tree and an integer, 5, which represents the count of the leaves in the tree. Why does this work? Because the operation `fun a -> fun s -> (a, s+1)` takes an `int` and wraps it in an `int state` monadic box that increments the state. When we give that same operations to our `tree_monadize` function, it then wraps an `int tree` in a box, one that does the same state-incrementing for each of its leaves. We can use the state monad to replace leaves with a number corresponding to that leave's ordinal position. When we do so, we reveal the order in which the monadic tree forces evaluation: # tree_monadize t1 (fun a -> fun s -> (s+1, s+1)) 0;; - : int tree * int = (Node (Node (Leaf 1, Leaf 2), Node (Leaf 3, Node (Leaf 4, Leaf 5))), 5) The key thing to notice is that instead of copying `a` into the monadic box, we throw away the `a` and put a copy of the state in instead. Reversing the order requires reversing the order of the state_bind operations. It's not obvious that this will type correctly, so think it through: let rec tree_monadize_rev (t : 'a tree) (f : 'a -> 'b state) : 'b tree state = match t with | Leaf a -> state_bind (f a) (fun b -> state_unit (Leaf b)) | Node (l, r) -> state_bind (tree_monadize r f) (fun r' -> (* R first *) state_bind (tree_monadize l f) (fun l'-> (* Then L *) state_unit (Node (l', r'))));; # tree_monadize_rev t1 (fun a -> fun s -> (s+1, s+1)) 0;; - : int tree * int = (Node (Node (Leaf 5, Leaf 4), Node (Leaf 3, Node (Leaf 2, Leaf 1))), 5) We will need below to depend on controlling the order in which nodes are visited when we use the continuation monad to solve the same-fringe problem. One more revealing example before getting down to business: replacing `state` everywhere in `tree_monadize` with `list` gives us # tree_monadize t1 (fun i -> [ [i; square i] ]);; - : 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. We might also have done this with a Reader monad, though then our environments would need to be of type `int -> int list`. Experiment with what happens if you supply the `tree_monadize` based on the List monad an operation like `fun i -> [2*i; 3*i]`. Use small trees for your experiment. [Why is the argument to `tree_monadize` `int -> int list list` instead of `int -> int list`? Well, as usual, the List monad bind operation will erase the outer list box, so if we want to replace the leaves with lists, we have to nest the replacement lists inside a disposable box.] 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 a = fun k -> k a;; let continuation_bind u f = fun k -> u (fun a -> f a k);; let rec tree_monadize (t : 'a tree) (f : 'a -> ('b, 'r) continuation) : ('b tree, 'r) continuation = match t with | Leaf a -> continuation_bind (f a) (fun b -> continuation_unit (Leaf b)) | Node (l, r) -> continuation_bind (tree_monadize l f) (fun l' -> continuation_bind (tree_monadize r f) (fun r' -> continuation_unit (Node (l', r'))));; We use the Continuation monad described above, and insert the `continuation` type in the appropriate place in the `tree_monadize` code. Then if we give the `tree_monadize` function an operation that converts `int`s into `'b`-wrapping Continuation monads, it will give us back a way to turn `int tree`s into corresponding `'b tree`-wrapping Continuation monads. So for example, we compute: # tree_monadize t1 (fun a k -> a :: k ()) (fun _ -> []);; - : int list = [2; 3; 5; 7; 11] We have found a way of collapsing a tree into a list of its leaves. Can you trace how this is working? Think first about what the operation `fun a k -> a :: k a` does when you apply it to a plain `int`, and the continuation `fun _ -> []`. Then given what we've said about `tree_monadize`, what should we expect `tree_monadize (fun a -> fun k -> a :: k a` to do? Soon we'll return to the same-fringe problem. Since the simple but inefficient way to solve it is to map each tree to a list of its leaves, this transformation is on the path to a more efficient solution. We'll just have to figure out how to postpone computing the tail of the list until it's needed... 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 `continuation_unit` as our first argument to `tree_monadize`, and then apply the result to the identity function: # tree_monadize t1 continuation_unit (fun t -> t);; - : 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 `tree_monadize`: (* Simulating the tree reader: distributing a operation over the leaves *) # tree_monadize t1 (fun a -> fun k -> k (square a)) (fun t -> t);; - : int tree = Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) (* Simulating the int list tree list *) # tree_monadize t1 (fun a -> fun k -> k [a; square a]) (fun t -> t);; - : int list tree = Node (Node (Leaf [2; 4], Leaf [3; 9]), Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121]))) (* Counting leaves *) # tree_monadize t1 (fun a -> fun k -> 1 + k a) (fun t -> 0);; - : int = 5 [To be fixed: exactly which kind of monad each of these computations simulates.] We could simulate the tree state example too by setting the relevant type to `('a, 'state -> 'result) continuation`. In fact, Andre Filinsky has suggested that the continuation monad is able to simulate any other monad (Google for "mother of all monads"). We would eventually want to generalize the continuation type to type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;; If you want to see how to parameterize the definition of the `tree_monadize` function, so that you don't have to keep rewriting it for each new monad, see [this code](/code/tree_monadize.ml). The idea of using continuations to characterize natural language meaning ------------------------------------------------------------------------ We might a philosopher or a linguist be interested in continuations, especially if efficiency of computation is usually not an issue? Well, the application of continuations to the same-fringe problem shows that continuations can manage order of evaluation in a well-controlled manner. In a series of papers, one of us (Barker) and Ken Shan have argued that a number of phenomena in natural langauge semantics are sensitive to the order of evaluation. We can't reproduce all of the intricate arguments here, but we can give a sense of how the analyses use continuations to achieve an analysis of natural language meaning. **Quantification and default quantifier scope construal**. We saw in the copy-string example and in the same-fringe example that local properties of a tree (whether a character is `S` or not, which integer occurs at some leaf position) can control global properties of the computation (whether the preceeding string is copied or not, whether the computation halts or proceeds). Local control of surrounding context is a reasonable description of in-situ quantification. (1) John saw everyone yesterday. This sentence means (roughly) forall x . yesterday(saw x) john That is, the quantifier *everyone* contributes a variable in the direct object position, and a universal quantifier that takes scope over the whole sentence. If we have a lexical meaning function like the following:
```let lex (s:string) k = match s with
| "everyone" -> Node (Leaf "forall x", k "x")
| "someone" -> Node (Leaf "exists y", k "y")
| _ -> k s;;

let sentence1 = Node (Leaf "John",
Node (Node (Leaf "saw",
Leaf "everyone"),
Leaf "yesterday"));;
```
Then we can crudely approximate quantification as follows:
```# tree_monadize sentence1 lex (fun x -> x);;
- : string tree =
Node
(Leaf "forall x",
Node (Leaf "John", Node (Node (Leaf "saw", Leaf "x"), Leaf "yesterday")))
```
In order to see the effects of evaluation order, observe what happens when we combine two quantifiers in the same sentence:
```# let sentence2 = Node (Leaf "everyone", Node (Leaf "saw", Leaf "someone"));;
# tree_monadize sentence2 lex (fun x -> x);;
- : string tree =
Node
(Leaf "forall x",
Node (Leaf "exists y", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
```
The universal takes scope over the existential. If, however, we replace the usual tree_monadizer with tree_monadizer_rev, we get inverse scope:
```# tree_monadize_rev sentence2 lex (fun x -> x);;
- : string tree =
Node
(Leaf "exists y",
Node (Leaf "forall x", Node (Leaf "x", Node (Leaf "saw", Leaf "y"))))
```
There are many crucially important details about quantification that are being simplified here, and the continuation treatment here is not scalable for a number of reasons. Nevertheless, it will serve to give an idea of how continuations can provide insight into the behavior of quantifiers. The Binary Tree monad --------------------- Of course, by now you may have realized that we have discovered a new monad, the Binary Tree monad. Just as mere lists are in fact a monad, so are trees. Here is the type constructor, unit, and bind: type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);; let tree_unit (a: 'a) : 'a tree = Leaf a;; let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree = match u with | Leaf a -> f a | 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 = f a 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 `f a`: . . __|__ __|__ | | | | a1 . f a1 . _|__ __|__ | | | | . a5 . f a5 bind _|__ f = __|__ | | | | . a4 . f a4 __|__ __|___ | | | | a2 a3 f a2 f a3 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 (f a) g) we'll give an example that will show how an inductive proof would proceed. Let `f a = Node (Leaf a, Leaf a)`. Then . ____|____ . . | | bind __|__ f = __|_ = . . | | | | __|__ __|__ a1 a2 f a1 f a2 | | | | a1 a1 a1 a1 Now when we bind this tree to `g`, we get . _____|______ | | . . __|__ __|__ | | | | g a1 g a1 g a1 g a1 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. What's this have to do with tree\_monadize? -------------------------------------------- So we've defined a Tree monad: type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);; let tree_unit (a: 'a) : 'a tree = Leaf a;; let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree = match u with | Leaf a -> f a | Node (l, r) -> Node (tree_bind l f, tree_bind r f);; What's this have to do with the `tree_monadize` functions we defined earlier? let rec tree_monadize (t : 'a tree) (f : 'a -> 'b reader) : 'b tree reader = match t with | Leaf a -> reader_bind (f a) (fun b -> reader_unit (Leaf b)) | Node (l, r) -> reader_bind (tree_monadize l f) (fun l' -> reader_bind (tree_monadize r f) (fun r' -> reader_unit (Node (l', r'))));; ... and so on for different monads? Well, notice that `tree\_monadizer` takes arguments whose types resemble that of a monadic `bind` function. Here's a schematic bind function compared with `tree\_monadizer`: bind (u:'a Monad) (f: 'a -> 'b Monad): 'b Monad tree\_monadizer (u:'a Tree) (f: 'a -> 'b Monad): 'b Tree Monad Comparing these types makes it clear that `tree\_monadizer` provides a way to distribute an arbitrary monad M across the leaves of any tree to form a new tree inside an M box. The more general answer is that each of those `tree\_monadize` functions is adding a Tree monad *layer* to a pre-existing Reader (and so on) monad. We discuss that further here: [[Monad Transformers]].