X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=manipulating_trees_with_monads.mdwn;h=0d9e33df3425822ec12e1b4b8aab11fa2594475c;hp=16bbd384885b3b088bab49dfdf3f0d758a8d9a63;hb=HEAD;hpb=8431bdd30785d6fe5f551219823d41c640918df4 diff --git a/manipulating_trees_with_monads.mdwn b/manipulating_trees_with_monads.mdwn deleted file mode 100644 index 16bbd384..00000000 --- a/manipulating_trees_with_monads.mdwn +++ /dev/null @@ -1,702 +0,0 @@ -[[!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 (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree = - match t with - | Leaf i -> Leaf (leaf_modifier i) - | Node (l, r) -> Node (tree_map leaf_modifier l, - tree_map leaf_modifier r);; - -`tree_map` 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;; - tree_map 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 `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 square t1;; - - : 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 `λ 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 (f : 'a -> 'b reader) (t : 'a tree) : '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 f l) (fun l' -> - reader_bind (tree_monadize f r) (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 int_readerize t1 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 -int_readerize t1`) to a different `int -> int` function---say, the -squaring function, `fun i -> i * i`---we get an entirely different -result: - - # tree_monadize int_readerize t1 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 (f : 'a -> 'b state) (t : 'a tree) : '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 f l) (fun l' -> - state_bind (tree_monadize f r) (fun r' -> - state_unit (Node (l', r'))));; - -Then we can count the number of leaves in the tree: - - # tree_monadize (fun a -> fun s -> (a, s+1)) t1 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 (fun a -> fun s -> (s+1, s+1)) t1 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 (f : 'a -> 'b state) (t : 'a tree) : '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 f r) (fun r' -> (* R first *) - state_bind (tree_monadize f l) (fun l'-> (* Then L *) - state_unit (Node (l', r'))));; - - # tree_monadize_rev (fun a -> fun s -> (s+1, s+1)) t1 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 (fun i -> [ [i; square i] ]) 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. 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 (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('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 f l) (fun l' -> - continuation_bind (tree_monadize f r) (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 (fun a -> fun k -> a :: k 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. Can you trace how this is working? Think first about what the -operation `fun a -> fun 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? - -In a moment, 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 its 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 continuation_unit t1 (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 (fun a -> fun k -> k (square a)) t1 (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 (fun a -> fun k -> k [a; square a]) t1 (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 (fun a -> fun k -> 1 + k a) t1 (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). - -Using continuations to solve the same fringe problem ----------------------------------------------------- - -We've seen two solutions to the same fringe problem so far. -The problem, recall, is to take two trees and decide whether they have -the same leaves in the same order. - -
- ta            tb          tc
- .             .           .
-_|__          _|__        _|__
-|  |          |  |        |  |
-1  .          .  3        1  .
-  _|__       _|__           _|__
-  |  |       |  |           |  |
-  2  3       1  2           3  2
-
-let ta = Node (Leaf 1, Node (Leaf 2, Leaf 3));;
-let tb = Node (Node (Leaf 1, Leaf 2), Leaf 3);;
-let tc = Node (Leaf 1, Node (Leaf 3, Leaf 2));;
-
- -So `ta` and `tb` are different trees that have the same fringe, but -`ta` and `tc` are not. - -The simplest solution is to map each tree to a list of its leaves, -then compare the lists. But because we will have computed the entire -fringe before starting the comparison, if the fringes differ in an -early position, we've wasted our time examining the rest of the trees. - -The second solution was to use tree zippers and mutable state to -simulate coroutines (see [[coroutines and aborts]]). In that -solution, we pulled the zipper on the first tree until we found the -next leaf, then stored the zipper structure in the mutable variable -while we turned our attention to the other tree. Because we stopped -as soon as we find the first mismatched leaf, this solution does not -have the flaw just mentioned of the solution that maps both trees to a -list of leaves before beginning comparison. - -Since zippers are just continuations reified, we expect that the -solution in terms of zippers can be reworked using continuations, and -this is indeed the case. Before we can arrive at a solution, however, -we must define a data structure called a stream: - - type 'a stream = End | Next of 'a * (unit -> 'a stream);; - -A stream is like a list in that it contains a series of objects (all -of the same type, here, type `'a`). The first object in the stream -corresponds to the head of a list, which we pair with a stream -representing the rest of a the list. There is a special stream called -`End` that represents a stream that contains no (more) elements, -analogous to the empty list `[]`. - -Actually, we pair each element not with a stream, but with a thunked -stream, that is, a function from the unit type to streams. The idea -is that the next element in the stream is not computed until we forced -the thunk by applying it to the unit: - -
-# let rec make_int_stream i = Next (i, fun () -> make_int_stream (i + 1));;
-val make_int_stream : int -> int stream = 
-# let int_stream = make_int_stream 1;;
-val int_stream : int stream = Next (1, )         (* First element: 1 *)
-# match int_stream with Next (i, rest) -> rest;;      
-- : unit -> int stream =                         (* Rest: a thunk *)
-
-(* Force the thunk to compute the second element *)
-# (match int_stream with Next (i, rest) -> rest) ();;
-- : int stream = Next (2, )      
-
- -You can think of `int_stream` as a functional object that provides -access to an infinite sequence of integers, one at a time. It's as if -we had written `[1;2;...]` where `...` meant "continue indefinitely". - -So, with streams in hand, we need only rewrite our continuation tree -monadizer so that instead of mapping trees to lists, it maps them to -streams. Instead of - - # tree_monadize (fun a k -> a :: k a) t1 (fun t -> []);; - - : int list = [2; 3; 5; 7; 11] - -as above, we have - - # tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End);; - - : int stream = Next (2, ) - -We can see the first element in the stream, the first leaf (namely, -2), but in order to see the next, we'll have to force a thunk. - -Then to complete the same-fringe function, we simply convert both -trees into leaf-streams, then compare the streams element by element. -The code is enitrely routine, but for the sake of completeness, here it is: - -
-let rec compare_streams stream1 stream2 =
-    match stream1, stream2 with 
-    | End, End -> true (* Done!  Fringes match. *)
-    | Next (next1, rest1), Next (next2, rest2) when next1 = next2 -> compare_streams (rest1 ()) (rest2 ())
-    | _ -> false;;
-
-let same_fringe t1 t2 =
-  let stream1 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t1 (fun _ -> End) in 
-  let stream2 = tree_monadize (fun i k -> Next (i, fun () -> k ())) t2 (fun _ -> End) in 
-  compare_streams stream1 stream2;;
-
- -Notice the forcing of the thunks in the recursive call to -`compare_streams`. So indeed: - -
-# same_fringe ta tb;;
-- : bool = true
-# same_fringe ta tc;;
-- : bool = false
-
- -Now, this implementation is a bit silly, since in order to convert the -trees to leaf streams, our tree_monadizer function has to visit every -node in the tree. But if we needed to compare each tree to a large -set of other trees, we could arrange to monadize each tree only once, -and then run compare_streams on the monadized trees. - -By the way, what if you have reason to believe that the fringes of -your trees are more likely to differ near the right edge than the left -edge? If we reverse evaluation order in the tree_monadizer function, -as shown above when we replaced leaves with their ordinal position, -then the resulting streams would produce leaves from the right to the -left. - -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 lex sentence1 (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 lex sentence2 (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 lex sentence2 (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\_mondadize? --------------------------------------------- - -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 (f : 'a -> 'b reader) (t : 'a tree) : '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 f l) (fun l' -> - reader_bind (tree_monadize f r) (fun r' -> - reader_unit (Node (l', r'))));; - -... and so on for different monads? - -The 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]]. -