Manipulating trees with monads
------------------------------
-This topic develops an idea based on a detailed suggestion of Ken
-Shan's. We'll build a series of functions that operate on trees,
-doing various things, including replacing leaves, counting nodes, and
-converting a tree to a list of leaves. The end result will be an
-application for continuations.
+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
+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)
+ 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))))))
+ let t1 = Node (Node (Leaf 2, Leaf 3),
+ Node (Leaf 5, Node (Leaf 7,
+ Leaf 11)))
.
___|___
| |
. .
- _|__ _|__
+ _|_ _|__
| | | |
2 3 5 .
_|__
Our first task will be to replace each leaf with its double:
- let rec treemap (newleaf : 'a -> 'b) (t : 'a tree) : 'b tree =
+ let rec tree_map (t : 'a tree) (leaf_modifier : 'a -> 'b): 'b tree =
match t with
- | Leaf x -> Leaf (newleaf x)
- | Node (l, r) -> Node ((treemap newleaf l),
- (treemap newleaf r));;
+ | Leaf i -> Leaf (leaf_modifier i)
+ | Node (l, r) -> Node (tree_map l leaf_modifier,
+ tree_map r leaf_modifier);;
-`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:
+`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;;
- treemap double t1;;
+ tree_map t1 double;;
- : int tree =
Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
| |
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`:
+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 x = x * x;;
- treemap square t1;;
- - : int tree =ppp
+ 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 `treemap` does is take some global, contextual
+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, `treemap` has the
-behavior of a reader monad. Let's make that explicit.
+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 treemap function more and
+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 treemap function return a (monadized)
-tree that is ready to accept any `int->int` function and produce the
+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 .
- ____|____
- | |
- . .
- __|__ __|__
- | | | |
- f2 f3 f5 .
- __|___
- | |
- f7 f11
+ \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 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)`.
+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
-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`.
+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.
- 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;;
+But we can do this:
-It's easy to figure out how to turn an `int` into an `int reader`:
+ 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'))));;
- let int2int_reader (x : 'a): 'b reader = fun (op : 'a -> 'b) -> op x;;
- int2int_reader 2 (fun i -> i + i);;
- - : int = 4
+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:
-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.
+ ------------
+ 1 ---> | 1 |
+ ------------
- 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))));;
+then I'll give you back the ability to do this:
-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.
+ ____________
+ . | . |
+ __|___ ---> | __|___ |
+ | | | | | |
+ 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:
- # treemonadizer int2int_reader t1 (fun i -> i + i);;
+ ____________
+ . | . | .
+ __|___ ---> | __|___ | 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, `treemonadizer
-int2int_reader t1`) to a different `int->int` function---say, the
+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:
- # treemonadizer int2int_reader t1 (fun i -> i * i);;
+ # 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 monad as a
+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 nodes in
+
+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 x i = (x, i+.5);;
- let state_bind u f i = let (a, i') = u i in f a (i'+.5);;
+ 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 `treemonadizer` function without any
+Gratifyingly, we can use the `tree_monadize` function without any
modification whatsoever, except for replacing the (parametric) type
-`reader` with `state`:
+`'b reader` with `'b state`, and substituting in the appropriate unit and bind:
- let rec treemonadizer (f : 'a -> 'b state) (t : 'a tree) : 'b tree state =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> 'b state) : '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))));;
+ | 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 nodes in the tree:
+Then we can count the number of leaves in the tree:
- # treemonadizer state_unit t1 0;;
+ # 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))), 13)
+ (Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))), 5)
.
___|___
| |
. .
- _|__ _|__
+ _|__ _|__ , 5
| | | |
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.
+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 `treemonadizer` with `list` gives us
+`state` everywhere in `tree_monadize` with `list` gives us
- # treemonadizer (fun x -> [ [x; square x] ]) t1;;
+ # tree_monadize t1 (fun i -> [ [i; square i] ]);;
- : int list tree list =
[Node
(Node (Leaf [2; 4], Leaf [3; 9]),
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.
+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 x c = c x;;
- let continuation_bind u f c = u (fun a -> f a c);;
+ let continuation_unit a = fun k -> k a;;
+ let continuation_bind u f = fun k -> u (fun a -> f a k);;
- let rec treemonadizer (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation =
+ let rec tree_monadize (t : 'a tree) (f : 'a -> ('b, 'r) continuation) : ('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))));;
+ | 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.
-We use the continuation monad described above, and insert the
-`continuation` type in the appropriate place in the `treemonadizer` code.
-We then compute:
+So for example, we compute:
- # treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
+ # 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.
+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?
-The continuation monad is amazingly flexible; we can use it to
+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 the
-continuation unit as our first argument to `treemonadizer`, and then
+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:
- # treemonadizer continuation_unit t1 (fun x -> x);;
+ # 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 `treemonadizer`:
+interesting functions for the first argument of `tree_monadize`:
(* Simulating the tree reader: distributing a operation over the leaves *)
- # treemonadizer (fun a c -> c (square a)) t1 (fun x -> x);;
+ # 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 *)
- # treemonadizer (fun a c -> c [a; square a]) t1 (fun x -> x);;
+ # 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 *)
- # treemonadizer (fun a c -> 1 + c a) t1 (fun x -> 0);;
+ # tree_monadize t1 (fun a -> fun k -> 1 + k a) (fun t -> 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
+[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:
+
+<pre>
+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"));;
+</pre>
+
+Then we can crudely approximate quantification as follows:
+
+<pre>
+# tree_monadize sentence1 lex (fun x -> x);;
+- : string tree =
+Node
+ (Leaf "forall x",
+ Node (Leaf "John", Node (Node (Leaf "saw", Leaf "x"), Leaf "yesterday")))
+</pre>
+
+In order to see the effects of evaluation order,
+observe what happens when we combine two quantifiers in the same
+sentence:
+
+<pre>
+# 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"))))
+</pre>
+
+The universal takes scope over the existential. If, however, we
+replace the usual tree_monadizer with tree_monadizer_rev, we get
+inverse scope:
+
+<pre>
+# 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"))))
+</pre>
+
+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:
+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 (x: 'a) = Leaf x;;
+ 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 x -> f x
- | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));;
+ | 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 = fa
+ 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 `fa`:
-
-\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5)))
+except that each leaf `a` has been replaced with `f a`:
. .
__|__ __|__
| | | |
- a1 . fa1 .
+ a1 . f a1 .
_|__ __|__
| | | |
- . a5 . fa5
+ . a5 . f a5
bind _|__ f = __|__
| | | |
- . a4 . fa4
+ . a4 . f a4
__|__ __|___
| | | |
- a2 a3 fa2 fa3
+ a2 a3 f a2 f a3
Given this equivalence, the right identity law
As for the associative law,
- Associativity: bind (bind u f) g = bind u (\a. bind (fa) g)
+ 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
-\tree (. (. (. (. (a1)(a2)))))
-\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
-
.
____|____
. . | |
bind __|__ f = __|_ = . .
| | | | __|__ __|__
- a1 a2 fa1 fa2 | | | |
+ a1 a2 f a1 f a2 | | | |
a1 a1 a1 a1
Now when we bind this tree to `g`, we get
- .
- ____|____
- | |
- . .
- __|__ __|__
- | | | |
- ga1 ga1 ga1 ga1
+ .
+ _____|______
+ | |
+ . .
+ __|__ __|__
+ | | | |
+ 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
[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]].