Then our unit is naturally:
- let s_unit (x : 'a) : ('a state) = fun (s : store) -> (x, s)
+ let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s)
And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box:
looks like this:
type 'a list = ['a];;
- l_unit (x : 'a) = [x];;
+ l_unit (a : 'a) = [a];;
l_bind u f = List.concat (List.map f u);;
Recall that `List.map` take a function and a list and returns the
into OCaml lists soon. We don't need to fully grasp the role of the `'b`'s
in order to proceed to build a monad:
- l'_unit (x : 'a) : ('a, 'b) list = fun x -> fun f z -> f x z
+ l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z
No problem. Arriving at bind is a little more complicated, but
exactly the same principles apply, you just have to be careful and
Let's write a general function that will map individuals into their
corresponding generalized quantifier:
- gqize (x : e) = fun (p : e -> t) -> p x
+ gqize (a : e) = fun (p : e -> t) -> p a
This function wraps up an individual in a fancy box. That is to say,
we are in the presence of a monad. The type constructor, the unit and
similar to the List monad just given:
type 'a continuation = ('a -> 'b) -> 'b
- c_unit (x : 'a) = fun (p : 'a -> 'b) -> p x
+ c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a
c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd =
- fun (k : 'a -> 'b) -> u (fun (x : 'a) -> f x k)
+ fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k)
How similar is it to the List monad? Let's examine the type
constructor and the terms from the list monad derived above:
type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b
- l'_unit x = fun f -> f x
- l'_bind u f = fun k -> u (fun x -> f x k)
+ l'_unit a = fun f -> f a
+ l'_bind u f = fun k -> u (fun a -> f a k)
(We performed a sneaky but valid eta reduction in the unit term.)
Refunctionalizing zippers
-------------------------
+Manipulating trees with monads
+------------------------------
+
+This thread 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.
+
+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 during the remainder of the
+course. Here's a type constructor for binary trees:
+
+ type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
+
+These are trees in which the internal nodes do not have labels. [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.,
+
+
+<pre>
+let t1 = Node ((Node ((Leaf 2), (Leaf 3))),
+ (Node ((Leaf 5),(Node ((Leaf 7),
+ (Leaf 11))))))
+
+ .
+ ___|___
+ | |
+ . .
+_|__ _|__
+| | | |
+2 3 5 .
+ _|__
+ | |
+ 7 11
+</pre>
+
+Our first task will be to replace each leaf with its double:
+
+<pre>
+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));;
+</pre>
+`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:
+
+<pre>
+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
+</pre>
+
+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`:
+
+<pre>
+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)))
+</pre>
+
+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))))
+<pre>
+\f .
+ ____|____
+ | |
+ . .
+__|__ __|__
+| | | |
+f2 f3 f5 .
+ __|___
+ | |
+ f7 f11
+</pre>
+
+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`.
+
+<pre>
+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;;
+</pre>
+
+It's easy to figure out how to turn an `int` into an `int reader`:
+
+<pre>
+let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
+int2int_reader 2 (fun i -> i + i);;
+- : int = 4
+</pre>
+
+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.
+
+<pre>
+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))));;
+</pre>
+
+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.
+
+<pre>
+# treemonadizer int2int_reader t1 (fun i -> i + i);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22)))
+</pre>
+
+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:
+
+<pre>
+# treemonadizer int2int_reader t1 (fun i -> i * i);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
+</pre>
+
+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.
+
+<pre>
+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);;
+</pre>
+
+Gratifyingly, we can use the `treemonadizer` function without any
+modification whatsoever, except for replacing the (parametric) type
+`reader` with `state`:
+
+<pre>
+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))));;
+</pre>
+
+Then we can count the number of nodes in the tree:
+
+<pre>
+# 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
+</pre>
+
+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
+
+<pre>
+# 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])))]
+</pre>
+
+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?
+
+<pre>
+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))));;
+</pre>
+
+We use the continuation monad described above, and insert the
+`continuation` type in the appropriate place in the `treemonadizer` code.
+We then compute:
+
+<pre>
+# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);;
+- : int list = [2; 3; 5; 7; 11]
+</pre>
+
+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:
+
+<pre>
+# treemonadizer continuation_unit t1 (fun x -> x);;
+- : int tree =
+Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11)))
+</pre>
+
+That is, nothing happens. But we can begin to substitute more
+interesting functions for the first argument of `treemonadizer`:
+
+<pre>
+(* 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
+</pre>
+
+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 tree monad
+--------------
+
+Of course, by now you may have realized that we have discovered a new
+monad, the tree monad:
+
+<pre>
+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));;
+</pre>
+
+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)))
+<pre>
+ . .
+ __|__ __|__
+ | | | |
+ a1 . fa1 .
+ _|__ __|__
+ | | | |
+ . a5 . fa5
+ bind _|__ f = __|__
+ | | | |
+ . a4 . fa4
+ __|__ __|___
+ | | | |
+ a2 a3 fa2 fa3
+</pre>
+
+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
+have to proceed. Let `f a = Node (Leaf a, Leaf a)`. Then
+
+\tree (. (. (. (. (a1)(a2)))))
+\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) ))
+<pre>
+ .
+ ____|____
+ . . | |
+bind __|__ f = __|_ = . .
+ | | | | __|__ __|__
+ a1 a2 fa1 fa2 | | | |
+ a1 a1 a1 a1
+</pre>
+
+Now when we bind this tree to `g`, we get
+
+<pre>
+ .
+ ____|____
+ | |
+ . .
+ __|__ __|__
+ | | | |
+ ga1 ga1 ga1 ga1
+</pre>
+
+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.