+The binary tree monad
+---------------------
+
+Of course, by now you may have realized that we have discovered a new
+monad, the binary 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
+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.
+
+
+Refunctionalizing zippers: from lists to continuations
+------------------------------------------------------
+
+Let's work with lists of chars for a change. To maximize readability, we'll
+indulge in an abbreviatory convention that "abc" abbreviates the
+list `['a'; 'b'; 'c']`.
+
+Task 1: replace each occurrence of 'S' with a copy of the string up to
+that point.
+
+Expected behavior:
+
+<pre>
+t1 "abSe" ~~> "ababe"
+</pre>
+
+
+In linguistic terms, this is a kind of anaphora
+resolution, where `'S'` is functioning like an anaphoric element, and
+the preceding string portion is the antecedent.
+
+This deceptively simple task gives rise to some mind-bending complexity.
+Note that it matters which 'S' you target first (the position of the *
+indicates the targeted 'S'):
+
+<pre>
+ t1 "aSbS"
+ *
+~~> t1 "aabS"
+ *
+~~> "aabaab"
+</pre>
+
+versus
+
+<pre>
+ t1 "aSbS"
+ *
+~~> t1 "aSbaSb"
+ *
+~~> t1 "aabaSb"
+ *
+~~> "aabaaabab"
+</pre>
+
+versus
+
+<pre>
+ t1 "aSbS"
+ *
+~~> t1 "aSbaSb"
+ *
+~~> t1 "aSbaaSbab"
+ *
+~~> t1 "aSbaaaSbaabab"
+ *
+~~> ...
+</pre>
+
+Aparently, this task, as simple as it is, is a form of computation,
+and the order in which the `'S'`s get evaluated can lead to divergent
+behavior.
+
+For now, as usual, we'll agree to always evaluate the leftmost `'S'`.
+
+This is a task well-suited to using a zipper.
+
+<pre>
+type 'a list_zipper = ('a list) * ('a list);;
+
+let rec t1 (z:char list_zipper) =
+ match z with (sofar, []) -> List.rev(sofar) (* Done! *)
+ | (sofar, 'S'::rest) -> t1 ((List.append sofar sofar), rest)
+ | (sofar, fst::rest) -> t1 (fst::sofar, rest);; (* Move zipper *)
+
+# t1 ([], ['a'; 'b'; 'S'; 'e']);;
+- : char list = ['a'; 'b'; 'a'; 'b'; 'e']
+
+# t1 ([], ['a'; 'S'; 'b'; 'S']);;
+- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
+</pre>
+
+Note that this implementation enforces the evaluate-leftmost rule.
+Task 1 completed.
+
+One way to see exactly what is going on is to watch the zipper in
+action by tracing the execution of `t1`. By using the `#trace`
+directive in the Ocaml interpreter, the system will print out the
+arguments to `t1` each time it is (recurcively) called:
+
+<pre>
+# #trace t1;;
+t1 is now traced.
+# t1 ([], ['a'; 'b'; 'S'; 'e']);;
+t1 <-- ([], ['a'; 'b'; 'S'; 'e'])
+t1 <-- (['a'], ['b'; 'S'; 'e'])
+t1 <-- (['b'; 'a'], ['S'; 'e'])
+t1 <-- (['b'; 'a'; 'b'; 'a'], ['e'])
+t1 <-- (['e'; 'b'; 'a'; 'b'; 'a'], [])
+t1 --> ['a'; 'b'; 'a'; 'b'; 'e']
+t1 --> ['a'; 'b'; 'a'; 'b'; 'e']
+t1 --> ['a'; 'b'; 'a'; 'b'; 'e']
+t1 --> ['a'; 'b'; 'a'; 'b'; 'e']
+t1 --> ['a'; 'b'; 'a'; 'b'; 'e']
+- : char list = ['a'; 'b'; 'a'; 'b'; 'e']
+</pre>
+
+The nice thing about computations involving lists is that it's so easy
+to visualize them as a data structure. Eventually, we want to get to
+a place where we can talk about more abstract computations. In order
+to get there, we'll first do the exact same thing we just did with
+concrete zipper using procedures.
+
+Think of a list as a procedural recipe: `['a'; 'b'; 'c']` means (1)
+start with the empty list `[]`; (2) make a new list whose first
+element is 'c' and whose tail is the list construted in the previous
+step; (3) make a new list whose first element is 'b' and whose tail is
+the list constructed in the previous step; and (4) make a new list
+whose first element is 'a' and whose tail is the list constructed in
+the previous step.
+
+What is the type of each of these steps? Well, it will be a function
+from the result of the previous step (a list) to a new list: it will
+be a function of type `char list -> char list`. We'll call each step
+a **continuation** of the recipe. So in this context, a continuation
+is a function of type `char list -> char list`.
+
+This means that we can now represent the sofar part of our zipper--the
+part we've already unzipped--as a continuation, a function describing
+how to finish building the list:
+
+<pre>
+let rec t1c (l: char list) (c: (char list) -> (char list)) =
+ match l with [] -> c []
+ | 'S'::rest -> t1c rest (fun x -> c (c x))
+ | a::rest -> t1c rest (fun x -> List.append (c x) [a]);;
+
+# t1c ['a'; 'b'; 'S'] (fun x -> x);;
+- : char list = ['a'; 'b'; 'a'; 'b']
+
+# t1c ['a'; 'S'; 'b'; 'S'] (fun x -> x);;
+- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b']
+</pre>
+
+Note that we don't need to do any reversing.
+