```-# fun f z -> z;;
-- : 'a -> 'b -> 'b =
-# fun f z -> f 1 z;;
-- : (int -> 'a -> 'b) -> 'a -> 'b =
-# fun f z -> f 2 (f 1 z);;
-- : (int -> 'a -> 'a) -> 'a -> 'a =
-# fun f z -> f 3 (f 2 (f 1 z))
-- : (int -> 'a -> 'a) -> 'a -> 'a =
-```
- -Finally, we're getting consistent principle types, so we can stop. -These types should remind you of the simply-typed lambda calculus -types for Church numerals (`(o -> o) -> o -> o`) with one extra bit -thrown in (in this case, and int). +and so on. To save time, we'll let the OCaml interpreter infer the +principle types of these functions (rather than inferring what the +types should be ourselves): + + # fun f z -> z;; + - : 'a -> 'b -> 'b = + # fun f z -> f 1 z;; + - : (int -> 'a -> 'b) -> 'a -> 'b = + # fun f z -> f 2 (f 1 z);; + - : (int -> 'a -> 'a) -> 'a -> 'a = + # fun f z -> f 3 (f 2 (f 1 z)) + - : (int -> 'a -> 'a) -> 'a -> 'a = + +We can see what the consistent, general principle types are at the end, so we +can stop. These types should remind you of the simply-typed lambda calculus +types for Church numerals (`(o -> o) -> o -> o`) with one extra type +thrown in, the type of the element a the head of the list +(in this case, an int). So here's our type constructor for our hand-rolled lists: - type 'a list' = (int -> 'a -> 'a) -> 'a -> 'a + type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b Generalizing to lists that contain any kind of element (not just ints), we have @@ -166,63 +189,110 @@ ints), we have So an `('a, 'b) list'` is a list containing elements of type `'a`, where `'b` is the type of some part of the plumbing. This is more -general than an ordinary Ocaml list, but we'll see how to map them -into Ocaml lists soon. We don't need to grasp the role of the `'b`'s +general than an ordinary OCaml list, but we'll see how to map them +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 systematic about it. - l'_bind (u:('a,'b) list') (f:'a -> ('c, 'd) list'): ('c, 'd) list' = ... + l'_bind (u : ('a,'b) list') (f : 'a -> ('c, 'd) list') : ('c, 'd) list' = ... -Unfortunately, we'll need to spell out the types: +Unpacking the types gives: - l'_bind (u: ('a -> 'b -> 'b) -> 'b -> 'b) - (f: 'a -> ('c -> 'd -> 'd) -> 'd -> 'd) + l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) + (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd) : ('c -> 'd -> 'd) -> 'd -> 'd = ... -It's a rookie mistake to quail before complicated types. You should +Perhaps a bit intimiating. +But it's a rookie mistake to quail before complicated types. You should be no more intimiated by complex types than by a linguistic tree with deeply embedded branches: complex structure created by repeated application of simple rules. +[This would be a good time to try to build your own term for the types +just given. Doing so (or attempting to do so) will make the next +paragraph much easier to follow.] + As usual, we need to unpack the `u` box. Examine the type of `u`. -This time, `u` will only deliver up its contents if we give `u` as an -argument a function expecting an `'a`. Once that argument is applied -to an object of type `'a`, we'll have what we need. Thus: +This time, `u` will only deliver up its contents if we give `u` an +argument that is a function expecting an `'a` and a `'b`. `u` will +fold that function over its type `'a` members, and that's how we'll get the `'a`s we need. Thus: - .... u (fun (x:'a) -> ... (f a) ... ) ... + ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ... -In order for `u` to have the kind of argument it needs, we have to -adjust `(f a)` (which has type `('c -> 'd -> 'd) -> 'd -> 'd`) in -order to deliver something of type `'b -> 'b`. The easiest way is to -alias `'d` to `'b`, and provide `(f a)` with an argument of type `'c --> 'b -> 'b`. Thus: +In order for `u` to have the kind of argument it needs, the `... (f a) ...` has to evaluate to a result of type `'b`. The easiest way to do this is to collapse (or "unify") the types `'b` and `'d`, with the result that `f a` will have type `('c -> 'b -> 'b) -> 'b -> 'b`. Let's postulate an argument `k` of type `('c -> 'b -> 'b)` and supply it to `(f a)`: - l'_bind (u: ('a -> 'b -> 'b) -> 'b -> 'b) - (f: 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) - : ('c -> 'b -> 'b) -> 'b -> 'b = - .... u (fun (x:'a) -> f a k) ... + ... u (fun (a : 'a) (b : 'b) -> ... f a k ... ) ... + +Now we have an argument `b` of type `'b`, so we can supply that to `(f a) k`, getting a result of type `'b`, as we need: -[Excercise: can you arrive at a fully general bind for this type -constructor, one that does not collapse `'d`'s with `'b`'s?] + ... u (fun (a : 'a) (b : 'b) -> f a k b) ... -As usual, we have to abstract over `k`, but this time, no further -adjustments are needed: +Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over it: - l'_bind (u: ('a -> 'b -> 'b) -> 'b -> 'b) - (f: 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) + fun (k : 'c -> 'b -> 'b) -> u (fun (a : 'a) (b : 'b) -> f a k b) + +This whole expression has type `('c -> 'b -> 'b) -> 'b -> 'b`, which is exactly the type of a `('c, 'b) list'`. So we can hypothesize that we our bind is: + + l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) + (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) : ('c -> 'b -> 'b) -> 'b -> 'b = - fun (k:'c -> 'b -> 'b) -> u (fun (x:'a) -> f a k) + fun k -> u (fun a b -> f a k b) + +That is a function of the right type for our bind, but to check whether it works, we have to verify it (with the unit we chose) against the monad laws, and reason whether it will have the right behavior. + +Here's a way to persuade yourself that it will have the right behavior. First, it will be handy to eta-expand our `fun k -> u (fun a b -> f a k b)` to: + + fun k z -> u (fun a b -> f a k b) z + +Now let's think about what this does. It's a wrapper around `u`. In order to behave as the list which is the result of mapping `f` over each element of `u`, and then joining (`concat`ing) the results, this wrapper would have to accept arguments `k` and `z` and fold them in just the same way that the list which is the result of mapping `f` and then joining the results would fold them. Will it? + +Suppose we have a list' whose contents are `[1; 2; 4; 8]`---that is, our list' will be `fun f z -> f 1 (f 2 (f 4 (f 8 z)))`. We call that list' `u`. Suppose we also have a function `f` that for each `int` we give it, gives back a list of the divisors of that `int` that are greater than 1. Intuitively, then, binding `u` to `f` should give us: + + concat (map f u) = + concat [[]; [2]; [2; 4]; [2; 4; 8]] = + [2; 2; 4; 2; 4; 8] + +Or rather, it should give us a list' version of that, which takes a function `k` and value `z` as arguments, and returns the right fold of `k` and `z` over those elements. What does our formula + + fun k z -> u (fun a b -> f a k b) z + +do? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists: + + [] + [2] + [2; 4] + [2; 4; 8] + +(or rather, their list' versions). Then it takes the accumulated result `b` of previous steps in the fold, and it folds `k` and `b` over the list generated by `f a`. The result of doing so is passed on to the next step as the accumulated result so far. -You should carefully check to make sure that this term is consistent -with the typing. +So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed: -Our theory is that this monad should be capable of exactly -replicating the behavior of the standard List monad. Let's test: + 0 ==> + right-fold + and 0 over [2; 4; 8] = 2+4+8+0 ==> + right-fold + and 2+4+8+0 over [2; 4] = 2+4+2+4+8+0 ==> + right-fold + and 2+4+2+4+8+0 over [2] = 2+2+4+2+4+8+0 ==> + right-fold + and 2+2+4+2+4+8+0 over [] = 2+2+4+2+4+8+0 + +which indeed is the result of right-folding + and 0 over `[2; 2; 4; 2; 4; 8]`. If you trace through how this works, you should be able to persuade yourself that our formula: + + fun k z -> u (fun a b -> f a k b) z + +will deliver just the same folds, for arbitrary choices of `k` and `z` (with the right types), and arbitrary list's `u` and appropriately-typed `f`s, as + + fun k z -> List.fold_right k (concat (map f u)) z + +would. + +For future reference, we might make two eta-reductions to our formula, so that we have instead: + + let l'_bind = fun k -> u (fun a -> f a k);; + +Let's make some more tests: l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] @@ -230,29 +300,17 @@ replicating the behavior of the standard List monad. Let's test: l'_bind (fun f z -> f 1 (f 2 z)) (fun i -> fun f z -> f i (f (i+1) z)) ~~> -Sigh. Ocaml won't show us our own list. So we have to choose an `f` -and a `z` that will turn our hand-crafted lists into standard Ocaml +Sigh. OCaml won't show us our own list. So we have to choose an `f` +and a `z` that will turn our hand-crafted lists into standard OCaml lists, so that they will print out. -
```-# let cons h t = h :: t;;  (* Ocaml is stupid about :: *)
-# l'_bind (fun f z -> f 1 (f 2 z))
-          (fun i -> fun f z -> f i (f (i+1) z)) cons [];;
-- : int list = [1; 2; 2; 3]
-```
```+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 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));;
+```
+`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: + +
```+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
+```
+ +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`: + +
```+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)))
+```
+ +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)))) +
```+\f    .
+  ____|____
+  |       |
+  .       .
+__|__   __|__
+|   |   |   |
+f2  f3  f5  .
+          __|___
+          |    |
+          f7  f11
+```
+ +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`. + +
```+type 'a reader = (int->int) -> 'a;;  (* mnemonic: e for environment *)
+```
+ +It's easy to figure out how to turn an `int` into an `int reader`: + +
```+let int2int_reader (x:'a): 'b reader = fun (op:'a -> 'b) -> op x;;
+int2int_reader 2 (fun i -> i + i);;
+- : int = 4
+```
+ +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. + +
```+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 ->
+```
+ +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. + +
```+# treemonadizer int2int_reader t1 (fun i -> i + i);;
+- : 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 +squaring function, `fun i -> i * i`---we get an entirely different +result: + +
```+# treemonadizer int2int_reader t1 (fun i -> i * i);;
+- : int tree =
+Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121)))
+```
+ +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. + +
```+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);;
+```
+ +Gratifyingly, we can use the `treemonadizer` function without any +modification whatsoever, except for replacing the (parametric) type +`reader` with `state`: + +
```+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))));;
+```
+ +Then we can count the number of nodes in the tree: + +
```+# 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
+```
+ +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 + +
```+# 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])))]
+```
+ +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? + +
```+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))));;
+```
+ +We use the continuation monad described above, and insert the +`continuation` type in the appropriate place in the `treemonadizer` code. +We then compute: + +
```+# treemonadizer (fun a c -> a :: (c 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. + +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: + +
```+# treemonadizer continuation_unit t1 (fun x -> x);;
+- : 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`: + +
```+(* 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
+```
+ +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 +--------------------- + +Of course, by now you may have realized that we have discovered a new +monad, the binary tree monad: + +
```+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));;
+```
+ +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))) +
```+                .                         .
+              __|__                     __|__
+              |   |                     |   |
+              a1  .                    fa1  .
+                 _|__                     __|__
+                 |  |                     |   |
+                 .  a5                    .  fa5
+   bind         _|__       f   =        __|__
+                |  |                    |   |
+                .  a4                   .  fa4
+              __|__                   __|___
+              |   |                   |    |
+              a2  a3                 fa2  fa3
+```
+ +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))) )) +
```+                                           .
+                                       ____|____
+          .               .            |       |
+bind    __|__   f  =    __|_    =      .       .
+        |   |           |   |        __|__   __|__
+        a1  a2         fa1 fa2       |   |   |   |
+                                     a1  a1  a1  a1
+```
+ +Now when we bind this tree to `g`, we get + +
```+           .
+       ____|____
+       |       |
+       .       .
+     __|__   __|__
+     |   |   |   |
+    ga1 ga1 ga1 ga1
+```
+ +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: + +
```+t1 "abSe" ~~> "ababe"
+```
+ + +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'): + +
```+    t1 "aSbS"
+         *
+~~> t1 "aabS"
+           *
+~~> "aabaab"
+```
+ +versus + +
```+    t1 "aSbS"
+           *
+~~> t1 "aSbaSb"
+         *
+~~> t1 "aabaSb"
+            *
+~~> "aabaaabab"
+```
+ +versus + +
```+    t1 "aSbS"
+           *
+~~> t1 "aSbaSb"
+            *
+~~> t1 "aSbaaSbab"
+             *
+~~> t1 "aSbaaaSbaabab"
+              *
+~~> ...
+```
+ +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. + +
```+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']
+```
+ +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: + +
```+# #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']
+```
+ +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: + +
```+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']
+```
+ +Note that we don't need to do any reversing.