+t "abSd" ~~> "ababd" ++ + +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'): + +

+ t "aSbS" + * +~~> t "aabS" + * +~~> "aabaab" ++ +versus + +

+ t "aSbS" + * +~~> t "aSbaSb" + * +~~> t "aabaSb" + * +~~> "aabaaabab" ++ +versus + +

+ t "aSbS" + * +~~> t "aSbaSb" + * +~~> t "aSbaaSbab" + * +~~> t "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, we'll agree to always evaluate the leftmost `'S'`, which +guarantees termination, and a final string without any `'S'` in it. + +This is a task well-suited to using a zipper. We'll define a function +`tz` (for task with zippers), which accomplishes the task by mapping a +char list zipper to a char list. We'll call the two parts of the +zipper `unzipped` and `zipped`; we start with a fully zipped list, and +move elements to the zipped part by pulling the zipped down until the +entire list has been unzipped (and so the zipped half of the zipper is empty). + +

+type 'a list_zipper = ('a list) * ('a list);; + +let rec tz (z:char list_zipper) = + match z with (unzipped, []) -> List.rev(unzipped) (* Done! *) + | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) + | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) + +# tz ([], ['a'; 'b'; 'S'; 'd']);; +- : char list = ['a'; 'b'; 'a'; 'b'; 'd'] + +# tz ([], ['a'; 'S'; 'b'; 'S']);; +- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] ++ +Note that this implementation enforces the evaluate-leftmost rule. +Task completed. + +One way to see exactly what is going on is to watch the zipper in +action by tracing the execution of `tz`. By using the `#trace` +directive in the Ocaml interpreter, the system will print out the +arguments to `tz` each time it is (recurcively) called. Note that the +lines with left-facing arrows (`<--`) show (recursive) calls to `tz`, +giving the value of its argument (a zipper), and the lines with +right-facing arrows (`-->`) show the output of each recursive call, a +simple list. + +

+# #trace tz;; +t1 is now traced. +# tz ([], ['a'; 'b'; 'S'; 'd']);; +tz <-- ([], ['a'; 'b'; 'S'; 'd']) +tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *) +tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *) +tz <-- (['b'; 'a'; 'b'; 'a'], ['d']) (* Special step *) +tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], []) (* Pull zipper *) +tz --> ['a'; 'b'; 'a'; 'b'; 'd'] (* Output reversed *) +tz --> ['a'; 'b'; 'a'; 'b'; 'd'] +tz --> ['a'; 'b'; 'a'; 'b'; 'd'] +tz --> ['a'; 'b'; 'a'; 'b'; 'd'] +tz --> ['a'; 'b'; 'a'; 'b'; 'd'] +- : char list = ['a'; 'b'; 'a'; 'b'; 'd'] ++ +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'; 'S'; 'd']` +is the result of the computation `a::(b::(S::(d::[])))` (or, in our old +style, `makelist a (makelist b (makelist S (makelist c empty)))`). +The recipe for constructing the list goes like this: + +

+(0) Start with the empty list [] +(1) make a new list whose first element is 'd' and whose tail is the list constructed in step (0) +(2) make a new list whose first element is 'S' and whose tail is the list constructed in step (1) +----------------------------------------- +(3) make a new list whose first element is 'b' and whose tail is the list constructed in step (2) +(4) make a new list whose first element is 'a' and whose tail is the list constructed in step (3) ++ +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`. For instance, the +continuation corresponding to the portion of the recipe below the +horizontal line is the function `fun (tail:char list) -> a::(b::tail)`. + +This means that we can now represent the unzipped part of our +zipper--the part we've already unzipped--as a continuation: a function +describing how to finish building the list. We'll write a new +function, `tc` (for task with continuations), that will take an input +list (not a zipper!) and a continuation and return a processed list. +The structure and the behavior will follow that of `tz` above, with +some small but interesting differences. We've included the orginal +`tz` to facilitate detailed comparison: + +

+let rec tz (z:char list_zipper) = + match z with (unzipped, []) -> List.rev(unzipped) (* Done! *) + | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) + | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) + +let rec tc (l: char list) (c: (char list) -> (char list)) = + match l with [] -> List.rev (c []) + | 'S'::zipped -> tc zipped (fun x -> c (c x)) + | target::zipped -> tc zipped (fun x -> target::(c x));; + +# tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);; +- : char list = ['a'; 'b'; 'a'; 'b'] + +# tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);; +- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] ++ +To emphasize the parallel, I've re-used the names `zipped` and +`target`. The trace of the procedure will show that these variables +take on the same values in the same series of steps as they did during +the execution of `tz` above. There will once again be one initial and +four recursive calls to `tc`, and `zipped` will take on the values +`"bSd"`, `"Sd"`, `"d"`, and `""` (and, once again, on the final call, +the first `match` clause will fire, so the the variable `zipper` will +not be instantiated). + +I have not called the functional argument `unzipped`, although that is +what the parallel would suggest. The reason is that `unzipped` is a +list, but `c` is a function. That's the most crucial difference, the +point of the excercise, and it should be emphasized. For instance, +you can see this difference in the fact that in `tz`, we have to glue +together the two instances of `unzipped` with an explicit `List.append`. +In the `tc` version of the task, we simply compose `c` with itself: +`c o c = fun x -> c (c x)`. + +Why use the identity function as the initial continuation? Well, if +you have already constructed the list "abSd", what's the next step in +the recipe to produce the desired result (which is the same list, +"abSd")? Clearly, the identity continuation. + +A good way to test your understanding is to figure out what the +continuation function `c` must be at the point in the computation when +`tc` is called with the first argument `"Sd"`. Two choices: is it +`fun x -> a::b::x`, or it is `fun x -> b::a::x`? +The way to see if you're right is to execute the following +command and see what happens: + + tc ['S'; 'd'] (fun x -> 'a'::'b'::x);; + +There are a number of interesting directions we can go with this task. +The task was chosen because the computation can be viewed as a +simplified picture of a computation using continuations, where `'S'` +plays the role of a control operator with some similarities to what is +often called `shift`. In the analogy, the list portrays a string of +functional applications, where `[f1; f2; f3; x]` represents `f1(f2(f3 +x))`. The limitation of the analogy is that it is only possible to +represent computations in which the applications are always +right-branching, i.e., the computation `((f1 f2) f3) x` cannot be +directly represented. + +One possibile development is that we could add a special symbol `'#'`, +and then the task would be to copy from the target `'S'` only back to +the closest `'#'`. This would allow the task to simulate delimited +continuations (for right-branching computations). + +The task is well-suited to the list zipper because the list monad has +an intimate connection with continuations. The following section +makes this connection. We'll return to the list task after talking +about generalized quantifiers below. + diff --git a/list_monad_as_continuation_monad.mdwn b/list_monad_as_continuation_monad.mdwn new file mode 100644 index 00000000..aa167a4b --- /dev/null +++ b/list_monad_as_continuation_monad.mdwn @@ -0,0 +1,303 @@ + + +Rethinking the list monad +------------------------- + +To construct a monad, the key element is to settle on a type +constructor, and the monad naturally follows from that. We'll remind +you of some examples of how monads follow from the type constructor in +a moment. This will involve some review of familair material, but +it's worth doing for two reasons: it will set up a pattern for the new +discussion further below, and it will tie together some previously +unconnected elements of the course (more specifically, version 3 lists +and monads). + +For instance, take the **Reader Monad**. Once we decide that the type +constructor is + + type 'a reader = env -> 'a + +then the choice of unit and bind is natural: + + let r_unit (a : 'a) : 'a reader = fun (e : env) -> a + +Since the type of an `'a reader` is `env -> 'a` (by definition), +the type of the `r_unit` function is `'a -> env -> 'a`, which is a +specific case of the type of the *K* combinator. So it makes sense +that *K* is the unit for the reader monad. + +Since the type of the `bind` operator is required to be + + r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader) + +We can reason our way to the correct `bind` function as follows. We +start by declaring the types determined by the definition of a bind operation: + + let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ... + +Now we have to open up the `u` box and get out the `'a` object in order to +feed it to `f`. Since `u` is a function from environments to +objects of type `'a`, the way we open a box in this monad is +by applying it to an environment: + + ... f (u e) ... + +This subexpression types to `'b reader`, which is good. The only +problem is that we invented an environment `e` that we didn't already have , +so we have to abstract over that variable to balance the books: + + fun e -> f (u e) ... + +This types to `env -> 'b reader`, but we want to end up with `env -> +'b`. Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment. So we end up as follows: + + r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = + f (u e) e + +And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does. + +[The bind we cite here is a condensed version of the careful `let a = u e in ...` +constructions we provided in earlier lectures. We use the condensed +version here in order to emphasize similarities of structure across +monads.] + +The **State Monad** is similar. Once we've decided to use the following type constructor: + + type 'a state = store -> ('a, store) + +Then our unit is naturally: + + 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: + + let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = + ... f (...) ... + +But unlocking the `u` box is a little more complicated. As before, we +need to posit a state `s` that we can apply `u` to. Once we do so, +however, we won't have an `'a`, we'll have a pair whose first element +is an `'a`. So we have to unpack the pair: + + ... let (a, s') = u s in ... (f a) ... + +Abstracting over the `s` and adjusting the types gives the result: + + let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = + fun (s : store) -> let (a, s') = u s in f a s' + +The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we +won't pause to explore it here, though conceptually its unit and bind +follow just as naturally from its type constructor. + +Our other familiar monad is the **List Monad**, which we were told +looks like this: + + type 'a list = ['a];; + l_unit (a : 'a) = [a];; + l_bind u f = List.concat (List.map f u);; + +Thinking through the list monad will take a little time, but doing so +will provide a connection with continuations. + +Recall that `List.map` takes a function and a list and returns the +result to applying the function to the elements of the list: + + List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]] + +and List.concat takes a list of lists and erases the embdded list +boundaries: + + List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3] + +And sure enough, + + l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] + +Now, why this unit, and why this bind? Well, ideally a unit should +not throw away information, so we can rule out `fun x -> []` as an +ideal unit. And units should not add more information than required, +so there's no obvious reason to prefer `fun x -> [x,x]`. In other +words, `fun x -> [x]` is a reasonable choice for a unit. + +As for bind, an `'a list` monadic object contains a lot of objects of +type `'a`, and we want to make some use of each of them (rather than +arbitrarily throwing some of them away). The only +thing we know for sure we can do with an object of type `'a` is apply +the function of type `'a -> 'a list` to them. Once we've done so, we +have a collection of lists, one for each of the `'a`'s. One +possibility is that we could gather them all up in a list, so that +`bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts +the object returned by the second argument of `bind` to always be of +type `'b list list`. We can elimiate that restriction by flattening +the list of lists into a single list: this is +just List.concat applied to the output of List.map. So there is some logic to the +choice of unit and bind for the list monad. + +Yet we can still desire to go deeper, and see if the appropriate bind +behavior emerges from the types, as it did for the previously +considered monads. But we can't do that if we leave the list type +as a primitive Ocaml type. However, we know several ways of implementing +lists using just functions. In what follows, we're going to use type +3 lists (the right fold implementation), though it's important to +wonder how things would change if we used some other strategy for +implementating lists. These were the lists that made lists look like +Church numerals with extra bits embdded in them: + + empty list: fun f z -> z + list with one element: fun f z -> f 1 z + list with two elements: fun f z -> f 2 (f 1 z) + list with three elements: fun f z -> f 3 (f 2 (f 1 z)) + +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 =

+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 *) +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;; ++ +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 -> + reader_bind (treemonadizer f r) (fun y -> + reader_unit (Node (x, y))));; ++ +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. + diff --git a/tree_and_list_zippers.mdwn b/tree_and_list_zippers.mdwn new file mode 100644 index 00000000..5d26ca97 --- /dev/null +++ b/tree_and_list_zippers.mdwn @@ -0,0 +1,253 @@ +[[!toc]] + +##List Zippers## + +Say you've got some moderately-complex function for searching through a list, for example: + + let find_nth (test : 'a -> bool) (n : int) (lst : 'a list) : (int * 'a) -> + let rec helper (position : int) n lst = + match lst with + | [] -> failwith "not found" + | x :: xs when test x -> (if n = 1 + then (position, x) + else helper (position + 1) (n - 1) xs + ) + | x :: xs -> helper (position + 1) n xs + in helper 0 n lst;; + +This searches for the `n`th element of a list that satisfies the predicate `test`, and returns a pair containing the position of that element, and the element itself. Good. But now what if you wanted to retrieve a different kind of information, such as the `n`th element matching `test`, together with its preceding and succeeding elements? In a real situation, you'd want to develop some good strategy for reporting when the target element doesn't have a predecessor and successor; but we'll just simplify here and report them as having some default value: + + let find_nth' (test : 'a -> bool) (n : int) (lst : 'a list) (default : 'a) : ('a * 'a * 'a) -> + let rec helper (predecessor : 'a) n lst = + match lst with + | [] -> failwith "not found" + | x :: xs when test x -> (if n = 1 + then (predecessor, x, match xs with [] -> default | y::ys -> y) + else helper x (n - 1) xs + ) + | x :: xs -> helper x n xs + in helper default n lst;; + +This duplicates a lot of the structure of `find_nth`; it just has enough different code to retrieve different information when the matching element is found. But now what if you wanted to retrieve yet a different kind of information...? + +Ideally, there should be some way to factor out the code to find the target element---the `n`th element of the list satisfying the predicate `test`---from the code that retrieves the information you want once the target is found. We might build upon the initial `find_nth` function, since that returns the *position* of the matching element. We could hand that result off to some other function that's designed to retrieve information of a specific sort surrounding that position. But suppose our list has millions of elements, and the target element is at position 600512. The search function will already have traversed 600512 elements of the list looking for the target, then the retrieval function would have to *start again from the beginning* and traverse those same 600512 elements again. It could go a bit faster, since it doesn't have to check each element against `test` as it traverses. It already knows how far it has to travel. But still, this should seem a bit wasteful. + +Here's an idea. What if we had some way of representing a list as "broken" at a specific point. For example, if our base list is: + + [10; 20; 30; 40; 50; 60; 70; 80; 90] + +we might imagine the list "broken" at position 3 like this (positions are numbered starting from 0): + + 40; + 30; 50; + 20; 60; + [10; 70; + 80; + 90] + +Then if we move one step forward in the list, it would be "broken" at position 4: + + 50; + 40; 60; + 30; 70; + 20; 80; + [10; 90] + +If we had some convenient representation of these "broken" lists, then our search function could hand *that* off to the retrieval function, and the retrieval function could start right at the position where the list was broken, without having to start at the beginning and traverse many elements to get there. The retrieval function would also be able to inspect elements both forwards and backwards from the position where the list was "broken". + +The kind of data structure we're looking for here is called a **list zipper**. To represent our first broken list, we'd use two lists: (1) containing the elements in the left branch, preceding the target element, *in the order reverse to their appearance in the base list*. (2) containing the target element and the rest of the list, in normal order. So: + + 40; + 30; 50; + 20; 60; + [10; 70; + 80; + 90] + +would be represented as `([30; 20; 10], [40; 50; 60; 70; 80; 90])`. To move forward in the base list, we pop the head element `40` off of the head element of the second list in the zipper, and push it onto the first list, getting `([40; 30; 20; 10], [50; 60; 70; 80; 90])`. To move backwards again, we pop off of the first list, and push it onto the second. To reconstruct the base list, we just "move backwards" until the first list is empty. (This is supposed to evoke the image of zipping up a zipper; hence the data structure's name.) + +We had some discussio in seminar of the right way to understand the "zipper" metaphor. I think it's best to think of the tab of the zipper being here: + + t + a + b + 40; + 30; 50; + 20; 60; + [10; 70; + 80; + 90] + +And imagine that you're just seeing the left half of a real-zipper, rotated 60 degrees counter-clockwise. When the list is all "zipped up", we've "move backwards" to the state where the first element is targetted: + + ([], [10; 20; 30; 40; 50; 60; 70; 80; 90]) + +However you understand the "zipper" metaphor, this is a very handy data structure, and it will become even more handy when we translate it over to more complicated base structures, like trees. To help get a good conceptual grip on how to do that, it's useful to introduce a kind of symbolism for talking about zippers. This is just a metalanguage notation, for us theorists; we don't need our programs to interpret the notation. We'll use a specification like this: + + [10; 20; 30; *; 50; 60; 70; 80; 90], * filled by 40 + +to represent a list zipper where the break is at position 3, and the element occupying that position is 40. For a list zipper, this is implemented using the pairs-of-lists structure described above. + + +##Tree Zippers## + +Now how could we translate a zipper-like structure over to trees? What we're aiming for is a way to keep track of where we are in a tree, in the same way that the "broken" lists let us keep track of where we are in the base list. + +It's important to set some ground rules for what will follow. If you don't understand these ground rules you will get confused. First off, for many uses of trees one wants some of the nodes or leafs in the tree to be *labeled* with additional information. It's important not to conflate the label with the node itself. Numerically one and the same piece of information---for example, the same `int`---could label two nodes of the tree without those nodes thereby being identical, as here: + + root + / \ + / \ + / \ label 1 + / \ + label 1 label 2 + +The leftmost leaf and the rightmost leaf have the same label; but they are different leafs. The leftmost leaf has a sibling leaf with the label 2; the rightmost leaf has no siblings that are leafs. Sometimes when one is diagramming trees, one will annotate the nodes with the labels, as above. Other times, when one is diagramming trees, one will instead want to annotate the nodes with tags to make it easier to refer to particular parts of the tree. So for instance, I could diagram the same tree as above as: + + 1 + / \ + 2 \ + / \ 5 + / \ + 3 4 + +Here I haven't drawn what the labels are. The leftmost leaf, the node tagged "3" in this diagram, doesn't have the label `3`. It has the label 1, as we said before. I just haven't put that into the diagram. The node tagged "2" doesn't have the label `2`. It doesn't have any label. The tree in this example only has information labeling its leafs, not any of its inner nodes. The identity of its inner nodes is exhausted by their position in the tree. + +That is a second thing to note. In what follows, we'll only be working with *leaf-labeled* trees. In some uses of trees, one also wants labels on inner nodes. But we won't be discussing any such trees now. Our trees only have labels on their leafs. The diagrams below will tag all of the nodes, as in the second diagram above, and won't display what the leafs' labels are. + +Final introductory comment: in particular applications, you may only need to work with binary trees---trees where internal nodes always have exactly two subtrees. That is what we'll work with in the homework, for example. But to get the guiding idea of how tree zippers work, it's helpful first to think about trees that permit nodes to have many subtrees. So that's how we'll start. + +Suppose we have the following tree: + + 9200 + / | \ + / | \ + / | \ + / | \ + / | \ + 500 920 950 + / | \ / | \ / | \ + 20 50 80 91 92 93 94 95 96 + 1 2 3 4 5 6 7 8 9 + +This is a leaf-labeled tree whose labels aren't displayed. The `9200` and so on are tags to make it easier for us to refer to particular parts of the tree. + +Suppose we want to represent that we're *at* the node marked `50`. We might use the following metalanguage notation to specify this: + + {parent = ...; siblings = [subtree 20; *; subtree 80]}, * filled by subtree 50 + +This is modeled on the notation suggested above for list zippers. Here `subtree 20` refers to the whole subtree rooted at node `20`: + + 20 + / | \ + 1 2 3 + +Similarly for `subtree 50` and `subtree 80`. We haven't said yet what goes in the `parent = ...` slot. Well, the parent of a subtree targetted on `node 50` should intuitively be a tree targetted on `node 500`: + + {parent = ...; siblings = [*; subtree 920; subtree 950]}, * filled by subtree 500 + +And the parent of that targetted subtree should intuitively be a tree targetted on `node 9200`: + + {parent = None; siblings = [*]}, * filled by tree 9200 + +This tree has no parents because it's the root of the base tree. Fully spelled out, then, our tree targetted on `node 50` would be: + + { + parent = { + parent = { + parent = None; + siblings = [*] + }, * filled by tree 9200; + siblings = [*; subtree 920; subtree 950] + }, * filled by subtree 500; + siblings = [subtree 20; *; subtree 80] + }, * filled by subtree 50 + +In fact, there's some redundancy in this structure, at the points where we have `* filled by tree 9200` and `* filled by subtree 500`. Since node 9200 doesn't have any label attached to it, the subtree rooted in it is determined by the rest of this structure; and so too with `subtree 500`. So we could really work with: + + { + parent = { + parent = { + parent = None; + siblings = [*] + }, + siblings = [*; subtree 920; subtree 950] + }, + siblings = [subtree 20; *; subtree 80] + }, * filled by subtree 50 + + +We still do need to keep track of what fills the outermost targetted position---`* filled by subtree 50`---because that contain a subtree of arbitrary complexity, that is not determined by the rest of this data structure. + +For simplicity, I'll continue to use the abbreviated form: + + {parent = ...; siblings = [subtree 20; *; subtree 80]}, * filled by subtree 50 + +But that should be understood as standing for the more fully-spelled-out structure. Structures of this sort are called **tree zippers**. They should already seem intuitively similar to list zippers, at least in what we're using them to represent. I think it may also be helpful to call them **targetted trees**, though, and so will be switching back and forth between these different terms. + +Moving left in our targetted tree that's targetted on `node 50` would be a matter of shifting the `*` leftwards: + + {parent = ...; siblings = [*; subtree 50; subtree 80]}, * filled by subtree 20 + +and similarly for moving right. If the sibling list is implemented as a list zipper, you should already know how to do that. If one were designing a tree zipper for a more restricted kind of tree, however, such as a binary tree, one would probably not represent siblings with a list zipper, but with something more special-purpose and economical. + +Moving downward in the tree would be a matter of constructing a tree targetted on some child of `node 20`, with the first part of the targetted tree above as its parent: + + { + parent = {parent = ...; siblings = [*; subtree 50; subtree 80]}; + siblings = [*; leaf 2; leaf 3] + }, * filled by leaf 1 + +How would we move upward in a tree? Well, we'd build a regular, untargetted tree with a root node---let's call it `20'`---and whose children are given by the outermost sibling list in the targetted tree above, after inserting the targetted subtree into the `*` position: + + node 20' + / | \ + / | \ + leaf 1 leaf 2 leaf 3 + +We'll call this new untargetted tree `subtree 20'`. The result of moving upward from our previous targetted tree, targetted on `leaf 1`, would be the outermost `parent` element of that targetted tree, with `subtree 20'` being the subtree that fills that parent's target position `*`: + + { + parent = ...; + siblings = [*; subtree 50; subtree 80] + }, * filled by subtree 20' + +Or, spelling that structure out fully: + + { + parent = { + parent = { + parent = None; + siblings = [*] + }, + siblings = [*; subtree 920; subtree 950] + }, + siblings = [*; subtree 50; subtree 80] + }, * filled by subtree 20' + +Moving upwards yet again would get us: + + { + parent = { + parent = None; + siblings = [*] + }, + siblings = [*; subtree 920; subtree 950] + }, * filled by subtree 500' + +where `subtree 500'` refers to a tree built from a root node whose children are given by the list `[*; subtree 50; subtree 80]`, with `subtree 20'` inserted into the `*` position. Moving upwards yet again would get us: + + { + parent = None; + siblings = [*] + }, * filled by tree 9200' + +where the targetted element is the root of our base tree. Like the "moving backward" operation for the list zipper, this "moving upward" operation is supposed to be reminiscent of closing a zipper, and that's why these data structures are called zippers. + +We haven't given you a real implementation of the tree zipper, but only a suggestive notation. We have however told you enough that you should be able to implement it yourself. Or if you're lazy, you can read: + +* [[!wikipedia Zipper (data structure)]] +* Huet, Gerard. ["Functional Pearl: The Zipper"](http://www.st.cs.uni-sb.de/edu/seminare/2005/advanced-fp/docs/huet-zipper.pdf) Journal of Functional Programming 7 (5): 549-554, September 1997. +* As always, [Oleg](http://okmij.org/ftp/continuations/Continuations.html#zipper) takes this a few steps deeper. + + diff --git a/week11.mdwn b/week11.mdwn index 357711a0..ceece2e7 100644 --- a/week11.mdwn +++ b/week11.mdwn @@ -3,1751 +3,13 @@ The material here benefited from many discussions with Ken Shan. [[!toc]] -##List Zippers## +##[[Tree and List Zippers]]## -Say you've got some moderately-complex function for searching through a list, for example: +##[[Coroutines and Aborts]]## - let find_nth (test : 'a -> bool) (n : int) (lst : 'a list) : (int * 'a) -> - let rec helper (position : int) n lst = - match lst with - | [] -> failwith "not found" - | x :: xs when test x -> (if n = 1 - then (position, x) - else helper (position + 1) (n - 1) xs - ) - | x :: xs -> helper (position + 1) n xs - in helper 0 n lst;; +##[[From Lists to Continuations]]## -This searches for the `n`th element of a list that satisfies the predicate `test`, and returns a pair containing the position of that element, and the element itself. Good. But now what if you wanted to retrieve a different kind of information, such as the `n`th element matching `test`, together with its preceding and succeeding elements? In a real situation, you'd want to develop some good strategy for reporting when the target element doesn't have a predecessor and successor; but we'll just simplify here and report them as having some default value: - - let find_nth' (test : 'a -> bool) (n : int) (lst : 'a list) (default : 'a) : ('a * 'a * 'a) -> - let rec helper (predecessor : 'a) n lst = - match lst with - | [] -> failwith "not found" - | x :: xs when test x -> (if n = 1 - then (predecessor, x, match xs with [] -> default | y::ys -> y) - else helper x (n - 1) xs - ) - | x :: xs -> helper x n xs - in helper default n lst;; - -This duplicates a lot of the structure of `find_nth`; it just has enough different code to retrieve different information when the matching element is found. But now what if you wanted to retrieve yet a different kind of information...? - -Ideally, there should be some way to factor out the code to find the target element---the `n`th element of the list satisfying the predicate `test`---from the code that retrieves the information you want once the target is found. We might build upon the initial `find_nth` function, since that returns the *position* of the matching element. We could hand that result off to some other function that's designed to retrieve information of a specific sort surrounding that position. But suppose our list has millions of elements, and the target element is at position 600512. The search function will already have traversed 600512 elements of the list looking for the target, then the retrieval function would have to *start again from the beginning* and traverse those same 600512 elements again. It could go a bit faster, since it doesn't have to check each element against `test` as it traverses. It already knows how far it has to travel. But still, this should seem a bit wasteful. - -Here's an idea. What if we had some way of representing a list as "broken" at a specific point. For example, if our base list is: - - [10; 20; 30; 40; 50; 60; 70; 80; 90] - -we might imagine the list "broken" at position 3 like this (positions are numbered starting from 0): - - 40; - 30; 50; - 20; 60; - [10; 70; - 80; - 90] - -Then if we move one step forward in the list, it would be "broken" at position 4: - - 50; - 40; 60; - 30; 70; - 20; 80; - [10; 90] - -If we had some convenient representation of these "broken" lists, then our search function could hand *that* off to the retrieval function, and the retrieval function could start right at the position where the list was broken, without having to start at the beginning and traverse many elements to get there. The retrieval function would also be able to inspect elements both forwards and backwards from the position where the list was "broken". - -The kind of data structure we're looking for here is called a **list zipper**. To represent our first broken list, we'd use two lists: (1) containing the elements in the left branch, preceding the target element, *in the order reverse to their appearance in the base list*. (2) containing the target element and the rest of the list, in normal order. So: - - 40; - 30; 50; - 20; 60; - [10; 70; - 80; - 90] - -would be represented as `([30; 20; 10], [40; 50; 60; 70; 80; 90])`. To move forward in the base list, we pop the head element `40` off of the head element of the second list in the zipper, and push it onto the first list, getting `([40; 30; 20; 10], [50; 60; 70; 80; 90])`. To move backwards again, we pop off of the first list, and push it onto the second. To reconstruct the base list, we just "move backwards" until the first list is empty. (This is supposed to evoke the image of zipping up a zipper; hence the data structure's name.) - -We had some discussio in seminar of the right way to understand the "zipper" metaphor. I think it's best to think of the tab of the zipper being here: - - t - a - b - 40; - 30; 50; - 20; 60; - [10; 70; - 80; - 90] - -And imagine that you're just seeing the left half of a real-zipper, rotated 60 degrees counter-clockwise. When the list is all "zipped up", we've "move backwards" to the state where the first element is targetted: - - ([], [10; 20; 30; 40; 50; 60; 70; 80; 90]) - -However you understand the "zipper" metaphor, this is a very handy datastructure, and it will become even more handy when we translate it over to more complicated base structures, like trees. To help get a good conceptual grip on how to do that, it's useful to introduce a kind of symbolism for talking about zippers. This is just a metalanguage notation, for us theorists; we don't need our programs to interpret the notation. We'll use a specification like this: - - [10; 20; 30; *; 50; 60; 70; 80; 90], * filled by 40 - -to represent a list zipper where the break is at position 3, and the element occupying that position is 40. For a list zipper, this is implemented using the pairs-of-lists structure described above. - - -##Tree Zippers## - -Now how could we translate a zipper-like structure over to trees? What we're aiming for is a way to keep track of where we are in a tree, in the same way that the "broken" lists let us keep track of where we are in the base list. - -It's important to set some ground rules for what will follow. If you don't understand these ground rules you will get confused. First off, for many uses of trees one wants some of the nodes or leafs in the tree to be *labeled* with additional information. It's important not to conflate the label with the node itself. Numerically one and the same piece of information---for example, the same `int`---could label two nodes of the tree without those nodes thereby being identical, as here: - - root - / \ - / \ - / \ label 1 - / \ - label 1 label 2 - -The leftmost leaf and the rightmost leaf have the same label; but they are different leafs. The leftmost leaf has a sibling leaf with the label 2; the rightmost leaf has no siblings that are leafs. Sometimes when one is diagramming trees, one will annotate the nodes with the labels, as above. Other times, when one is diagramming trees, one will instead want to annotate the nodes with tags to make it easier to refer to particular parts of the tree. So for instance, I could diagram the same tree as above as: - - 1 - / \ - 2 \ - / \ 5 - / \ - 3 4 - -Here I haven't drawn what the labels are. The leftmost leaf, the node tagged "3" in this diagram, doesn't have the label `3`. It has the label 1, as we said before. I just haven't put that into the diagram. The node tagged "2" doesn't have the label `2`. It doesn't have any label. The tree in this example only has information labeling its leafs, not any of its inner nodes. The identity of its inner nodes is exhausted by their position in the tree. - -That is a second thing to note. In what follows, we'll only be working with *leaf-labeled* trees. In some uses of trees, one also wants labels on inner nodes. But we won't be discussing any such trees now. Our trees only have labels on their leafs. The diagrams below will tag all of the nodes, as in the second diagram above, and won't display what the leafs' labels are. - -Final introductory comment: in particular applications, you may only need to work with binary trees---trees where internal nodes always have exactly two subtrees. That is what we'll work with in the homework, for example. But to get the guiding idea of how tree zippers work, it's helpful first to think about trees that permit nodes to have many subtrees. So that's how we'll start. - -Suppose we have the following tree: - - 9200 - / | \ - / | \ - / | \ - / | \ - / | \ - 500 920 950 - / | \ / | \ / | \ - 20 50 80 91 92 93 94 95 96 - 1 2 3 4 5 6 7 8 9 - -This is a leaf-labeled tree whose labels aren't displayed. The `9200` and so on are tags to make it easier for us to refer to particular parts of the tree. - -Suppose we want to represent that we're *at* the node marked `50`. We might use the following metalanguage notation to specify this: - - {parent = ...; siblings = [subtree 20; *; subtree 80]}, * filled by subtree 50 - -This is modeled on the notation suggested above for list zippers. Here `subtree 20` refers to the whole subtree rooted at node `20`: - - 20 - / | \ - 1 2 3 - -Similarly for `subtree 50` and `subtree 80`. We haven't said yet what goes in the `parent = ...` slot. Well, the parent of a subtree targetted on `node 50` should intuitively be a tree targetted on `node 500`: - - {parent = ...; siblings = [*; subtree 920; subtree 950]}, * filled by subtree 500 - -And the parent of that targetted subtree should intuitively be a tree targetted on `node 9200`: - - {parent = None; siblings = [*]}, * filled by tree 9200 - -This tree has no parents because it's the root of the base tree. Fully spelled out, then, our tree targetted on `node 50` would be: - - { - parent = { - parent = { - parent = None; - siblings = [*] - }, * filled by tree 9200; - siblings = [*; subtree 920; subtree 950] - }, * filled by subtree 500; - siblings = [subtree 20; *; subtree 80] - }, * filled by subtree 50 - -In fact, there's some redundancy in this structure, at the points where we have `* filled by tree 9200` and `* filled by subtree 500`. Since node 9200 doesn't have any label attached to it, the subtree rooted in it is determined by the rest of this structure; and so too with `subtree 500`. So we could really work with: - - { - parent = { - parent = { - parent = None; - siblings = [*] - }, - siblings = [*; subtree 920; subtree 950] - }, - siblings = [subtree 20; *; subtree 80] - }, * filled by subtree 50 - - -We still do need to keep track of what fills the outermost targetted position---`* filled by subtree 50`---because that contain a subtree of arbitrary complexity, that is not determined by the rest of this data structure. - -For simplicity, I'll continue to use the abbreviated form: - - {parent = ...; siblings = [subtree 20; *; subtree 80]}, * filled by subtree 50 - -But that should be understood as standing for the more fully-spelled-out structure. Structures of this sort are called **tree zippers**. They should already seem intuitively similar to list zippers, at least in what we're using them to represent. I think it may also be helpful to call them **targetted trees**, though, and so will be switching back and forth between these different terms. - -Moving left in our targetted tree that's targetted on `node 50` would be a matter of shifting the `*` leftwards: - - {parent = ...; siblings = [*; subtree 50; subtree 80]}, * filled by subtree 20 - -and similarly for moving right. If the sibling list is implemented as a list zipper, you should already know how to do that. If one were designing a tree zipper for a more restricted kind of tree, however, such as a binary tree, one would probably not represent siblings with a list zipper, but with something more special-purpose and economical. - -Moving downward in the tree would be a matter of constructing a tree targetted on some child of `node 20`, with the first part of the targetted tree above as its parent: - - { - parent = {parent = ...; siblings = [*; subtree 50; subtree 80]}; - siblings = [*; leaf 2; leaf 3] - }, * filled by leaf 1 - -How would we move upward in a tree? Well, we'd build a regular, untargetted tree with a root node---let's call it `20'`---and whose children are given by the outermost sibling list in the targetted tree above, after inserting the targetted subtree into the `*` position: - - node 20' - / | \ - / | \ - leaf 1 leaf 2 leaf 3 - -We'll call this new untargetted tree `subtree 20'`. The result of moving upward from our previous targetted tree, targetted on `leaf 1`, would be the outermost `parent` element of that targetted tree, with `subtree 20'` being the subtree that fills that parent's target position `*`: - - { - parent = ...; - siblings = [*; subtree 50; subtree 80] - }, * filled by subtree 20' - -Or, spelling that structure out fully: - - { - parent = { - parent = { - parent = None; - siblings = [*] - }, - siblings = [*; subtree 920; subtree 950] - }, - siblings = [*; subtree 50; subtree 80] - }, * filled by subtree 20' - -Moving upwards yet again would get us: - - { - parent = { - parent = None; - siblings = [*] - }, - siblings = [*; subtree 920; subtree 950] - }, * filled by subtree 500' - -where `subtree 500'` refers to a tree built from a root node whose children are given by the list `[*; subtree 50; subtree 80]`, with `subtree 20'` inserted into the `*` position. Moving upwards yet again would get us: - - { - parent = None; - siblings = [*] - }, * filled by tree 9200' - -where the targetted element is the root of our base tree. Like the "moving backward" operation for the list zipper, this "moving upward" operation is supposed to be reminiscent of closing a zipper, and that's why these data structures are called zippers. - -We haven't given you a real implementation of the tree zipper, but only a suggestive notation. We have however told you enough that you should be able to implement it yourself. Or if you're lazy, you can read: - -* [[!wikipedia Zipper (data structure)]] -* Huet, Gerard. ["Functional Pearl: The Zipper"](http://www.st.cs.uni-sb.de/edu/seminare/2005/advanced-fp/docs/huet-zipper.pdf) Journal of Functional Programming 7 (5): 549-554, September 1997. -* As always, [Oleg](http://okmij.org/ftp/continuations/Continuations.html#zipper) takes this a few steps deeper. - - -##Same-fringe using a zipper-based coroutine## - -Recall back in [[Assignment4]], we asked you to enumerate the "fringe" of a leaf-labeled tree. Both of these trees (here I *am* drawing the labels in the diagram): - - . . - / \ / \ - . 3 1 . - / \ / \ - 1 2 2 3 - -have the same fringe: `[1;2;3]`. We also asked you to write a function that determined when two trees have the same fringe. The way you approached that back then was to enumerate each tree's fringe, and then compare the two lists for equality. Today, and then again in a later class, we'll encounter new ways to approach the problem of determining when two trees have the same fringe. - - -Supposing you did work out an implementation of the tree zipper, then one way to determine whether two trees have the same fringe would be: go downwards (and leftwards) in each tree as far as possible. Compare the targetted leaves. If they're different, stop because the trees have different fringes. If they're the same, then for each tree, move rightward if possible; if it's not (because you're at the rightmost position in a sibling list), more upwards then try again to move rightwards. Repeat until you are able to move rightwards. Once you do move rightwards, go downwards (and leftwards) as far as possible. Then you'll be targetted on the next leaf in the tree's fringe. The operations it takes to get to "the next leaf" may be different for the two trees. For example, in these trees: - - . . - / \ / \ - . 3 1 . - / \ / \ - 1 2 2 3 - -you won't move upwards at the same steps. Keep comparing "the next leafs" until they are different, or you exhaust the leafs of only one of the trees (then again the trees have different fringes), or you exhaust the leafs of both trees at the same time, without having found leafs with different labels. In this last case, the trees have the same fringe. - -If your trees are very big---say, millions of leaves---you can imagine how this would be quicker and more memory-efficient than traversing each tree to construct a list of its fringe, and then comparing the two lists so built to see if they're equal. For one thing, the zipper method can abort early if the fringes diverge early, without needing to traverse or build a list containing the rest of each tree's fringe. - -Let's sketch the implementation of this. We won't provide all the details for an implementation of the tree zipper, but we will sketch an interface for it. - -First, we define a type for leaf-labeled, binary trees: - - type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree) - -Next, the interface for our tree zippers. We'll help ourselves to OCaml's **record types**. These are nothing more than tuples with a pretty interface. Instead of saying: - - # type blah = Blah of (int * int * (char -> bool));; - -and then having to remember which element in the triple was which: - - # let b1 = Blah (1, (fun c -> c = 'M'), 2);; - Error: This expression has type int * (char -> bool) * int - but an expression was expected of type int * int * (char -> bool) - # (* damnit *) - # let b1 = Blah (1, 2, (fun c -> c = 'M'));; - val b1 : blah = Blah (1, 2,

-t "abSd" ~~> "ababd" -- - -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'): - -

- t "aSbS" - * -~~> t "aabS" - * -~~> "aabaab" -- -versus - -

- t "aSbS" - * -~~> t "aSbaSb" - * -~~> t "aabaSb" - * -~~> "aabaaabab" -- -versus - -

- t "aSbS" - * -~~> t "aSbaSb" - * -~~> t "aSbaaSbab" - * -~~> t "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, we'll agree to always evaluate the leftmost `'S'`, which -guarantees termination, and a final string without any `'S'` in it. - -This is a task well-suited to using a zipper. We'll define a function -`tz` (for task with zippers), which accomplishes the task by mapping a -char list zipper to a char list. We'll call the two parts of the -zipper `unzipped` and `zipped`; we start with a fully zipped list, and -move elements to the zipped part by pulling the zipped down until the -entire list has been unzipped (and so the zipped half of the zipper is empty). - -

-type 'a list_zipper = ('a list) * ('a list);; - -let rec tz (z:char list_zipper) = - match z with (unzipped, []) -> List.rev(unzipped) (* Done! *) - | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) - | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) - -# tz ([], ['a'; 'b'; 'S'; 'd']);; -- : char list = ['a'; 'b'; 'a'; 'b'; 'd'] - -# tz ([], ['a'; 'S'; 'b'; 'S']);; -- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -- -Note that this implementation enforces the evaluate-leftmost rule. -Task completed. - -One way to see exactly what is going on is to watch the zipper in -action by tracing the execution of `tz`. By using the `#trace` -directive in the Ocaml interpreter, the system will print out the -arguments to `tz` each time it is (recurcively) called. Note that the -lines with left-facing arrows (`<--`) show (recursive) calls to `tz`, -giving the value of its argument (a zipper), and the lines with -right-facing arrows (`-->`) show the output of each recursive call, a -simple list. - -

-# #trace tz;; -t1 is now traced. -# tz ([], ['a'; 'b'; 'S'; 'd']);; -tz <-- ([], ['a'; 'b'; 'S'; 'd']) -tz <-- (['a'], ['b'; 'S'; 'd']) (* Pull zipper *) -tz <-- (['b'; 'a'], ['S'; 'd']) (* Pull zipper *) -tz <-- (['b'; 'a'; 'b'; 'a'], ['d']) (* Special step *) -tz <-- (['d'; 'b'; 'a'; 'b'; 'a'], []) (* Pull zipper *) -tz --> ['a'; 'b'; 'a'; 'b'; 'd'] (* Output reversed *) -tz --> ['a'; 'b'; 'a'; 'b'; 'd'] -tz --> ['a'; 'b'; 'a'; 'b'; 'd'] -tz --> ['a'; 'b'; 'a'; 'b'; 'd'] -tz --> ['a'; 'b'; 'a'; 'b'; 'd'] -- : char list = ['a'; 'b'; 'a'; 'b'; 'd'] -- -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'; 'S'; 'd']` -is the result of the computation `a::(b::(S::(d::[])))` (or, in our old -style, `makelist a (makelist b (makelist S (makelist c empty)))`). -The recipe for constructing the list goes like this: - -

-(0) Start with the empty list [] -(1) make a new list whose first element is 'd' and whose tail is the list constructed in step (0) -(2) make a new list whose first element is 'S' and whose tail is the list constructed in step (1) ------------------------------------------ -(3) make a new list whose first element is 'b' and whose tail is the list constructed in step (2) -(4) make a new list whose first element is 'a' and whose tail is the list constructed in step (3) -- -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`. For instance, the -continuation corresponding to the portion of the recipe below the -horizontal line is the function `fun (tail:char list) -> a::(b::tail)`. - -This means that we can now represent the unzipped part of our -zipper--the part we've already unzipped--as a continuation: a function -describing how to finish building the list. We'll write a new -function, `tc` (for task with continuations), that will take an input -list (not a zipper!) and a continuation and return a processed list. -The structure and the behavior will follow that of `tz` above, with -some small but interesting differences. We've included the orginal -`tz` to facilitate detailed comparison: - -

-let rec tz (z:char list_zipper) = - match z with (unzipped, []) -> List.rev(unzipped) (* Done! *) - | (unzipped, 'S'::zipped) -> tz ((List.append unzipped unzipped), zipped) - | (unzipped, target::zipped) -> tz (target::unzipped, zipped);; (* Pull zipper *) - -let rec tc (l: char list) (c: (char list) -> (char list)) = - match l with [] -> List.rev (c []) - | 'S'::zipped -> tc zipped (fun x -> c (c x)) - | target::zipped -> tc zipped (fun x -> target::(c x));; - -# tc ['a'; 'b'; 'S'; 'd'] (fun x -> x);; -- : char list = ['a'; 'b'; 'a'; 'b'] - -# tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);; -- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -- -To emphasize the parallel, I've re-used the names `zipped` and -`target`. The trace of the procedure will show that these variables -take on the same values in the same series of steps as they did during -the execution of `tz` above. There will once again be one initial and -four recursive calls to `tc`, and `zipped` will take on the values -`"bSd"`, `"Sd"`, `"d"`, and `""` (and, once again, on the final call, -the first `match` clause will fire, so the the variable `zipper` will -not be instantiated). - -I have not called the functional argument `unzipped`, although that is -what the parallel would suggest. The reason is that `unzipped` is a -list, but `c` is a function. That's the most crucial difference, the -point of the excercise, and it should be emphasized. For instance, -you can see this difference in the fact that in `tz`, we have to glue -together the two instances of `unzipped` with an explicit `List.append`. -In the `tc` version of the task, we simply compose `c` with itself: -`c o c = fun x -> c (c x)`. - -Why use the identity function as the initial continuation? Well, if -you have already constructed the list "abSd", what's the next step in -the recipe to produce the desired result (which is the same list, -"abSd")? Clearly, the identity continuation. - -A good way to test your understanding is to figure out what the -continuation function `c` must be at the point in the computation when -`tc` is called with the first argument `"Sd"`. Two choices: is it -`fun x -> a::b::x`, or it is `fun x -> b::a::x`? -The way to see if you're right is to execute the following -command and see what happens: - - tc ['S'; 'd'] (fun x -> 'a'::'b'::x);; - -There are a number of interesting directions we can go with this task. -The task was chosen because the computation can be viewed as a -simplified picture of a computation using continuations, where `'S'` -plays the role of a control operator with some similarities to what is -often called `shift`. In the analogy, the list portrays a string of -functional applications, where `[f1; f2; f3; x]` represents `f1(f2(f3 -x))`. The limitation of the analogy is that it is only possible to -represent computations in which the applications are always -right-branching, i.e., the computation `((f1 f2) f3) x` cannot be -directly represented. - -One possibile development is that we could add a special symbol `'#'`, -and then the task would be to copy from the target `'S'` only back to -the closest `'#'`. This would allow the task to simulate delimited -continuations (for right-branching computations). - -The task is well-suited to the list zipper because the list monad has -an intimate connection with continuations. The following section -makes this connection. We'll return to the list task after talking -about generalized quantifiers below. - - -Rethinking the list monad -------------------------- - -To construct a monad, the key element is to settle on a type -constructor, and the monad naturally follows from that. We'll remind -you of some examples of how monads follow from the type constructor in -a moment. This will involve some review of familair material, but -it's worth doing for two reasons: it will set up a pattern for the new -discussion further below, and it will tie together some previously -unconnected elements of the course (more specifically, version 3 lists -and monads). - -For instance, take the **Reader Monad**. Once we decide that the type -constructor is - - type 'a reader = env -> 'a - -then the choice of unit and bind is natural: - - let r_unit (a : 'a) : 'a reader = fun (e : env) -> a - -Since the type of an `'a reader` is `env -> 'a` (by definition), -the type of the `r_unit` function is `'a -> env -> 'a`, which is a -specific case of the type of the *K* combinator. So it makes sense -that *K* is the unit for the reader monad. - -Since the type of the `bind` operator is required to be - - r_bind : ('a reader) -> ('a -> 'b reader) -> ('b reader) - -We can reason our way to the correct `bind` function as follows. We -start by declaring the types determined by the definition of a bind operation: - - let r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = ... - -Now we have to open up the `u` box and get out the `'a` object in order to -feed it to `f`. Since `u` is a function from environments to -objects of type `'a`, the way we open a box in this monad is -by applying it to an environment: - - ... f (u e) ... - -This subexpression types to `'b reader`, which is good. The only -problem is that we invented an environment `e` that we didn't already have , -so we have to abstract over that variable to balance the books: - - fun e -> f (u e) ... - -This types to `env -> 'b reader`, but we want to end up with `env -> -'b`. Once again, the easiest way to turn a `'b reader` into a `'b` is to apply it to an environment. So we end up as follows: - - r_bind (u : 'a reader) (f : 'a -> 'b reader) : ('b reader) = - f (u e) e - -And we're done. This gives us a bind function of the right type. We can then check whether, in combination with the unit function we chose, it satisfies the monad laws, and behaves in the way we intend. And it does. - -[The bind we cite here is a condensed version of the careful `let a = u e in ...` -constructions we provided in earlier lectures. We use the condensed -version here in order to emphasize similarities of structure across -monads.] - -The **State Monad** is similar. Once we've decided to use the following type constructor: - - type 'a state = store -> ('a, store) - -Then our unit is naturally: - - 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: - - let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = - ... f (...) ... - -But unlocking the `u` box is a little more complicated. As before, we -need to posit a state `s` that we can apply `u` to. Once we do so, -however, we won't have an `'a`, we'll have a pair whose first element -is an `'a`. So we have to unpack the pair: - - ... let (a, s') = u s in ... (f a) ... - -Abstracting over the `s` and adjusting the types gives the result: - - let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = - fun (s : store) -> let (a, s') = u s in f a s' - -The **Option/Maybe Monad** doesn't follow the same pattern so closely, so we -won't pause to explore it here, though conceptually its unit and bind -follow just as naturally from its type constructor. - -Our other familiar monad is the **List Monad**, which we were told -looks like this: - - type 'a list = ['a];; - l_unit (a : 'a) = [a];; - l_bind u f = List.concat (List.map f u);; - -Thinking through the list monad will take a little time, but doing so -will provide a connection with continuations. - -Recall that `List.map` takes a function and a list and returns the -result to applying the function to the elements of the list: - - List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]] - -and List.concat takes a list of lists and erases the embdded list -boundaries: - - List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3] - -And sure enough, - - l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] - -Now, why this unit, and why this bind? Well, ideally a unit should -not throw away information, so we can rule out `fun x -> []` as an -ideal unit. And units should not add more information than required, -so there's no obvious reason to prefer `fun x -> [x,x]`. In other -words, `fun x -> [x]` is a reasonable choice for a unit. - -As for bind, an `'a list` monadic object contains a lot of objects of -type `'a`, and we want to make some use of each of them (rather than -arbitrarily throwing some of them away). The only -thing we know for sure we can do with an object of type `'a` is apply -the function of type `'a -> 'a list` to them. Once we've done so, we -have a collection of lists, one for each of the `'a`'s. One -possibility is that we could gather them all up in a list, so that -`bind' [1;2] (fun i -> [i;i]) ~~> [[1;1];[2;2]]`. But that restricts -the object returned by the second argument of `bind` to always be of -type `'b list list`. We can elimiate that restriction by flattening -the list of lists into a single list: this is -just List.concat applied to the output of List.map. So there is some logic to the -choice of unit and bind for the list monad. - -Yet we can still desire to go deeper, and see if the appropriate bind -behavior emerges from the types, as it did for the previously -considered monads. But we can't do that if we leave the list type -as a primitive Ocaml type. However, we know several ways of implementing -lists using just functions. In what follows, we're going to use type -3 lists (the right fold implementation), though it's important to -wonder how things would change if we used some other strategy for -implementating lists. These were the lists that made lists look like -Church numerals with extra bits embdded in them: - - empty list: fun f z -> z - list with one element: fun f z -> f 1 z - list with two elements: fun f z -> f 2 (f 1 z) - list with three elements: fun f z -> f 3 (f 2 (f 1 z)) - -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 =

-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 *) -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;; -- -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 -> - reader_bind (treemonadizer f r) (fun y -> - reader_unit (Node (x, y))));; -- -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. +##[[List Monad as Continuation Monad]]## +##[[Manipulating Trees with Monads]]## -- 2.11.0