-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 simple task gives rise to considerable 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" - * -~~> ... -- -Apparently, 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 from the zipped part to the unzipped part by pulling the -zipper down until the entire list has been unzipped (at which point -the zipped half of the zipper will be 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 (recursively) 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 zippers using procedures instead. - -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 -(or group of steps) 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 -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'; 'd'] - -# tc ['a'; 'S'; 'b'; 'S'] (fun x -> x);; -- : char list = ['a'; 'a'; 'b'; 'a'; 'a'; 'b'] -- -To emphasize the parallel, we'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` (in -`tz`) is a list, but `c` (in `tc`) is a function. ('c' stands for -'continuation', of course.) 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 (and -relatively computationally inefficient) `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 initial list `"abSd"`, what's the next -step in the recipe to produce the desired result, i.e, the very 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 reason this task was chosen is because it 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 input list portrays a -sequence 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 with embedded prompts. - -The reason the task is well-suited to the list zipper is in part -because the list monad has an intimate connection with continuations. -The following section explores this connection. We'll return to the -list task after talking about generalized quantifiers below. - - - diff --git a/hints/assignment_7_hint_6.mdwn b/hints/assignment_7_hint_6.mdwn index 38f6c2e0..e513284b 100644 --- a/hints/assignment_7_hint_6.mdwn +++ b/hints/assignment_7_hint_6.mdwn @@ -20,14 +20,14 @@ (* if one_dpm isn't already false at (r, h), we want to check its behavior when updated with phi bind_set (unit_set one_dpm) phi === phi one_dpm; do you remember why? *) - let (truth_value, _, _) = one_dpm (r, h) + let (truth_value, r', h') = one_dpm (r, h) in let truth_value' = truth_value && (truths (phi one_dpm) (r, h) = []) (* new_dpm must return a (bool, r, h) *) - in (truth_value', r, h) + in (truth_value', r', h') in unit_set new_dpm;; - **Note: Simon pointed out a subtle error in this code, which we will look into fixing. At the moment, the subtle error is still there.** + **Thanks to Simon Charlow** for catching a subtle error in previous versions of this function. Fixed 1 Dec. * Representing \[[and φ ψ]] is simple: @@ -38,6 +38,8 @@ * Here are `or` and `if`: + (These probably still manifest the bug Simon spotted.) + let or_op (phi : clause) (psi : clause) = fun one_dpm -> unit_set ( fun (r, h) -> diff --git a/index.mdwn b/index.mdwn index 5811ceee..1d090eb6 100644 --- a/index.mdwn +++ b/index.mdwn @@ -67,9 +67,11 @@ preloaded is available at [[assignment 3 evaluator]]. (30 Nov) Lecture notes for [[Week11]]; [[Assignment8]]. -> Topics: Zippers, Continuations +> Topics: [[Tree and List Zippers]]; [[Coroutines and Aborts]]; [[From List Zippers to Continuations]]. -(6 Dec) Lecture notes for Week12 +(6 Dec) Lecture notes for [[Week12]] + +> Topics: [[List Monad as Continuation Monad]]; [[Manipulating Trees with Monads]]; ... (13 Dec) Lecture notes for Week13 diff --git a/list_monad_as_continuation_monad.mdwn b/list_monad_as_continuation_monad.mdwn index 8e3551a9..438e2a80 100644 --- a/list_monad_as_continuation_monad.mdwn +++ b/list_monad_as_continuation_monad.mdwn @@ -1,50 +1,51 @@ [[!toc]] +We're going to come at continuations from three different directions, and each +time we're going to end up at the same place: a particular monad, which we'll +call the continuation monad. + Rethinking the list monad ------------------------- -To construct a monad, the key element is to settle on a type -constructor, and the monad more or less 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 +To construct a monad, the key element is to settle on how to implement its type, and +the monad more or less naturally follows from that. +We'll remind you of some examples of how monads follow from their types +in a moment. This will involve some review of familiar 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 +For instance, take the **Reader Monad**. Once we decide to define its type as: - type 'a reader = env -> 'a + 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 + let r_unit (a : 'a) : 'a reader = fun (e : env) -> a The reason this is a fairly natural choice is that because the type of an `'a reader` is `env -> 'a` (by definition), the type of the `r_unit` function is `'a -> env -> 'a`, which is an instance of the -type of the *K* combinator. So it makes sense that *K* is the unit +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) + r_bind : 'a reader -> ('a -> 'b reader) -> 'b reader We can reason our way to the traditional reader `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) = ... + 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 made use of an environment `e` that we didn't already have, @@ -59,9 +60,8 @@ and using the Curry-Howard labeling of the proof as our bind term.] 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 -+ r_bind (u : 'a reader) (f : 'a -> 'b reader) : 'b reader = + fun e -> 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. @@ -72,28 +72,28 @@ monads.] The **State Monad** is similar. Once we've decided to use the following type constructor: - type 'a state = store -> ('a, store) + type 'a state = store -> ('a, store) Then our unit is naturally: - let s_unit (a : 'a) : ('a state) = fun (s : store) -> (a, s) + let s_unit (a : 'a) : 'a state = fun (s : store) -> (a, s) And we can reason our way to the bind function in a way similar to the reasoning given above. First, we need to apply `f` to the contents of the `u` box: - let s_bind (u : 'a state) (f : 'a -> 'b state) : 'b state = - ... f (...) ... + 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 +need to posit a store `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) ... + ... 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' + 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 @@ -102,61 +102,61 @@ 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);; + (* 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: +result of applying the function to the elements of the list: - List.map (fun i -> [i;i+1]) [1;2] ~~> [[1; 2]; [2; 3]] + 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 +and `List.concat` takes a list of lists and erases one level of embedded list boundaries: - List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3] + List.concat [[1; 2]; [2; 3]] ~~> [1; 2; 2; 3] -And sure enough, +And sure enough, - l_bind [1;2] (fun i -> [i, i+1]) ~~> [1; 2; 2; 3] + 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 +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 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 +the function of type `'a -> 'b 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 +`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 +type `('something list) list`. We can eliminate 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. +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 +a primitive OCaml type. However, we know several ways of implementing +lists using just functions. In what follows, we're going to use version 3 lists, the right fold implementation (though it's important and intriguing 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: +strategy for implementing lists). These were the lists that made +lists look like Church numerals with extra bits embedded 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)) + 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 @@ -174,60 +174,59 @@ types should be ourselves): 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). +thrown in, the type of the element at the head of the list +(in this case, an `int`). So here's our type constructor for our hand-rolled lists: - type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b + type 'b list' = (int -> 'b -> 'b) -> 'b -> 'b Generalizing to lists that contain any kind of element (not just -ints), we have +`int`s), we have - type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b + type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b 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 fully grasp the role of the `'b`'s +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 (a : 'a) : ('a, 'b) list = fun a -> fun f z -> f a z + l'_unit (a : 'a) : ('a, 'b) list = fun a -> fun k z -> k 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. +Take an `'a` and return its v3-style singleton. 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' = ... Unpacking the types gives: - l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) - (f : 'a -> ('c -> 'd -> 'd) -> 'd -> 'd) - : ('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 = ... -Perhaps a bit intimiating. +Perhaps a bit intimidating. 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 +be no more intimidated 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 +[This would be a good time to try to reason your way to your own term having the type just specified. 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` 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: +argument that is a function expecting an `'a` and a `'b`. `u` will +fold that function over its type `'a` members, and that's where we can get at the `'a`s we need. Thus: ... u (fun (a : 'a) (b : 'b) -> ... f a ... ) ... -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)`: +In order for `u` to have the kind of argument it needs, the `fun a b -> ... f a ...` has to have type `'a -> 'b -> 'b`; so 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`: ... 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: +Now the function we're supplying to `u` also receives an argument `b` of type `'b`, so we can supply that to `f a k`, getting a result of type `'b`, as we need: ... u (fun (a : 'a) (b : 'b) -> f a k b) ... @@ -237,75 +236,77 @@ Now, we've used a `k` that we pulled out of nowhere, so we need to abstract over 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 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 -> u (fun a b -> f a k b) + l'_bind (u : ('a -> 'b -> 'b) -> 'b -> 'b) + (f : 'a -> ('c -> 'b -> 'b) -> 'b -> 'b) + : ('c -> 'b -> 'b) -> 'b -> 'b = + 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 + 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? +Now let's think about what this does. It's a wrapper around `u`. In order to behave as the list `v` 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 `v` would. +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: +Suppose we have a list' whose contents are `[1; 2; 4; 8]`---that is, our list' `u` will be `fun f z -> f 1 (f 2 (f 4 (f 8 z)))`. 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]] = + v = + List.concat (List.map f u) = + List.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 +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. 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: + fun k z -> u (fun a b -> f a k b) z + +do that? Well, for each element `a` in `u`, it applies `f` to that `a`, getting one of the lists: - [] + [] ; result of applying f to leftmost a [2] [2; 4] - [2; 4; 8] + [2; 4; 8] ; result of applying f to rightmost a -(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. +(or rather, their list' versions). Then it takes the accumulated result `b` of previous steps in the `k`,`z`-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 of the `k`,`z`-fold as the new accumulated result `b`. So if, for example, we let `k` be `+` and `z` be `0`, then the computation would proceed: 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 + 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: +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 + 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 +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 + fun k z -> List.fold_right k v z = + fun k z -> List.fold_right k (List.concat (List.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 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] - - l'_bind (fun f z -> f 1 (f 2 z)) - (fun i -> fun f z -> f i (f (i+1) z)) ~~>

-let t1 = Node ((Node ((Leaf 2), (Leaf 3))), - (Node ((Leaf 5),(Node ((Leaf 7), - (Leaf 11)))))) - - . - ___|___ - | | - . . -_|__ _|__ -| | | | -2 3 5 . - _|__ - | | - 7 11 -+ 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, + let rec tree_map (leaf_modifier : 'a -> 'b) (t : 'a tree) : 'b tree = + match t with + | Leaf i -> Leaf (leaf_modifier i) + | Node (l, r) -> Node (tree_map leaf_modifier l, + tree_map leaf_modifier r);; + +`tree_map` 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 + let double i = i + i;; + tree_map 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 `tree_map` +code. However, because we've left what to do to each leaf as 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 +`tree_map`. For instance, we can easily square each leaf instead by supplying the appropriate `int -> int` operation in place of `double`: -

-let square x = x * x;; -treemap square t1;; -- : int tree =ppp -Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) -+ let square i = i * i;; + tree_map 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 +Note that what `tree_map` 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. +to each subpart of the computation. In other words, `tree_map` has the +behavior of a reader monad. Let's make that explicit. -In general, we're on a journey of making our treemap function more and -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 +In general, we're on a journey of making our `tree_map` function more and +more flexible. So the next step---combining the tree transformer with +a reader monad---is to have the `tree_map` 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 -+\tree (. (. (f 2) (f 3)) (. (f 5) (. (f 7) (f 11)))) + + \f . + _____|____ + | | + . . + __|___ __|___ + | | | | + f 2 f 3 f 5 . + __|___ + | | + f 7 f 11 That is, we want to transform the ordinary tree `t1` (of type `int -tree`) into a reader object of type `(int->int)-> int tree`: something -that, when you apply it to an `int->int` function returns an `int -tree` in which each leaf `x` has been replaced with `(f x)`. +tree`) into a reader object of type `(int -> int) -> int tree`: something +that, when you apply it to an `int -> int` function `f` returns an `int +tree` in which each leaf `i` has been replaced with `f i`. 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))));; -+Jacobson-inspired link monad), etc. In the present case, it will be +enough to expect that our "environment" will be some function of type +`int -> int`. + + type 'a reader = (int -> int) -> 'a;; (* mnemonic: e for environment *) + let reader_unit (a : 'a) : 'a reader = fun _ -> a;; + let reader_bind (u: 'a reader) (f : 'a -> 'b reader) : 'b reader = fun e -> f (u e) e;; + +It would be a simple matter to turn an *integer* into an `int reader`: + + let int_readerize : int -> int reader = fun (a : int) -> fun (modifier : int -> int) -> modifier a;; + int_readerize 2 (fun i -> i + i);; + - : int = 4 + +But how do we do the analagous transformation when our `int`s are scattered over the leaves of a tree? How do we turn an `int tree` into a reader? +A tree is not the kind of thing that we can apply a +function of type `int -> int` to. + +But we can do this: + + let rec tree_monadize (f : 'a -> 'b reader) (t : 'a tree) : 'b tree reader = + match t with + | Leaf i -> reader_bind (f i) (fun i' -> reader_unit (Leaf i')) + | Node (l, r) -> reader_bind (tree_monadize f l) (fun x -> + reader_bind (tree_monadize 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 +something of type `'a` into an `'b reader`, and I'll show you how to +turn an `'a tree` into an `'b tree reader`. In more fanciful terms, +the `tree_monadize` function builds plumbing that connects all of the leaves of a tree into one connected monadic network; it threads the -monad through the leaves. +`'b reader` monad through the original tree's 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))) -+ # tree_monadize int_readerize t1 double;; + - : int tree = + Node (Node (Leaf 4, Leaf 6), Node (Leaf 10, Node (Leaf 14, Leaf 22))) Here, our environment is the doubling function (`fun i -> i + i`). If -we apply the very same `int tree reader` (namely, `treemonadizer -int2int_reader t1`) to a different `int->int` function---say, the +we apply the very same `int tree reader` (namely, `tree_monadize +int_readerize 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))) -+ # tree_monadize int_readerize t1 square;; + - : 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 +Now that we have a tree transformer that accepts a reader monad as a parameter, we can see what it would take to swap in a different monad. + + + For instance, we can use a state monad to count the number of nodes in 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);; -+ type 'a state = int -> 'a * int;; + let state_unit a = fun s -> (a, s);; + let state_bind_and_count u f = fun s -> let (a, s') = u s in f a (s' + 1);; -Gratifyingly, we can use the `treemonadizer` function without any +Gratifyingly, we can use the `tree_monadize` function without any modification whatsoever, except for replacing the (parametric) type -`reader` with `state`: +`'b reader` with `'b state`, and substituting in the appropriate unit and bind: -

-let rec treemonadizer (f:'a -> 'b state) (t:'a tree):('b tree) state = - 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))));; -+ let rec tree_monadize (f : 'a -> 'b state) (t : 'a tree) : 'b tree state = + match t with + | Leaf i -> state_bind_and_count (f i) (fun i' -> state_unit (Leaf i')) + | Node (l, r) -> state_bind_and_count (tree_monadize f l) (fun x -> + state_bind_and_count (tree_monadize 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 -+ # tree_monadize 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 +`state` everywhere in `tree_monadize` 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])))] -+ # tree_monadize (fun i -> [ [i; square i] ]) 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))));; -+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 a = fun k -> k a;; + let continuation_bind u f = fun k -> u (fun a -> f a k);; + + let rec tree_monadize (f : 'a -> ('b, 'r) continuation) (t : 'a tree) : ('b tree, 'r) continuation = + match t with + | Leaf i -> continuation_bind (f i) (fun i' -> continuation_unit (Leaf i')) + | Node (l, r) -> continuation_bind (tree_monadize f l) (fun x -> + continuation_bind (tree_monadize 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. +`continuation` type in the appropriate place in the `tree_monadize` code. We then compute: -

-# treemonadizer (fun a c -> a :: (c a)) t1 (fun t -> []);; -- : int list = [2; 3; 5; 7; 11] -+ # tree_monadize (fun a k -> a :: (k 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 +note that an interestingly uninteresting thing happens if we use +`continuation_unit` as our first argument to `tree_monadize`, and then apply the result to the identity function: -

-# treemonadizer continuation_unit t1 (fun x -> x);; -- : int tree = -Node (Node (Leaf 2, Leaf 3), Node (Leaf 5, Node (Leaf 7, Leaf 11))) -+ # tree_monadize continuation_unit t1 (fun i -> i);; + - : 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 -+interesting functions for the first argument of `tree_monadize`: + + (* Simulating the tree reader: distributing a operation over the leaves *) + # tree_monadize (fun a k -> k (square a)) t1 (fun i -> i);; + - : int tree = + Node (Node (Leaf 4, Leaf 9), Node (Leaf 25, Node (Leaf 49, Leaf 121))) + + (* Simulating the int list tree list *) + # tree_monadize (fun a k -> k [a; square a]) t1 (fun i -> i);; + - : int list tree = + Node + (Node (Leaf [2; 4], Leaf [3; 9]), + Node (Leaf [5; 25], Node (Leaf [7; 49], Leaf [11; 121]))) + + (* Counting leaves *) + # tree_monadize (fun a k -> 1 + k a) t1 (fun i -> 0);; + - : int = 5 We could simulate the tree state example too, but it would require -generalizing the type of the continuation monad to +generalizing the type of the continuation monad to - type ('a -> 'b -> 'c) continuation = ('a -> 'b) -> 'c;; + type ('a, 'b, 'c) continuation = ('a -> 'b) -> 'c;; The binary tree monad --------------------- @@ -307,79 +296,74 @@ 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));; -+ type 'a tree = Leaf of 'a | Node of ('a tree) * ('a tree);; + let tree_unit (a: 'a) = Leaf a;; + let rec tree_bind (u : 'a tree) (f : 'a -> 'b tree) : 'b tree = + match u with + | Leaf a -> f a + | Node (l, r) -> Node ((tree_bind l f), (tree_bind r f));; For once, let's check the Monad laws. The left identity law is easy: - Left identity: bind (unit a) f = bind (Leaf a) f = fa + Left identity: bind (unit a) f = bind (Leaf a) f = f a To check the other two laws, we need to make the following observation: it is easy to prove based on `tree_bind` by a simple induction on the structure of the first argument that the tree resulting from `bind u f` is a tree with the same strucure as `u`, -except that each leaf `a` has been replaced with `fa`: - -\tree (. (fa1) (. (. (. (fa2)(fa3)) (fa4)) (fa5))) -

- . . - __|__ __|__ - | | | | - a1 . fa1 . - _|__ __|__ - | | | | - . a5 . fa5 - bind _|__ f = __|__ - | | | | - . a4 . fa4 - __|__ __|___ - | | | | - a2 a3 fa2 fa3 -+except that each leaf `a` has been replaced with `f a`: + +\tree (. (f a1) (. (. (. (f a2) (f a3)) (f a4)) (f a5))) + + . . + __|__ __|__ + | | | | + a1 . f a1 . + _|__ __|__ + | | | | + . a5 . f a5 + bind _|__ f = __|__ + | | | | + . a4 . f a4 + __|__ __|___ + | | | | + a2 a3 f a2 f a3 Given this equivalence, the right identity law - Right identity: bind u unit = u + Right identity: bind u unit = u falls out once we realize that - bind (Leaf a) unit = unit a = Leaf a + 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) + Associativity: bind (bind u f) g = bind u (\a. bind (f a) g) we'll give an example that will show how an inductive proof would proceed. Let `f a = Node (Leaf a, Leaf a)`. Then -\tree (. (. (. (. (a1)(a2))))) -\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))) )) -

- . - ____|____ - . . | | -bind __|__ f = __|_ = . . - | | | | __|__ __|__ - a1 a2 fa1 fa2 | | | | - a1 a1 a1 a1 -+\tree (. (. (. (. (a1) (a2))))) +\tree (. (. (. (. (a1) (a1)) (. (a1) (a1))))) + + . + ____|____ + . . | | + bind __|__ f = __|_ = . . + | | | | __|__ __|__ + a1 a2 f a1 f a2 | | | | + a1 a1 a1 a1 Now when we bind this tree to `g`, we get -

- . - ____|____ - | | - . . - __|__ __|__ - | | | | - ga1 ga1 ga1 ga1 -+ . + _____|______ + | | + . . + __|__ __|__ + | | | | + g a1 g a1 g a1 g a1 At this point, it should be easy to convince yourself that using the recipe on the right hand side of the associative law will @@ -390,6 +374,5 @@ 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. +that is intended to represent non-deterministic computations as a tree. diff --git a/new_stuff.mdwn b/new_stuff.mdwn index ce753428..8843d37e 100644 --- a/new_stuff.mdwn +++ b/new_stuff.mdwn @@ -1,20 +1,5 @@ Page for Chris and Jim to see what each other is working on, but hasn't necessarily been posted to main wiki index yet. -[[Week11]] - -[[Tree and List Zippers]] - -[[Coroutines and Aborts]] - -[[From Lists to Continuations]] - -[[List Monad as Continuation Monad]] - -[[Manipulating Trees with Monads]] - -[[Zipper-Lists-Continuations]] -- now obsolete? - -[[Assignment8]] [[Curry-Howard]] diff --git a/tree_and_list_zippers.mdwn b/tree_and_list_zippers.mdwn index 5d26ca97..ee627a7e 100644 --- a/tree_and_list_zippers.mdwn +++ b/tree_and_list_zippers.mdwn @@ -5,28 +5,28 @@ 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;; + 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;; + 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...? @@ -66,7 +66,7 @@ The kind of data structure we're looking for here is called a **list zipper**. T 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: +We had some discussion 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 @@ -93,7 +93,7 @@ to represent a list zipper where the break is at position 3, and the element occ 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: +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 leaves 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 / \ @@ -102,7 +102,7 @@ It's important to set some ground rules for what will follow. If you don't under / \ 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: +The leftmost leaf and the rightmost leaf have the same label; but they are different leaves. The leftmost leaf has a sibling leaf with the label 2; the rightmost leaf has no siblings that are leaves. 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 / \ @@ -111,9 +111,9 @@ The leftmost leaf and the rightmost leaf have the same label; but they are diffe / \ 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. +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 leaves, 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. +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 leaves. The diagrams below will tag all of the nodes, as in the second diagram above, and won't display what the leaves' 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. diff --git a/week11.mdwn b/week11.mdwn index df004311..f2fb9dbd 100644 --- a/week11.mdwn +++ b/week11.mdwn @@ -5,83 +5,8 @@ The material here benefited from many discussions with Ken Shan. ##[[Coroutines and Aborts]]## -##[[From Lists to Continuations]]## +##[[From List Zippers to Continuations]]## -##[[List Monad as Continuation Monad]]## +continue with [[Week12]]... -##[[Manipulating Trees with Monads]]## - - --------------------------------------- - -In coming weeks, we'll learn about a different way to create threads, that relies on **continuations** rather than on those two tools. All of these tools are inter-related. As Oleg says, "Zipper can be viewed as a delimited continuation reified as a data structure." These different tools are also inter-related with monads. Many of these tools can be used to define the others. We'll explore some of the connections between them in the remaining weeks, but we encourage you to explore more. - - -##Introducing Continuations## - -A continuation is "the rest of the program." Or better: an **delimited continuation** is "the rest of the program, up to a certain boundary." An **undelimited continuation** is "the rest of the program, period." - -Even if you haven't read specifically about this notion (for example, even if you haven't read Chris and Ken's work on using continuations in natural language semantics), you'll have brushed shoulders with it already several times in this course. - -A naive semantics for atomic sentences will say the subject term is of type `e`, and the predicate of type `e -> t`, and that the subject provides an argument to the function expressed by the predicate. - -Monatague proposed we instead take subject terms to be of type `(e -> t) -> t`, and that now it'd be the predicate (still of type `e -> t`) that provides an argument to the function expressed by the subject. - -If all the subject did then was supply an `e` to the `e -> t` it receives as an argument, we wouldn't have gained anything we weren't already able to do. But of course, there are other things the subject can do with the `e -> t` it receives as an argument. For instance, it can check whether anything in the domain satisfies that `e -> t`; or whether most things do; and so on. - -This inversion of who is the argument and who is the function receiving the argument is paradigmatic of working with continuations. We did the same thing ourselves back in the early days of the seminar, for example in our implementation of pairs. In the untyped lambda calculus, we identified the pair `(x, y)` with a function: - - \handler. handler x y - -A pair-handling function would accept the two elements of a pair as arguments, and then do something with one or both of them. The important point here is that the handler was supplied as an argument to the pair. Eventually, the handler would itself be supplied with arguments. But only after it was supplied as an argument to the pair. This inverts the order you'd expect about what is the data or argument, and what is the function that operates on it. - -Consider a complex computation, such as: - - 1 + 2 * (1 - g (3 + 4)) - -Part of this computation---`3 + 4`---leads up to supplying `g` with an argument. The rest of the computation---`1 + 2 * (1 - ___)`---waits for the result of applying `g` to that argument and will go on to do something with it (inserting the result into the `___` slot). That "rest of the computation" can be regarded as a function: - - \result. 1 + 2 * (1 - result) - -This function will be applied to whatever is the result of `g (3 + 4)`. So this function can be called the *continuation* of that application of `g`. For some purposes, it's useful to be able to invert the function/argument order here, and rather than supplying the result of applying `g` to the continuation, we instead supply the continuation to `g`. Well, not to `g` itself, since `g` only wants a single `int` argument. But we might build some `g`-like function which accepts not just an `int` argument like `g` does, but also a continuation argument. - -Go back and read the material on "Aborting a Search Through a List" in [[Week4]] for an example of doing this. - -In very general terms, the strategy is to work with functions like this: - - let g' k (i : int) = - ... do stuff ... - ... if you want to abort early, supply an argument to k ... - ... do more stuff ... - ... normal result - in let gcon = fun result -> 1 + 2 * (1 - result) - in gcon (g' gcon (3 + 4)) - -It's a convention to use variables like `k` for continuation arguments. If the function `g'` never supplies an argument to its contination argument `k`, but instead just finishes evaluating to a normal result, that normal result will be delivered to `g'`'s continuation `gcon`, just as happens when we don't pass around any explicit continuation variables. - -The above snippet of OCaml code doesn't really capture what happens when we pass explicit continuation variables. For suppose that inside `g'`, we do supply an argument to `k`. That would go into the `result` parameter in `gcon`. But then what happens once we've finished evaluating the application of `gcon` to that `result`? In the OCaml snippet above, the final value would then bubble up through the context in the body of `g'` where `k` was applied, and eventually out to the final line of the snippet, where it once again supplied an argument to `gcon`. That's not what happens with a real continuation. A real continuation works more like this: - - let g' k (i : int) = - ... do stuff ... - ... if you want to abort early, supply an argument to k ... - ... do more stuff ... - ... normal result - in let gcon = fun result -> - let final_value = 1 + 2 * (1 - result) - in end_program_with final_value - in gcon (g' gcon (3 + 4)) - -So once we've finished evaluating the application of `gcon` to a `result`, the program is finished. (This is how undelimited continuations behave. We'll discuss delimited continuations later.) - -So now, guess what would be the result of doing the following: - - let g' k (i : int) = - 1 + k i - in let gcon = fun result -> - let final_value = (1, result) - in end_program_with final_value - in gcon (g' gcon (3 + 4)) - - - diff --git a/week12.mdwn b/week12.mdwn new file mode 100644 index 00000000..e59349b0 --- /dev/null +++ b/week12.mdwn @@ -0,0 +1,7 @@ +The material here benefited from many discussions with Ken Shan. + +##[[List Monad as Continuation Monad]]## + +##[[Manipulating Trees with Monads]]## + +more to come... diff --git a/zipper-lists-continuations.mdwn b/zipper-lists-continuations.mdwn deleted file mode 100644 index 7687e657..00000000 --- a/zipper-lists-continuations.mdwn +++ /dev/null @@ -1,915 +0,0 @@ - -[[!toc]] - -Today we're going to encounter continuations. We're going to come at -them from three different directions, and each time we're going to end -up at the same place: a particular monad, which we'll call the -continuation monad. - -Much of this discussion benefited from detailed comments and -suggestions from Ken Shan. - - -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. - - -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. -