-# fun f z -> z;; -- : 'a -> 'b -> 'b =- -Finally, we're getting consistent principle types, so we can stop. -These types should remind you of the simply-typed lambda calculus -types for Church numerals (`(o -> o) -> o -> o`) with one extra bit -thrown in (in this case, and int). +and so on. To save time, we'll let the OCaml interpreter infer the +principle types of these functions (rather than inferring what the +types should be ourselves): + + # fun f z -> z;; + - : 'a -> 'b -> 'b =-# fun f z -> f 1 z;; -- : (int -> 'a -> 'b) -> 'a -> 'b = -# fun f z -> f 2 (f 1 z);; -- : (int -> 'a -> 'a) -> 'a -> 'a = -# fun f z -> f 3 (f 2 (f 1 z)) -- : (int -> 'a -> 'a) -> 'a -> 'a = -

-# let cons h t = h :: t;; (* Ocaml is stupid about :: *) -# l'_bind (fun f z -> f 1 (f 2 z)) - (fun i -> fun f z -> f i (f (i+1) z)) cons [];; -- : int list = [1; 2; 2; 3] -+ # let cons h t = h :: t;; (* OCaml is stupid about :: *) + # l'_bind (fun f z -> f 1 (f 2 z)) + (fun i -> fun f z -> f i (f (i+1) z)) cons [];; + - : int list = [1; 2; 2; 3] Ta da! -Just for mnemonic purposes (sneaking in an instance of eta reduction -to the definition of unit), we can summarize the result as follows: - - type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b - l'_unit x = fun f -> f x - l'_bind u f = fun k -> u (fun x -> f x k) - -To bad this digression, though it ties together various -elements of the course, has *no relevance whatsoever* to the topic of -continuations. Montague's PTQ treatment of DPs as generalized quantifiers ---------------------------------------------------------- @@ -270,46 +328,566 @@ generalized quantifier `fun pred -> pred j` of type `(e -> t) -> t`. Let's write a general function that will map individuals into their corresponding generalized quantifier: - gqize (x:e) = fun (p:e->t) -> p x + gqize (a : e) = fun (p : e -> t) -> p a + +This function is what Partee 1987 calls LIFT, and it would be +reasonable to use it here, but we will avoid that name, given that we +use that word to refer to other functions. -This function wraps up an individual in a fancy box. That is to say, +This function wraps up an individual in a box. That is to say, we are in the presence of a monad. The type constructor, the unit and the bind follow naturally. We've done this enough times that we won't belabor the construction of the bind function, the derivation is -similar to the List monad just given: +highly similar to the List monad just given: - type 'a continuation = ('a -> 'b) -> 'b - c_unit (x:'a) = fun (p:'a -> 'b) -> p x - c_bind (u:('a -> 'b) -> 'b) (f: 'a -> ('c -> 'd) -> 'd): ('c -> 'd) -> 'd = - fun (k:'a -> 'b) -> u (fun (x:'a) -> f x k) + type 'a continuation = ('a -> 'b) -> 'b + c_unit (a : 'a) = fun (p : 'a -> 'b) -> p a + c_bind (u : ('a -> 'b) -> 'b) (f : 'a -> ('c -> 'd) -> 'd) : ('c -> 'd) -> 'd = + fun (k : 'a -> 'b) -> u (fun (a : 'a) -> f a k) -How similar is it to the List monad? Let's examine the type -constructor and the terms from the list monad derived above: +Note that `c_bind` is exactly the `gqize` function that Montague used +to lift individuals into the continuation monad. + +That last bit in `c_bind` looks familiar---we just saw something like +it in the List monad. How similar is it to the List monad? Let's +examine the type constructor and the terms from the list monad derived +above: type ('a, 'b) list' = ('a -> 'b -> 'b) -> 'b -> 'b - l'_unit x = fun f -> f x - l'_bind u f = fun k -> u (fun x -> f x k) + l'_unit a = fun f -> f a + l'_bind u f = fun k -> u (fun a -> f a k) (We performed a sneaky but valid eta reduction in the unit term.) The unit and the bind for the Montague continuation monad and the homemade List monad are the same terms! In other words, the behavior of the List monad and the behavior of the continuations monad are -parallel in a deep sense. To emphasize the parallel, we can -instantiate the type of the list' monad using the Ocaml list type: - - type 'a c_list = ('a -> 'a list) -> 'a list - let c_list_unit x = fun f -> f x;; - let c_list_bind u f = fun k -> u (fun x -> f x k);; - -Have we really discovered that lists are secretly continuations? -Or have we merely found a way of simulating lists using list -continuations? Both perspectives are valid, and we can use our -intuitions about the list monad to understand continuations, and vice -versa. The connections will be expecially relevant when we consider -indefinites and Hamblin semantics on the linguistic side, and -non-determinism on the list monad side. - -Refunctionalizing zippers -------------------------- +parallel in a deep sense. + +Have we really discovered that lists are secretly continuations? Or +have we merely found a way of simulating lists using list +continuations? Well, strictly speaking, what we have done is shown +that one particular implementation of lists---the right fold +implementation---gives rise to a continuation monad fairly naturally, +and that this monad can reproduce the behavior of the standard list +monad. But what about other list implementations? Do they give rise +to monads that can be understood in terms of continuations? + +Manipulating trees with monads +------------------------------ + +This thread develops an idea based on a detailed suggestion of Ken +Shan's. We'll build a series of functions that operate on trees, +doing various things, including replacing leaves, counting nodes, and +converting a tree to a list of leaves. The end result will be an +application for continuations. + +From an engineering standpoint, we'll build a tree transformer that +deals in monads. We can modify the behavior of the system by swapping +one monad for another. (We've already seen how adding a monad can add +a layer of funtionality without disturbing the underlying system, for +instance, in the way that the reader monad allowed us to add a layer +of intensionality to an extensional grammar, but we have not yet seen +the utility of replacing one monad with other.) + +First, we'll be needing a lot of trees during the remainder of the +course. Here's a type constructor for binary trees: + + type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree) + +These are trees in which the internal nodes do not have labels. [How +would you adjust the type constructor to allow for labels on the +internal nodes?] + +We'll be using trees where the nodes are integers, e.g., + + +

+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. + +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.