[[!toc]] # Continuations Last week we saw how to turn a list zipper into a continuation-based list processor. The function computed something we called "the task", which was a simplified langauge involving control operators. abSdS ~~> ababdS ~~> ababdababd The task is to process the list from left to right, and at each "S", double the list so far. Here, "S" is a kind of control operator, and captures the entire previous computation. We also considered a variant in which '#' delimited the portion of the list to be copied: ab#deSfg ~~> abededfg In this variant, "S" and "#" correspond to `shift` and `reset`, which provide access to delimited continuations. The expository logic of starting with this simplified task is the notion that as lists are to trees, so is this task to full-blown continuations. So to the extent that, say, list zippers are easier to grasp than tree zippers, the task is easier to grasp than full continuations. We then presented CPS transforms, and demonstrated how they provide an order-independent analysis of order of evaluation. In order to continue to explore continuations, we will proceed in the followin fashion. ## The continuation monad Let's take a look at some of our favorite monads from the point of view of types. Here, `==>` is the Kleisli arrow. Reader monad: types: Mα ==> β -> α ⇧: \ae.a : α -> Mα compose: \fghe.f(ghe)e : (Q->MR)->(P->MQ)->(P->MR) gloss: copy environment and distribute it to f and g State monad: types: α ==> β -> (α x β) ⇧: \ae.(a,e) : α -> Mα compose: \fghe.let (x,s) = ghe in fxs thread the state through g, then through f List monad: types: α ==> [α] ⇧: \a.[a] : α -> Mα compose: \fgh.concat(map f (gh)) gloss: compose f and g pointwise Maybe monad: types: α ==> Nothing | Just α ⇧: \a. Just a compose: \fgh. case gh of Nothing -> Nothing | Just x -> case fx of Nothing -> Nothing | Just y -> Just y gloss: strong Kline Now we need a type for a continuation. A continuized term is one that expects its continuation as an argument. The continuation of a term is a function from the normal value of that term to a result. So if the term has type continuized term has type α, the continuized version has type (α -> ρ) -> ρ: Continuation monad: types: Mα => (α -> ρ) -> ρ ⇧: \ak.ka compose: \fghk.f(\f'.g h (\g'.f(g'h) gloss: first give the continuation to f, then build a continuation out of the result to give to The first thing we should do is demonstrate that this monad is suitable for accomplishing the task. We lift the computation `("a" ++ ("b" ++ ("c" ++ "d")))` into the monad as follows: t1 = (map1 ((++) "a") (map1 ((++) "b") (map1 ((++) "c") (mid "d")))) Here, `(++) "a"` is a function of type [Char] -> [Char] that prepends the string "a", so `map1 ((++) "a")` takes a string continuation k and returns a new string continuation that takes a string s returns "a" ++ k(s). So `t1 (\k->k) == "abcd"`. We can now define a shift operator to perform the work of "S": shift u k = u(\s.k(ks)) Shift takes two arguments: a string continuation u of type (String -> String) -> String, and a string continuation k of type String -> String. Since u is the result returned by the argument to shift, it represents the tail of the list after the shift operator. Then k is the continuation of the expression headed by `shift`. So in order to execute the task, shift needs to invoke k twice. Note that the shift operator constructs a new continuation by composing its second argument with itself (i.e., the new doubled continuation is \s.k(ks)). Once it has constructed this new continuation, it delivers it as an argument to the remaining part of the computation! (map1 ((++) "a") (map1 ((++) "b") (shift (mid "d")))) (\k->k) == "ababd" Let's just make sure that we have the left-to-right evaluation we were hoping for by evaluating "abSdeSf": t6 = map1 ((++) "a") (map1 ((++) "b") (shift (map1 ((++) "d") (map1 ((++) "e") (shift (mid "f")))))) t6 (\k->k) == "ababdeababdef" As expected. In order to add a reset operator #, we can have # u k = k(u(\k.k)) ab#deSf ~~> abdedef Note that the lifting of the original unmonadized computation treated prepending "a" as a one-place operation. If we decompose this operation into a two-place operation of appending combined with a string "a", an interesting thing happens. map2 f u v k = u(\u' -> v (\v' -> k(f u' v'))) shift k = k (k "") t2 = map2 (++) (mid "a") (map2 (++) (mid "b") (map2 (++) shift (map2 (++) (mid "d") (mid [])))) t2 (\k->k) == "ababdd" First, we need map2 instead of map1. Second, the type of the shift operator will be a string continuation, rather than a function from string continuations to string continuations. But here's the interesting part: from the point of view of the shift operator, the continuation that it will be fed will be the entire rest of the computation. This includes not only what has come before, but what will come after it as well. That means when the continuation is doubled (self-composed), the result duplicates not only the prefix `(ab ~~> abab)`, but also the suffix `(d ~~> dd)`. In some sense, then, continuations allow functions to see into the future! What do you expect will be the result of executing "aScSe" under the second perspective? The answer to this question is not determined by the informal specification of the task that we've been using, since under the new perspective, we're copying complete (bi-directional) contexts rather than just left contexts. It might be natural to assume that what execution does is choose an S, and double its context, temporarily treating all other shift operators as dumb letters, then choosing a remaining S to execute. If that was our interpreation of the task, then no string with more than one S would ever terminate, since the number S's after each reduction step would always be 2(n-1), where n is the number before reduction. But there is another equally natural way to answer the question. Assume the leftmost S is executed first. What will the value of its continuation k be? It will be a function that maps a string s to the result of computing `ascSe`, which will be `ascascee`. So `k(k "")` will be `k(acacee)`, which will result in `a(acacee)ca(acacee)cee` (the parentheses are added to show stages in the construction of the final result). So the continuation monad behaves in a way that we expect a continuation to behave. But where did the continuation monad come from? Let's consider some ways of deriving the continuation monad. ## Viewing Montague's PTQ as CPS Montague's conception of determiner phrases as generalized quantifiers is a limited form of continuation-passing. (See, e.g., chapter 4 of Barker and Shan 2014.) Start by assuming that ordinary DPs such as proper names denote objects of type `e`. Then verb phrases denote functions from individuals to truth values, i.e., functions of type `e -> t`. The meaning of extraordinary DPs such as *every woman* or *no dog* can't be expressed as a simple individual. As Montague argued, it works much better to view them as predicates on verb phrase meanings, i.e., as having type `(e->t)->t`. Then *no woman left* is true just in case the property of leaving is true of no woman: no woman: \k.not \exists x . (woman x) & kx left: \x.left x (no woman) (left) = not \exists x . woman x & left x Montague also proposed that all determiner phrases should have the same type. After all, we can coordinate proper names with quantificational DPs, as in *John and no dog left*. Then generalized quantifier corresponding to the proper name *John* is the quantifier `\k.kj`. At this point, we have a type for our Kliesli arrow and a value for our mid. Given some result type (such as `t` in the Montague application), α ==> (α -> ρ) -> ρ ⇧a = \k.ka It remains only to find a suitable value for bind. Montague didn't provide one, but it's easy to find: bind :: Mα -> (α -> Mβ) -> Mβ given variables of the following types u :: Mα == (α -> ρ) -> ρ f :: α -> Mβ k :: β -> ρ x :: α We have bind u f = \k.u(\x.fxk) Let's carefully make sure that this types out: bind u f = \k. u (\x . f x k) -------- -- α -> Mβ α ------------ ------ Mβ β -> ρ -- -------------------- α ρ ------------- ------------------------ (α -> ρ) -> ρ α -> ρ --------------------------------- ρ ----------------------- (β -> ρ) -> ρ Yep! Is there any other way of building a bind operator? Well, the challenge is getting hold of the "a" that is buried inside of `u`. In the Reader monad, we could get at the a inside of the box by applying the box to an environment. In the State monad, we could get at the a by applying to box to a state and deconstructing the resulting pair. In the continuation case, the only way to do it is to apply the boxed a (i.e., u) to a function that takes an a as an argument. That means that f must be invoked inside the scope of the functional argument to u. So we've deduced the structure ... u (\x. ... f x ...) ... At this point, in order to provide u with an argument of the appropriate type, the argument must not only take objects of type α as an argument, it must return a result of type ρ. That means that we must apply fx to an object of type β -> ρ. We can hypothesize such an object, as long as we eliminate that hypothesis later (by binding it), and we have the complete bind operation. The way in which the value of type α that is needed in order to unlock the function f is hidden inside of u is directly analogous to the concept of "data hiding" in object-oriented programming. See Pierce's discussion of how to extend system F with existential type quantification by encoding the existential using continuations. So the Kliesli type pretty much determines the bind operation. ## What continuations are doing Ok, we have a continuation monad. We derived it from first principles, and we have seen that it behaves at least in some respects as we expect a continuation monad to behave (in that it allows us to give a good implementation of the task). ## How continuations can simulate other monads Because the continuation monad allows the result type ρ to be any type, we can choose ρ in clever ways that allow us to simulate other monads. Reader: ρ = env -> α State: ρ = s -> (α, s) Maybe: ρ = Just α | Nothing You see how this is going to go. Let's see an example by adding an abort operator to our task language, which represents what we want to have happen if we divide by zero, where what we want to do is return Nothing. abort k = Nothing mid a k = k a map2 f u v k = u(\u' -> v (\v' -> k(f u' v'))) t13 = map2 (++) (mid "a") (map2 (++) (mid "b") (map2 (++) (mid "c") (mid "d"))) t13 (\k->Just k) == Just "abcd" t14 = map2 (++) (mid "a") (map2 (++) abort (map2 (++) (mid "c") (mid "d"))) t14 (\k->Just k) == Nothing Super cool. ## Delimited versus undelimited continuations ## Natural language requries delimited continuations