an order-independent analysis of order of evaluation.
In order to continue to explore continuations, we will proceed in the
-followin fashion.
+following fashion: we introduce the traditional continuation monad,
+and show how it solves the task, then generalize the task to
+include doubling of both the left and the right context.
## 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"`.
+In order to build a monad, we start with a Kleisli arrow.
+
+ Continuation monad: types: given some ρ, Mα => (α -> ρ) -> ρ
+ ⇧ == \ak.ka : a -> Ma
+ bind == \ufk. u(\x.fxk)
+
+We'll first show that this monad solves the task, then we'll consider
+the monad in more detail.
+
+The unmonadized computation (without the shifty "S" operator) is
+
+ t1 = + a (+ b (+ c d)) ~~> abcd
+
+where "+" is string concatenation and the symbol a is shorthand for
+the string "a".
+
+In order to use the continuation monad to solve the list task,
+we choose α = ρ = [Char]. So "abcd" is a list of characters, and
+a boxed list has type M[Char] == ([Char] -> [Char]) -> [Char].
+
+Writing ¢ in between its arguments, t1 corresponds to the following
+monadic computation:
+
+ mt1 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (⇧+ ¢ ⇧c ¢ ⇧d))
+
+We have to lift each functor (+) and each object (e.g., "b") into the
+monad using mid (`⇧`), then combine them using monadic function
+application, where
+
+ ¢ mf mx = \k -> mf (\f -> mx (\a -> k(f x)))
+
+for the continuation monad.
+
+The way in which we extract a value from a continuation box is by
+applying it to a continuation; often, it is convenient to supply the
+trivial continuation, the identity function \k.k = I. So in fact,
+
+ t1 = mt1 I
+
+That is, the original computation is the monadic version applied to
+the trivial continuation.
+
+We can now imagine replacing the third element ("c") with a shifty
+operator. We would like to replace just the one element, and we will
+do just that in a moment; but in order to simulate the original task,
+we'll have to take a different strategy initially. We'll start by
+imagining a shift operator that combined direction with the tail of
+the list, like this:
+
+ mt2 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ ⇧d))
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.
+Shift takes two arguments: a string continuation u of type M[Char],
+and a string continuation k of type [Char] -> [Char]. Since u is the
+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!
+ mt2 I == "ababd"
- (map1 ((++) "a") (map1 ((++) "b") (shift (mid "d")))) (\k->k) == "ababd"
+just as desired.
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"))))))
+ mt3 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ (⇧+ ¢ ⇧d ¢ (⇧+ ¢ ⇧e ¢ (shift ⇧f)))))
+
+Then
- t6 (\k->k) == "ababdeababdef"
+ mt3 I = "ababdeababdef" -- structure: (ababde)(ababde)f
+
As expected.
-In order to add a reset operator #, we can have
+For a reset operator #, we can have
- # u k = k(u(\k.k))
- ab#deSf ~~> abdedef
+ # u k = k(u(\k.k)) -- ex.: 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.
+So the continuation monad solves the list task using continuations in
+a way that conforms to our by-now familiar strategy of lifting a
+computation into a monad, and then writing a few key functions (shift,
+reset) that exploit the power of the monad.
+Now we should consider what happens when we write a shift operator
+that takes the place of a single letter.
- map2 f u v k = u(\u' -> v (\v' -> k(f u' v')))
- shift k = k (k "")
+ mt2 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (shift ¢ ⇧d))
+ mt4 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ ⇧b ¢ (⇧+ ¢ shift' ¢ ⇧d))
- t2 = map2 (++) (mid "a")
- (map2 (++) (mid "b")
- (map2 (++) shift
- (map2 (++) (mid "d")
- (mid []))))
- t2 (\k->k) == "ababdd"
+Instead of mt2, we have mt4. So now the type of "c" (a boxed string,
+type M[Char]) is the same as the type of the new shift operator, shift'.
+
+ shift' = \k.k(k"")
+
+This shift operator takes a continuation k of type [Char]->[Char], and
+invokes it twice. Since k requires an argument of type [Char], we
+need to use the first invocation of k to construction a [Char]; we do
+this by feeding it a string. Since the task does not replace the
+shift operator with any marker, we give the empty string "" as the
+argument.
+
+But now the new shift operator captures more than just the preceeding
+part of the construction---it captures the entire context, including
+the portion of the sequence that follows it. That is,
+
+ mt4 I = "ababdd"
+
+We have replaced "S" in "abSd" with "ab_d", where the underbar will be
+replaced with the empty string supplied in the definition of shift'.
+Crucially, not only is the prefix "ab" duplicated, so is the suffix
+"d".
+
+Things get interesting when we have more than one operator in the
+initial list. What should we expect if we start with "aScSe"?
+If we assume that when we evaluate each S, all the other S's become
+temporarily inert, we expect a reduction path like
+
+ aScSe ~~> aacSecSe
+
+But note that the output has just as many S's as the input--if that is
+what our reduction strategy delivers, then any initial string with
+more than one S will never reach a normal form.
+
+But that's not what the continuation operator shift' delivers.
+
+ mt5 = ⇧+ ¢ ⇧a ¢ (⇧+ ¢ shift' ¢ (⇧+ ¢ ⇧c ¢ (⇧+ ¢ shift' ¢ "e")))
+
+ mt5 I = "aacaceecaacaceecee" -- structure: "aacaceecaacaceecee"
+
+Huh?
+
+This is considerably harder to understand than the original list task.
+The key is figuring out in each case what function the argument k to
+the shift operator gets bound to.
+
+Let's go back to a simple one-shift example, "aSc". Let's trace what
+the shift' operator sees as its argument k by replacing ⇧ and ¢ with
+their definitions:
+
+ ⇧+ ¢ ⇧a ¢ (⇧+ ¢ shift' ¢ ⇧c) I
+ = \k.⇧+(\f.⇧a(\x.k(fx))) ¢ (⇧+ ¢ shift' ¢ ⇧c) I
+ = \k.(\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(\x.k(fx))) I
+ ~~> (\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(\x.I(fx)))
+ ~~> (\k.⇧+(\f.⇧a(\x.k(fx))))(\f.(⇧+ ¢ shift' ¢ ⇧c)(f))
+ ~~> ⇧+(\f.⇧a(\x.(\f.(⇧+ ¢ shift' ¢ ⇧c)(f))(fx))))
+ ~~> ⇧+(\f.⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(fx)))
+ = (\k.k+)(\f.⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(fx)))
+ ~~> ⇧a(\x.(⇧+ ¢ shift' ¢ ⇧c)(+x))
+ = (\k.ka)(\x.(⇧+ ¢ shift' ¢ ⇧c)(+x))
+ ~~> (⇧+ ¢ shift' ¢ ⇧c)(+a)
+ = (\k.⇧+(\f.shift(\x.k(fx)))) ¢ ⇧c (+a)
+ = (\k.(\k.⇧+(\f.shift(\x.k(fx))))(\f.⇧c(\x.k(fx))))(+a)
+ ~~> (\k.⇧+(\f.shift(\x.k(fx))))(\f'.⇧c(\x'.(+a)(f'x')))
+ ~~> ⇧+(\f.shift(\x.(\f'.⇧c(\x'.(+a)(f'x')))(fx)))
+ ~~> ⇧+(\f.shift(\x.⇧c(\x'.(+a)((fx)x'))))
+ = (\k.k+)(\f.shift(\x.⇧c(\x'.(+a)((fx)x'))))
+ ~~> shift(\x.⇧c(\x'.(+a)((+x)x'))))
+ = shift(\x.(\k.kc)(\x'.(+a)((+x)x'))))
+ ~~> shift(\x.(+a)((+x)c))
+
+So now we see what the argument of shift will be: a function k from
+strings x to the string asc. So shift k will be k(k "") = aacc.
+
+Ok, this is ridiculous. We need a way to get ahead of this deluge of
+lambda conversion. We'll adapt the notational strategy developed in
+Barker and Shan 2014:
+
+Instead of writing
+
+ \k.g(kf): (α -> ρ) -> ρ
+
+we'll write
+
+ g[] ρ
+ --- : ---
+ f α
+
+Then
+ []
+ mid(x) = --
+ x
+
+and
+
+ g[] ρ h[] ρ g[h[]] ρ
+ --- : ---- ¢ --- : --- = ------ : ---
+ f α->β x α fx β
+
+Here's the justification:
+
+ (\FXk.F(\f.X(\x.k(fx)))) (\k.g(kf)) (\k.h(kx))
+ ~~> (\Xk.(\k.g(kf))(\f.X(\x.k(fx)))) (\k.h(kx))
+ ~~> \k.(\k.g(kf))(\f.(\k.h(kx))(\x.k(fx)))
+ ~~> \k.g((\f.(\k.h(kx))(\x.k(fx)))f)
+ ~~> \k.g((\k.h(kx))(\x.k(fx)))
+ ~~> \k.g(h(\x.k(fx))x)
+ ~~> \k.g(h(k(fx)))
+
+Then
+ (\ks.k(ks))[]
+ shift = \k.k(k("")) = -------------
+ ""
+
+Let 2 == \ks.k(ks).
+
+so aSc lifted into the monad is
+
+ [] 2[] []
+ -- ¢ ( --- ¢ --- ) =
+ a "" c
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.
+(\k.k(k1))(\s.(\k.k(k2))(\r.sr))
+(\k.k(k1))(\s.(\r.sr)((\r.sr)2))
+(\k.k(k1))(\s.(\r.sr)(s2))
+(\k.k(k1))(\s.s(s2))
+(\s.s(s2))((\s.s(s2))1)
+(\s.s(s2))(1(12))
+(1(12))((1(12)2))
+
+
+
+
+
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,
+`(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?
+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.
-This depends on which S is executed first. Assume the first 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 parenthesese are added to
-show stages in the construction of the final result).
+## 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).
-Note that this is a different result than assuming that what execution
-does is choose an S, and double its context, 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 (on the bi-directional continuation
-perspective).
+## 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.
-## Viewing Montague's PTQ as CPS
+ 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.
-## How continuations can simulate other monads
## Delimited versus undelimited continuations
+"*the* continuation monad"