-;; call-by-value CPS
-; see Dancy and Filinski, "Representing control: a study of the CPS transformation" (1992)
-; and Sabry, "Note on axiomatizing the semantics of control operators" (1996)
-
-; [x] = var x
-let var = \x (\k. k x) in
-; [\x. body] = lam (\x. [body])
-let lam = \x_body (\k. k (\x. x_body x)) in
-; [M N] = app [M] [N]
-let app = \m n. (\k. m (\m. n (\n. m n k))) in
-
-; helpers
-let app3 = \a b c. app (app a b) c in
-let app4 = \a b c d. app (app (app a b) c) d in
-; [succ] = op1 succ
-let op1 = \op. \u. u (\a k. k (op a)) in
-; [plus] = op2 plus
-let op2 = \op. \u. u (\a v. v (\b k. k (op a b))) in
-let op3 = \op. \u. u (\a v. v (\b w. w (\c k. k (op a b c)))) in
-
-;; continuation operators
-; [let/cc k M] = letcc (\k. [M])
-let callcc = \k. k (\f u. (\j. f j u) (\y w. u y)) in
-let letcc = \x_body. app callcc (lam x_body) in
-let letcc = \k_body. \k. (\j. (k_body j) k) (\y w. k y) in
-
-; [abort M] = abort [M]
-let abort = \body. \k. body (\m m) in
-; [prompt M] = prompt [M]
-let prompt = \body. \k. k (body (\m m)) in
-; [shift k M] = shift (\k. [M])
-let shift = \k_body. \k. (\j. (k_body j) (\m m)) (\y w. w (k y)) in
-
-;; examples
-; (+ 100 (let/cc k (+ 10 1))) ~~> 111
-; app3 (op2 plus) (var hundred) (letcc (\k. app3 (op2 plus) (var ten) (var one)))
-
-; (+ 100 (let/cc k (+ 10 (k 1)))) ~~> 101
-; app3 (op2 plus) (var hundred) (letcc (\k. app3 (op2 plus) (var ten) (app (var k) (var one))))
-
-; (+ 100 (+ 10 (abort 1))) ~~> 1
-; app3 (op2 plus) (var hundred) (app3 (op2 plus) (var ten) (abort (var one)))
-
-; (+ 100 (prompt (+ 10 (abort 1)))) ~~> 101
-; app3 (op2 plus) (var hundred) (prompt (app3 (op2 plus) (var ten) (abort (var one))))
-
-; (+ 1000 (prompt (+ 100 (shift k (+ 10 1))))) ~~> 1011
-; app3 (op2 plus) (var thousand) (prompt (app3 (op2 plus) (var hundred) (shift (\k. ((op2 plus) (var ten) (var one))))))
-
-; (+ 1000 (prompt (+ 100 (shift k (k (+ 10 1)))))) ~~> 1111
-; app3 (op2 plus) (var thousand) (prompt (app3 (op2 plus) (var hundred) (shift (\k. (app (var k) ((op2 plus) (var ten) (var one)))))))
-
-; (+ 1000 (prompt (+ 100 (shift k (+ 10 (k 1)))))) ~~> 1111 but added differently
-; app3 (op2 plus) (var thousand) (prompt (app3 (op2 plus) (var hundred) (shift (\k. ((op2 plus) (var ten) (app (var k) (var one)))))))
-
-; (+ 100 ((prompt (+ 10 (shift k k))) 1)) ~~> 111
-; app3 (op2 plus) (var hundred) (app (prompt (app3 (op2 plus) (var ten) (shift (\k. (var k))))) (var one))
-
-; (+ 100 (prompt (+ 10 (shift k (k (k 1)))))) ~~> 121
-; app3 (op2 plus) (var hundred) (prompt (app3 (op2 plus) (var ten) (shift (\k. app (var k) (app (var k) (var one))))))
+Gaining control over order of evaluation
+----------------------------------------
+We know that evaluation order matters. We're beginning to learn how
+to gain some control over order of evaluation (think of Jim's abort handler).
+We continue to reason about order of evaluation.
+
+A lucid discussion of evaluation order in the
+context of the lambda calculus can be found here:
+[Sestoft: Demonstrating Lambda Calculus Reduction](http://www.itu.dk/~sestoft/papers/mfps2001-sestoft.pdf).
+Sestoft also provides a lovely on-line lambda evaluator:
+[Sestoft: Lambda calculus reduction workbench](http://www.itu.dk/~sestoft/lamreduce/index.html),
+which allows you to select multiple evaluation strategies,
+and to see reductions happen step by step.
+
+Evaluation order matters
+------------------------
+
+We've seen this many times. For instance, consider the following
+reductions. It will be convenient to use the abbreviation `w =
+\x.xx`. I'll
+indicate which lambda is about to be reduced with a * underneath:
+
+<pre>
+(\x.y)(ww)
+ *
+y
+</pre>
+
+Done! We have a normal form. But if we reduce using a different
+strategy, things go wrong:
+
+<pre>
+(\x.y)(ww) =
+(\x.y)((\x.xx)w) =
+ *
+(\x.y)(ww) =
+(\x.y)((\x.xx)w) =
+ *
+(\x.y)(ww)
+</pre>
+
+Etc.
+
+As a second reminder of when evaluation order matters, consider using
+`Y = \f.(\h.f(hh))(\h.f(hh))` as a fixed point combinator to define a recursive function:
+
+<pre>
+Y (\f n. blah) =
+(\f.(\h.f(hh))(\h.f(hh))) (\f n. blah)
+ *
+(\f.f((\h.f(hh))(\h.f(hh)))) (\f n. blah)
+ *
+(\f.f(f((\h.f(hh))(\h.f(hh))))) (\f n. blah)
+ *
+(\f.f(f(f((\h.f(hh))(\h.f(hh)))))) (\f n. blah)
+</pre>
+
+And we never get the recursion off the ground.
+
+
+Using a Continuation Passing Style transform to control order of evaluation
+---------------------------------------------------------------------------
+
+We'll present a technique for controlling evaluation order by transforming a lambda term
+using a Continuation Passing Style transform (CPS), then we'll explore
+what the CPS is doing, and how.
+
+In order for the CPS to work, we have to adopt a new restriction on
+beta reduction: beta reduction does not occur underneath a lambda.
+That is, `(\x.y)z` reduces to `z`, but `\u.(\x.y)z` does not reduce to
+`\u.z`, because the `\u` protects the redex in the body from
+reduction. (In this context, a "redex" is a part of a term that matches
+the pattern `...((\xM)N)...`, i.e., something that can potentially be
+the target of beta reduction.)
+
+Start with a simple form that has two different reduction paths:
+
+reducing the leftmost lambda first: `(\x.y)((\x.z)u) ~~> y`
+
+reducing the rightmost lambda first: `(\x.y)((\x.z)u) ~~> (\x.y)z ~~> y`
+
+After using the following call-by-name CPS transform---and assuming
+that we never evaluate redexes protected by a lambda---only the first
+reduction path will be available: we will have gained control over the
+order in which beta reductions are allowed to be performed.
+
+Here's the CPS transform defined:
+
+ [x] = x
+ [\xM] = \k.k(\x[M])
+ [MN] = \k.[M](\m.m[N]k)
+
+Here's the result of applying the transform to our simple example:
+
+ [(\x.y)((\x.z)u)] =
+ \k.[\x.y](\m.m[(\x.z)u]k) =
+ \k.(\k.k(\x.[y]))(\m.m(\k.[\x.z](\m.m[u]k))k) =
+ \k.(\k.k(\x.y))(\m.m(\k.(\k.k(\x.z))(\m.muk))k)
+
+Because the initial `\k` protects (i.e., takes scope over) the entire
+transformed term, we can't perform any reductions. In order to watch
+the computation unfold, we have to apply the transformed term to a
+trivial continuation, usually the identity function `I = \x.x`.
+
+ [(\x.y)((\x.z)u)] I =
+ (\k.[\x.y](\m.m[(\x.z)u]k)) I
+ *
+ [\x.y](\m.m[(\x.z)u] I) =
+ (\k.k(\x.y))(\m.m[(\x.z)u] I)
+ * *
+ (\x.y)[(\x.z)u] I --A--
+ *
+ y I
+
+The application to `I` unlocks the leftmost functor. Because that
+functor (`\x.y`) throws away its argument (consider the reduction in the
+line marked (A)), we never need to expand the
+CPS transform of the argument. This means that we never bother to
+reduce redexes inside the argument.
+
+Compare with a call-by-value xform:
+
+ {x} = \k.kx
+ {\aM} = \k.k(\a{M})
+ {MN} = \k.{M}(\m.{N}(\n.mnk))
+
+This time the reduction unfolds in a different manner:
+
+ {(\x.y)((\x.z)u)} I =
+ (\k.{\x.y}(\m.{(\x.z)u}(\n.mnk))) I
+ *
+ {\x.y}(\m.{(\x.z)u}(\n.mnI)) =
+ (\k.k(\x.{y}))(\m.{(\x.z)u}(\n.mnI))
+ * *
+ {(\x.z)u}(\n.(\x.{y})nI) =
+ (\k.{\x.z}(\m.{u}(\n.mnk)))(\n.(\x.{y})nI)
+ *
+ {\x.z}(\m.{u}(\n.mn(\n.(\x.{y})nI))) =
+ (\k.k(\x.{z}))(\m.{u}(\n.mn(\n.(\x.{y})nI)))
+ * *
+ {u}(\n.(\x.{z})n(\n.(\x.{y})nI)) =
+ (\k.ku)(\n.(\x.{z})n(\n.(\x.{y})nI))
+ * *
+ (\x.{z})u(\n.(\x.{y})nI) --A--
+ *
+ {z}(\n.(\x.{y})nI) =
+ (\k.kz)(\n.(\x.{y})nI)
+ * *
+ (\x.{y})zI
+ *
+ {y}I =
+ (\k.ky)I
+ *
+ I y
+
+In this case, the argument does get evaluated: consider the reduction
+in the line marked (A).
+
+Both xforms make the following guarantee: as long as redexes
+underneath a lambda are never evaluated, there will be at most one
+reduction available at any step in the evaluation.
+That is, all choice is removed from the evaluation process.
+
+Now let's verify that the CBN CPS avoids the infinite reduction path
+discussed above (remember that `w = \x.xx`):
+
+ [(\x.y)(ww)] I =
+ (\k.[\x.y](\m.m[ww]k)) I
+ *
+ [\x.y](\m.m[ww]I) =
+ (\k.k(\x.y))(\m.m[ww]I)
+ * *
+ (\x.y)[ww]I
+ *
+ y I
+
+
+Questions and exercises:
+
+1. Prove that {(\x.y)(ww)} does not terminate.
+
+2. Why is the CBN xform for variables `[x] = x' instead of something
+involving kappas?
+
+3. Write an Ocaml function that takes a lambda term and returns a
+CPS-xformed lambda term. You can use the following data declaration:
+
+ type form = Var of char | Abs of char * form | App of form * form;;
+
+4. The discussion above talks about the "leftmost" redex, or the
+"rightmost". But these words apply accurately only in a special set
+of terms. Characterize the order of evaluation for CBN (likewise, for
+CBV) more completely and carefully.
+
+5. What happens (in terms of evaluation order) when the application
+rule for CBV CPS is changed to `{MN} = \k.{N}(\n.{M}(\m.mnk))`?
+
+
+Thinking through the types
+--------------------------
+
+This discussion is based on [Meyer and Wand 1985](http://citeseer.ist.psu.edu/viewdoc/download?doi=10.1.1.44.7943&rep=rep1&type=pdf).
+
+Let's say we're working in the simply-typed lambda calculus.
+Then if the original term is well-typed, the CPS xform will also be
+well-typed. But what will the type of the transformed term be?
+
+The transformed terms all have the form `\k.blah`. The rule for the
+CBN xform of a variable appears to be an exception, but instead of
+writing `[x] = x`, we can write `[x] = \k.xk`, which is
+eta-equivalent. The `k`'s are continuations: functions from something
+to a result. Let's use σ as the result type. The each `k` in
+the transform will be a function of type ρ --> σ for some
+choice of ρ.
+
+We'll need an ancilliary function ': for any ground type a, a' = a;
+for functional types a->b, (a->b)' = ((a' -> σ) -> σ) -> (b' -> σ) -> σ.
+
+ Call by name transform
+
+ Terms Types
+
+ [x] = \k.xk [a] = (a'->o)->o
+ [\xM] = \k.k(\x[M]) [a->b] = ((a->b)'->o)->o
+ [MN] = \k.[M](\m.m[N]k) [b] = (b'->o)->o
+
+Remember that types associate to the right. Let's work through the
+application xform and make sure the types are consistent. We'll have
+the following types:
+
+ M:a->b
+ N:a
+ MN:b
+ k:b'->o
+ [N]:(a'->o)->o
+ m:((a'->o)->o)->(b'->o)->o
+ m[N]:(b'->o)->o
+ m[N]k:o
+ [M]:((a->b)'->o)->o = ((((a'->o)->o)->(b'->o)->o)->o)->o
+ [M](\m.m[N]k):o
+ [MN]:(b'->o)->o
+
+Be aware that even though the transform uses the same symbol for the
+translation of a variable (i.e., `[x] = x`), in general the variable
+in the transformed term will have a different type than in the source
+term.
+
+Excercise: what should the function ' be for the CBV xform? Hint:
+see the Meyer and Wand abstract linked above for the answer.