[lambda.git] / cps.mdwn

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
`\w.z`, because the `\w` 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)w)  ~~> y`

reducing the rightmost lambda first: `(\x.y)((\x.z)w)  ~~> (\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 problem term:

    [(\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
     *
    y I

The application to `I` unlocks the leftmost functor.  Because that
functor (`\x.y`) throws away its argument, we never need to expand the
CPS transform of 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)w)} 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)
     *
    {z}(\n.(\x.{y})nI) =
    (\k.kz)(\n.(\x.{y})nI)
     *      *
    (\x.{y})zI
     *
    {y}I =
    (\k.ky)I
     *
    I y

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 &sigma; as the result type.  The each `k` in
the transform will be a function of type &rho; --> &sigma; for some
choice of &rho;.

We'll need an ancilliary function ': for any ground type a, a' = a;
for functional types a->b, (a->b)' = ((a' -> &sigma;) -> &sigma;) -> (b' -> &sigma;) -> &sigma;.

    Call by name transform

    Terms                            Types

    [x] = \k.xk                      [a] = (a'->&sigma;)->&sigma;
    [\xM] = \k.k(\x[M])              [a->b] = ((a->b)'->&sigma;)->&sigma;
    [MN] = \k.[M](\m.m[N]k)          [b] = (b'->&sigma;)->&sigma;

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'->&sigma;
    [N]:(a'->&sigma;)->&sigma;
    m:((a'->&sigma;)->&sigma;)->(b'->&sigma;)->&sigma;
    m[N]:(b'->&sigma;)->&sigma;
    m[N]k:&sigma; 
    [M]:((a->b)'->&sigma;)->&sigma; = ((((a'->&sigma;)->&sigma;)->(b'->&sigma;)->&sigma;)->&sigma;)->&sigma;
    [M](\m.m[N]k):&sigma;
    [MN]:(b'->&sigma;)->&sigma;

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.