changes

[lambda.git] / cps.mdwn

diff --git a/cps.mdwn b/cps.mdwn
index e1f7f58..6668b48 100644 (file)
--- a/cps.mdwn
+++ b/cps.mdwn
@@ -18,8 +18,8 @@ Evaluation order matters
 
 We've seen this many times.  For instance, consider the following
 reductions.  It will be convenient to use the abbreviation `w =
 
 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:
+\x.xx`.  I'll
+indicate which lambda is about to be reduced with a * underneath:

 
 <pre>
 (\x.y)(ww)
 
 <pre>
 (\x.y)(ww)


@@ -68,101 +68,133 @@ 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.
 
 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 `\w.(\x.y)z` does not, because
-the `\w` protects the redex in the body from reduction.  
-(A redex is a subform ...(\xM)N..., i.e., something that can be the
-target of beta reduction.)
+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:
 
 
 Start with a simple form that has two different reduction paths:
 

-reducing the leftmost lambda first: `(\x.y)((\x.z)w)  ~~> y`
+reducing the leftmost lambda first: `(\x.y)((\x.z)u)  ~~> y`

 
 

-reducing the rightmost lambda first: `(\x.y)((\x.z)w)  ~~> (x.y)z ~~> 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.
 
 
 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:
+Here's the CPS transform defined:

 
 

-    [x] => x
-    [\xM] => \k.k(\x[M])
-    [MN] => \k.[M](\m.m[N]k)
+    [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:
+Here's the result of applying the transform to our simple example:

 
 

-    [(\x.y)((\x.z)w)]
-    \k.[\x.y](\m.m[(\x.z)w]k)
-    \k.(\k.k(\x.[y]))(\m.m(\k.[\x.z](\m.m[w]k))k)
-    \k.(\k.k(\x.y))(\m.m(\k.(\k.k(\x.z))(\m.mwk))k)
+    [(\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 the entire transformed term, 
-we can't perform any reductions.  In order to see the computation
-unfold, we have to apply the transformed term to a trivial
-continuation, usually the identity function `I = \x.x`.
+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)w)] I
-    \k.[\x.y](\m.m[(\x.z)w]k) I
-    [\x.y](\m.m[(\x.z)w] I)
-    (\k.k(\x.y))(\m.m[(\x.z)w] I)
-    (\x.y)[(\x.z)w] I
+    [(\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
     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.
+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:
 
 
 Compare with a call-by-value xform:
 

-    <x> => \k.kx
-    <\aM> => \k.k(\a<M>)
-    <MN> => \k.<M>(\m.<N>(\n.mnk))
+    {x} = \k.kx
+    {\aM} = \k.k(\a{M})
+    {MN} = \k.{M}(\m.{N}(\n.mnk))

 
 This time the reduction unfolds in a different manner:
 
 
 This time the reduction unfolds in a different manner:
 

-    <(\x.y)((\x.z)w)> I
-    (\k.<\x.y>(\m.<(\x.z)w>(\n.mnk))) I
-    <\x.y>(\m.<(\x.z)w>(\n.mnI))
-    (\k.k(\x.<y>))(\m.<(\x.z)w>(\n.mnI))
-    <(\x.z)w>(\n.(\x.<y>)nI)
-    (\k.<\x.z>(\m.<w>(\n.mnk)))(\n.(\x.<y>)nI)
-    <\x.z>(\m.<w>(\n.mn(\n.(\x.<y>)nI)))
-    (\k.k(\x.<z>))(\m.<w>(\n.mn(\n.(\x.<y>)nI)))
-    <w>(\n.(\x.<z>)n(\n.(\x.<y>)nI))
-    (\k.kw)(\n.(\x.<z>)n(\n.(\x.<y>)nI))
-    (\x.<z>)w(\n.(\x.<y>)nI)
-    <z>(\n.(\x.<y>)nI)
-    (\k.kz)(\n.(\x.<y>)nI)
-    (\x.<y>)zI
-    <y>I
+    {(\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
     (\k.ky)I

+     *

     I y
 
     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.
 
 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.
 

-Questions and excercises:
+Now let's verify that the CBN CPS avoids the infinite reduction path
+discussed above (remember that `w = \x.xx`):

 
 

-1. Why is the CBN xform for variables `[x] = x' instead of something
+    [(\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?  
 
 involving kappas?  
 

-2. Write an Ocaml function that takes a lambda term and returns a
+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;;
 
 CPS-xformed lambda term.  You can use the following data declaration:
 
     type form = Var of char | Abs of char * form | App of form * form;;
 

-3. What happens (in terms of evaluation order) when the application
-rule for CBN CPS is changed to `[MN] = \k.[N](\n.[M]nk)`?  Likewise,
-What happens when the application rule for CBV CPS is changed to 
-`<MN> = \k.<N>(\n.<M>(\m.mnk))`?
-
-4. What happens when the application rules for the CPS xforms are changed to
+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.

 
 

-   [MN] = \k.<M>(\m.m<N>k)
-   <MN> = \k.[M](\m.[N](\n.mnk))
+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
 
 
 Thinking through the types


@@ -176,22 +208,22 @@ 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
 
 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
+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;
 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' -> (b' -> o) -> o.
+for functional types a->b, (a->b)' = ((a' -> &sigma;) -> &sigma;) -> (b' -> &sigma;) -> &sigma;.

 
     Call by name transform
 
     Terms                            Types
 
 
     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
+    [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
 
 Remember that types associate to the right.  Let's work through the
 application xform and make sure the types are consistent.  We'll have


@@ -201,15 +233,18 @@ the following types:
     N:a
     MN:b 
     k:b'->o
     N:a
     MN:b 
     k:b'->o

-    [N]:a'
-    m:a'->(b'->o)->o
+    [N]:(a'->o)->o
+    m:((a'->o)->o)->(b'->o)->o

     m[N]:(b'->o)->o
     m[N]k:o 
     m[N]:(b'->o)->o
     m[N]k:o 

-    [M]:((a->b)'->o)->o = ((a'->(b'->o)->o)->o)->o
+    [M]:((a->b)'->o)->o = ((((a'->o)->o)->(b'->o)->o)->o)->o

     [M](\m.m[N]k):o
     [MN]:(b'->o)->o
 
     [M](\m.m[N]k):o
     [MN]:(b'->o)->o
 

-Note that even though the transform uses the same symbol for the
-translation of a variable, in general it will have a different type in
-the transformed term.
+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.