From 339a5442b568742f36ddab4a44f77cfb26b609a2 Mon Sep 17 00:00:00 2001
From: Chris Barker <barker@kappa.linguistics.fas.nyu.edu>
Date: Fri, 10 Jun 2011 15:07:39 -0400
Subject: [PATCH 1/1] changes

---
 cps.mdwn | 25 +++++++++++++++----------
 1 file changed, 15 insertions(+), 10 deletions(-)

diff --git a/cps.mdwn b/cps.mdwn
index a3f04595..6668b48c 100644
--- a/cps.mdwn
+++ b/cps.mdwn
@@ -69,16 +69,16 @@ 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
+`\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)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
@@ -91,7 +91,7 @@ Here's the CPS transform defined:
     [\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)u)] =
     \k.[\x.y](\m.m[(\x.z)u]k) =
@@ -109,13 +109,15 @@ trivial continuation, usually the identity function `I = \x.x`.
     [\x.y](\m.m[(\x.z)u] I) =
     (\k.k(\x.y))(\m.m[(\x.z)u] I)
      *           *
-    (\x.y)[(\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, 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:
 
@@ -125,7 +127,7 @@ Compare with a call-by-value xform:
 
 This time the reduction unfolds in a different manner:
 
-    {(\x.y)((\x.z)w)} 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)) =
@@ -140,7 +142,7 @@ This time the reduction unfolds in a different manner:
     {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)
+    (\x.{z})u(\n.(\x.{y})nI)       --A--
      *
     {z}(\n.(\x.{y})nI) =
     (\k.kz)(\n.(\x.{y})nI)
@@ -152,6 +154,9 @@ This time the reduction unfolds in a different manner:
      *
     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.
-- 
2.11.0