Date: Wed, 18 Mar 2015 23:33:52 0400
Subject: [PATCH] light editing, rename functions for clarity

topics/week7__95__combinatory__95__evaluator.mdwn  223 ++++++++++++
1 file changed, 117 insertions(+), 106 deletions()
diff git a/topics/week7__95__combinatory__95__evaluator.mdwn b/topics/week7__95__combinatory__95__evaluator.mdwn
index 8349a40c..9c3ac87a 100644
 a/topics/week7__95__combinatory__95__evaluator.mdwn
+++ b/topics/week7__95__combinatory__95__evaluator.mdwn
@@ 2,15 +2,17 @@
[[!toc levels=2]]
# Reasoning about evaluation order in Combinatory Logic
+## Reasoning about evaluation order in Combinatory Logic
We've discussed [[evaluation ordertopics/week3_evaluation_order]]
before, primarily in connection with the untyped lambda calculus.
Whenever a term contains more than one redex, we have to choose which
one to reduce, and this choice can make a difference. For instance,
recall that
+recall that:
 Î© == ÏÏ == (\x.xx)(\x.xx), so
+ Î© == ÏÏ == (\x.xx)(\x.xx)
+
+so:
((\x.I)Î©) == ((\x.I)((\x.xx)(\x.xx)))
* *
@@ 18,12 +20,11 @@ recall that
There are two redexes in this term; we've marked the operative lambdas
with a star. If we reduce the leftmost redex first, the term reduces
to the normal form `I` in one step. But if we reduce the rightmost
redex instead, the "reduced" form is `(\x.I)Î©` again, and we are in
danger of entering an infinite loop.
+redex instead, the "reduced" form is `(\x.I)Î©` again, and we're starting off on an infinite loop.
Thanks to the introduction of sum types (disjoint union) in the last lecture, we
are now in a position to gain a deeper understanding of evaluation
order by writing a program that allows us to reasoning explicitly about evaluation.
+order by writing a program that allows us to reason explicitly about evaluation.
One thing we'll see is that it is all too easy for the evaluation
order properties of an evaluator to depend on the evaluation order
@@ 37,7 +38,7 @@ progress.
The first evaluator we will develop will evaluate terms in Combinatory
Logic. This significantly simplifies the discussion, since we won't
need to worry about variables or substitution. As we develop and
extend our evaluator in future weeks, we'll switch to lambdas, but for
+extend our evaluator in homework and other lectures, we'll switch to lambdas, but for
now, working with the simplicity of Combinatory Logic will make it
easier to highlight evaluation order issues.
@@ 69,11 +70,11 @@ contain more than one redex:
KIÎ© == KI(SII(SII))
* *
we can choose to reduce the leftmost redex by applying the reduction
+We can choose to reduce the leftmost redex by applying the reduction
rule for `K`, in which case the term reduces to the normal form `I` in
one step; or we can choose to reduce the Skomega part, by applying the
reduction rule `S`, in which case we do not get a normal form, and
we're headed towards an infinite loop.
+we're headed into an infinite loop.
With sum types, we can define CL terms in OCaml as follows:
@@ 82,28 +83,31 @@ With sum types, we can define CL terms in OCaml as follows:
let skomega = App (App (App (S, I), I), App (App (S, I), I))
This type definition says that a term in CL is either one of the three
simple expressions (`I`, `K`, or `S`), or else a pair of CL
+atomic expressions (`I`, `K`, or `S`), or else a pair of CL
expressions. `App` stands for Functional Application. With this type
definition, we can encode Skomega, as well as other terms whose
reduction behavior we want to try to control.
+reduction behavior we want to try to control. We can *control* it because
+the `App` variant of our datatype merely *encodes* the application of the
+head to the argument, and doesn't actually *perform* that application.
+We have to explicitly model the application ourself.
Using pattern matching, it is easy to code the onestep reduction
rules for CL:
 let reduce_one_step (t:term):term = match t with
 App(I,a) > a
+ let reduce_if_redex (t:term) : term = match t with
+  App(I,a) > a
 App(App(K,a),b) > a
 App(App(App(S,a),b),c) > App(App(a,c),App(b,c))
 _ > t
 # reduce_one_step (App(App(K,S),I));;
+ # reduce_if_redex (App(App(K,S),I));;
 : term = S
 # reduce_one_step skomega;;
+ # reduce_if_redex skomega;;
 : term = App (App (I, App (App (S, I), I)), App (I, App (App (S, I), I)))
The definition of `reduce_one_step` explicitly says that it expects
its input argument `t` to have type `term`, and the second `:term`
says that the type of the output the function delivers as a result will also be of
+The definition of `reduce_if_redex` explicitly says that it expects
+its input argument `t` to have type `term`, and the second `: term`
+says that the result the function delivers will also be of
type `term`.
The type constructor `App` obscures things a bit, but it's still
@@ 112,9 +116,12 @@ reduction rules for CL. The OCaml interpreter responses given above show us tha
function faithfully recognizes that `KSI ~~> S`, and that `Skomega ~~>
I(SII)(I(SII))`.
We can now say precisely what it means to be a redex in CL.
+As you would expect, a term in CL is in normal form when it contains
+no redexes (analogously for head normal form, weak head normal form, etc.)
 let is_redex (t:term):bool = not (t = reduce_one_step t)
+How can we tell whether a term is a redex? Here's one way:
+
+ let is_redex (t:term):bool = not (t = reduce_if_redex t)
# is_redex K;;
 : bool = false
@@ 125,23 +132,23 @@ We can now say precisely what it means to be a redex in CL.
# is_redex skomega;;
 : book = true
Warning: this definition relies on the accidental fact that the
onestep reduction of a CL term is never identical to the original
term. This would not work for the untyped lambda calculus, since
`((\x.xx)(\x.xx)) ~~> ((\x.xx)(\x.xx))` in one step.

Note that in order to decide whether two terms are equal, OCaml has to
+In order to decide whether two terms are equal, OCaml has to
recursively compare the elements of complex CL terms. It is able to
figure out how to do this because we provided an explicit definition
of the datatype `term`.
As you would expect, a term in CL is in normal form when it contains
no redexes (analogously for head normal form, weak head normal form, etc.)

In order to fully reduce a term, we need to be able to reduce redexes
that are not at the top level of the term.
Because we need to process subparts, and because the result after
processing a subpart may require further processing, the recursive
+Warning: this method for telling whether a term is a redex relies on the accidental fact that the
+onestep reduction of a CL term is never identical to the original
+term. This would not work for the untyped lambda calculus, since
+`((\x.xx)(\x.xx)) ~~> ((\x.xx)(\x.xx))` in one step. Neither would it work if
+we had chosen some other combinators as primitives (`W W1 W2` reduces to
+`W1 W2 W2`, so if they are all `W`s we'd be in trouble.) We will discuss
+some alternative strategies in other notes.
+
+So far, we've only asked whether a term _is_ a redex, not whether it _contains_ other redexes as subterms. But
+in order to fully reduce a term, we need to be able to reduce redexes
+that are not at the top level of the term. Because we need to process subterms, and because the result after
+processing a subterm may require further processing, the recursive
structure of our evaluation function has to be somewhat subtle. To
truly understand, we will need to do some sophisticated thinking
about how recursion works.
@@ 149,13 +156,14 @@ about how recursion works.
We'll develop our full reduction function in two stages. Once we have
it working, we'll then consider a variant.
 let rec reduce_stage1 (t:term):term =
 if (is_redex t) then reduce_stage1 (reduce_one_step t)
+ let rec reduce_try1 (t:term) : term =
+ if (is_redex t) then let t' = reduce_if_redex t
+ in reduce_try1 t'
else t
If the input is a redex, we ship it off to `reduce_one_step` for
+If the input is a redex, we ship it off to `reduce_if_redex` for
processing. But just in case the result of the onestep reduction is
itself a redex, we recursively call `reduce_stage1`. The recursion
+itself a redex, we recursively apply `reduce_try1` to the result. The recursion
will continue until the result is no longer a redex. We're aiming at
allowing the evaluator to recognize that
@@ 165,24 +173,27 @@ When trying to understand how recursive functions work, it can be
extremely helpful to examine an execution trace of inputs and
outputs.
 # #trace reduce_stage1;;
 reduce_stage1 is now traced.
 # reduce_stage1 (App (I, App (I, K)));;
 reduce_stage1 < App (I, App (I, K))
 reduce_stage1 < App (I, K)
 reduce_stage1 < K
 reduce_stage1 > K
 reduce_stage1 > K
 reduce_stage1 > K
+ # #trace reduce_try1;;
+ reduce_try1 is now traced.
+
+The first `#` there is OCaml's prompt. The text beginning `#trace ...` is what we typed. Now OCaml will report on all the input to, and results from, the `reduce_try1` function. Watch:
+
+ # reduce_try1 (App (I, App (I, K)));;
+ reduce_try1 < App (I, App (I, K))
+ reduce_try1 < App (I, K)
+ reduce_try1 < K
+ reduce_try1 > K
+ reduce_try1 > K
+ reduce_try1 > K
 : term = K
In the trace, "`<`" shows the input argument to a call to
`reduce_stage1`, and "`>`" shows the output result.
+`reduce_try1`, and "`>`" shows the output result.
Since the initial input (`I(IK)`) is a redex, the result after the
onestep reduction is `IK`. We recursively call `reduce_stage1` on
+onestep reduction is `IK`. We recursively call `reduce_try1` on
this input. Since `IK` is itself a redex, the result after onestep
reduction is `K`. We recursively call `reduce_stage1` on this input. Since
+reduction is `K`. We recursively call `reduce_try1` on this input. Since
`K` is not a redex, the recursion bottoms out, and we return the
result.
@@ 193,24 +204,24 @@ the following reduction path:
But the reduction function as written above does not deliver this result:
 # reduce_stage1 (App (App (I, I), K));;
+ # reduce_try1 (App (App (I, I), K));;
 : term = App (App (I, I), K)
The reason is that the toplevel term is not a redex to start with,
so `reduce_stage1` returns it without any evaluation.
+so `reduce_try1` returns it without any evaluation.
What we want is to
evaluate the subparts of a complex term. We'll do this by evaluating
the subparts of the toplevel expression.
+What we want is to evaluate the subterms of a complex term. We'll do this by pattern matching our
+toplevel term to see when it _has_ subterms:
 let rec reduce (t:term):term = match t with
 I > I
+ let rec reduce_try2 (t : term) : term = match t with
+  I > I
 K > K
 S > S
 App (a, b) >
 let t' = App (reduce a, reduce b) in
 if (is_redex t') then reduce 2 (reduce_one_step t')
 else t'
+ let t' = App (reduce_try2 a, reduce_try2 b) in
+ if (is_redex t') then let t'' = reduce_if_redex t'
+ in reduce_try2 t''
+ else t'
Since we need access to the subterms, we do pattern matching on the
input. If the input is simple (the first three `match` cases), we
@@ 221,24 +232,24 @@ at the top level.
To understand how this works, follow the trace
carefully:
 # reduce (App(App(I,I),K));;
 reduce < App (App (I, I), K)
+ # reduce_try2 (App(App(I,I),K));;
+ reduce_try2 < App (App (I, I), K)
 reduce < K ; first main recursive call
 reduce > K
+ reduce_try2 < K ; first main recursive call
+ reduce_try2 > K
 reduce < App (I, I) ; second main recursive call
 reduce < I
 reduce > I
 reduce < I
 reduce > I
 reduce < I
 reduce > I
 reduce > I
+ reduce_try2 < App (I, I) ; second main recursive call
+ reduce_try2 < I
+ reduce_try2 > I
+ reduce_try2 < I
+ reduce_try2 > I
+ reduce_try2 < I
+ reduce_try2 > I
+ reduce_try2 > I
 reduce < K ; third
 reduce > K
 reduce > K
+ reduce_try2 < K ; third
+ reduce_try2 > K
+ reduce_try2 > K
 : term = K
Ok, there's a lot going on here. Since the input is complex, the
@@ 259,12 +270,12 @@ In any case, in the second main recursive call, we evaluate `II`. The
result is `I`.
At this point, we have constructed `t' == App(I,K)`. Since that's a
redex, we ship it off to reduce_one_step, getting the term `K` as a
+redex, we ship it off to reduce_if_redex, getting the term `K` as a
result. The third recursive call checks that there is no more
reduction work to be done (there isn't), and that's our final result.
You can see in more detail what is going on by tracing both reduce
and reduce_one_step, but that makes for some long traces.
+and reduce_if_redex, but that makes for some long traces.
So we've solved our first problem: `reduce` now recognizes that `IIK ~~>
K`, as desired.
@@ 273,7 +284,7 @@ Because the OCaml interpreter evaluates each subexpression in the
course of building `t'`, however, it will always evaluate the right
hand subexpression, whether it needs to or not. And sure enough,
 # reduce (App(App(K,I),skomega));;
+ # reduce_try2 (App(App(K,I),skomega));;
Cc CcInterrupted.
Running the evaluator with this input leads to an infinite loop, and
@@ 297,25 +308,26 @@ type term = I  S  K  App of (term * term) data Term = I  S
let skomega = App (App (App (S,I), I), App (App (S,I), I)) skomega = (App (App (App S I) I) (App (App S I) I))
 reduce_one_step :: Term > Term
let reduce_one_step (t:term):term = match t with reduce_one_step t = case t of
 App(I,a) > a App I a > a
+ reduce_if_redex :: Term > Term
+let reduce_if_redex (t:term):term = match t with reduce_if_redex t = case t of
+  App(I,a) > a App I a > a
 App(App(K,a),b) > a App (App K a) b > a
 App(App(App(S,a),b),c) > App(App(a,c),App(b,c)) App (App (App S a) b) c > App (App a c) (App b c)
 _ > t _ > t
is_redex :: Term > Bool
let is_redex (t:term):bool = not (t = reduce_one_step t) is_redex t = not (t == reduce_one_step t)
+let is_redex (t:term):bool = not (t = reduce_if_redex t) is_redex t = not (t == reduce_if_redex t)
 reduce :: Term > Term
let rec reduce (t:term):term = match t with reduce t = case t of
 I > I I > I
+ reduce_try2 :: Term > Term
+let rec reduce_try2 (t : term) : term = match t with reduce_try2 t = case t of
+  I > I I > I
 K > K K > K
 S > S S > S
  App (a, b) > App a b >
 let t' = App (reduce a, reduce b) in let t' = App (reduce a) (reduce b) in
 if (is_redex t') then reduce (reduce_one_step t') if (is_redex t') then reduce (reduce_one_step t')
 else t' else t'
+  App (a, b) > App a b >
+ let t' = App (reduce_try2 a, reduce_try2 b) in let t' = App (reduce_try2 a) (reduce_try2 b) in
+ if (is_redex t') then let t'' = reduce_if_redex t' if (is_redex t') then reduce_try2 (reduce_if_redex t')
+ in reduce_try2 t'' else t'
+ else t'
There are some differences in the way types are made explicit, and in
@@ 324,16 +336,16 @@ Haskell). But the two programs are essentially identical.
Yet the Haskell program finds the normal form for `KIÎ©`:
 *Main> reduce (App (App K I) skomega)
+ *Main> reduce_try2 (App (App K I) skomega)
I
Woa! First of all, this is wierd. Haskell's evaluation strategy is
called "lazy". Apparently, Haskell is so lazy that even after we've
asked it to construct t' by evaluating `reduce a` and `reduce b`, it
doesn't bother computing `reduce b`. Instead, it waits to see if we
+asked it to construct `t'` by evaluating `reduce_try2 a` and `reduce_try2 b`, it
+doesn't bother computing `reduce_try2 b`. Instead, it waits to see if we
ever really need to use the result.
So the program as written does NOT fully determine evaluation order
+So the program as written does _not_ fully determine evaluation order
behavior. At this stage, we have defined an evaluation order that
still depends on the evaluation order of the underlying interpreter.
@@ 353,20 +365,19 @@ wait until we have Continuation Passing Style transforms.
The answer to the first question (Can we adjust the OCaml evaluator to
exhibit lazy behavior?) is quite simple:
let rec reduce_lazy (t:term):term = match t with
 I > I
  K > K
  S > S
  App (a, b) >
 let t' = App (reduce_lazy a, b) in
 if (is_redex t') then reduce_lazy (reduce_one_step t')
 else t'


There is only one small difference: instead of setting `t'` to `App
(reduce a, reduce b)`, we omit one of the recursive calls, and have
`App (reduce a, b)`. That is, we don't evaluate the righthand
+ let rec reduce_lazy (t : term) : term = match t with
+  I > I
+  K > K
+  S > S
+  App (a, b) >
+ let t' = App (reduce_lazy a, b) in
+ if (is_redex t') then let t'' = reduce_if_redex t'
+ in reduce_lazy t''
+ else t'
+
+There is only one small difference from `reduce_try2`: instead of setting `t'` to `App
+(reduce_try3 a, reduce_try3 b)`, we omit one of the recursive calls, and have
+`App (reduce_try3 a, b)`. That is, we don't evaluate the righthand
subexpression at all. Ever! The only way to get evaluated is to
somehow get into functor position.

2.11.0