XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=week4.mdwn;h=58d6bc38204bd3bcbf4afabe6d9c4306b82735de;hp=5e219484b7cae7ff81e112bad9481044747c1204;hb=72cca371a48d35b0177427b35f23ae8cc704fdf2;hpb=7b3b1c578a8c6d50c71ab570a1a49a1c3385ca62
diff git a/week4.mdwn b/week4.mdwn
index 5e219484..58d6bc38 100644
 a/week4.mdwn
+++ b/week4.mdwn
@@ 6,8 +6,7 @@ A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
`T` has a fixed point, then there exists some `X` such that `X <~~>
TX` (that's what it means to *have* a fixed point).

let L = \x. T (x x) in
+let L = \x. T (x x) in
let X = L L in
X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
@@ 15,66 +14,74 @@ X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
Please slow down and make sure that you understand what justified each
of the equalities in the last line.
#Q: How do you know that for any term `T`, `YT` is a fixed point of `T`?#
+#Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?#
A: Note that in the proof given in the previous answer, we chose `T`
and then set `X = WW = (\x.T(xx))(\x.T(xx))`. If we abstract over
`T`, we get the Y combinator, `\T.(\x.T(xx))(\x.T(xx))`. No matter
what argument `T` we feed Y, it returns some `X` that is a fixed point
+and then set X ≡ L L ≡ (\x. T (x x)) (\x. T (x x))
. If we abstract over
+`T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`. No matter
+what argument `T` we feed `Y`, it returns some `X` that is a fixed point
of `T`, by the reasoning in the previous answer.
#Q: So if every term has a fixed point, even `Y` has fixed point.#
A: Right:
 let Y = \T.(\x.T(xx))(\x.T(xx)) in
 Y Y = \T.(\x.T(xx))(\x.T(xx)) Y
 = (\x.Y(xx))(\x.Y(xx))
 = Y((\x.Y(xx))(\x.Y(xx)))
 = Y(Y((\x.Y(xx))(\x.Y(xx))))
 = Y(Y(Y(...(Y(YY))...)))
+let Y = \T. (\x. T (x x)) (\x. T (x x)) in
+Y Y
+≡ \T. (\x. T (x x)) (\x. T (x x)) Y
+~~> (\x. Y (x x)) (\x. Y (x x))
+~~> Y ((\x. Y (x x)) (\x. Y (x x)))
+~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
+~~> Y (Y (Y (...(Y (Y Y))...)))
+
+
#Q: Ouch! Stop hurting my brain.#
A: Let's come at it from the direction of arithmetic. Recall that we
+A: Is that a question?
+
+Let's come at it from the direction of arithmetic. Recall that we
claimed that even `succ`the function that added one to any
numberhad a fixed point. How could there be an X such that X = X+1?
That would imply that
 X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
+ X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ X)) <~~> succ (... (succ X)...)
In other words, the fixed point of `succ` is a term that is its own
successor. Let's just check that `X = succ X`:
 let succ = \n s z. s (n s z) in
 let X = (\x.succ(xx))(\x.succ(xx)) in
 succ X
 = succ ((\x.succ(xx))(\x.succ(xx)))
 = succ (succ ((\x.succ(xx))(\x.succ(xx))))
 = succ (succ X)
+let succ = \n s z. s (n s z) in
+let X = (\x. succ (x x)) (\x. succ (x x)) in
+succ X
+≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
+~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x)) ))
+≡ succ (succ X)
+
+
+You should see the close similarity with `Y Y` here.
You should see the close similarity with YY here.
#Q. So `Y` applied to `succ` returns a number that is not finite!#
A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
numeral:
 [same definitions]
 succ X
 = (\n s z. s (n s z)) X
 = \s z. s (X s z)
 = succ (\s z. s (X s z)) ; using fixedpoint reasoning
 = \s z. s ([succ (\s z. s (X s z))] s z)
 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
 = \s z. s (s (succ (\s z. s (X s z))))
+[same definitions]
+succ X
+≡ (\n s z. s (n s z)) X
+~~> \s z. s (X s z)
+<~~> succ (\s z. s (X s z)) ; using fixedpoint reasoning
+≡ (\n s z. s (n s z)) (\s z. s (X s z))
+~~> \s z. s ((\s z. s (X s z)) s z)
+~~> \s z. s (s (X s z))
+
So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
and returns a sequence of nested applications of `s`...
You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
succ)(Y succ)`? What would you expect infinity minus infinity to be?
+likewise for `mul`, `sub`, `pow`. What happens if we try `sub (Y
+succ) (Y succ)`? What would you expect infinity minus infinity to be?
(Hint: choose your evaluation strategy so that you add two `s`s to the
first number for every `s` that you add to the second number.)
@@ 84,13 +91,14 @@ represents arithmetic infinity.
It's important to bear in mind the simplest term in question is not
infinite:
 Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
+ Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))
The way that infinity enters into the picture is that this term has
no normal form: no matter how many times we perform beta reduction,
there will always be an opportunity for more beta reduction. (Lather,
rinse, repeat!)
+
#Q. That reminds me, what about [[evaluation order]]?#
A. For a recursive function that has a wellbehaved base case, such as
@@ 100,70 +108,70 @@ which we have to make a choice about which beta reduction to perform
next: one choice leads to a normal form, the other choice leads to
endless reduction:
 let prefac = \f n. isZero n 1 (mult n (f (pred n))) in
 let fac = Y prefac in
 fac 2
 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
 = [(\x.prefac(xx))(\x.prefac(xx))] 2
 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
 = [(\f n. isZero n 1 (mult n (f (pred n))))
 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
 = [\n. isZero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
 = isZero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
 ...
 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
 = mult 2 (mult 1 (isZero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
 = mult 2 (mult 1 1)
 = mult 2 1
 = 2
+let prefact = \f n. iszero n 1 (mul n (f (pred n))) in
+let fact = Y prefact in
+fact 2
+≡ [(\f. (\x. f (x x)) (\x. f (x x))) prefact] 2
+~~> [(\x. prefact (x x)) (\x. prefact (x x))] 2
+~~> [prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 2
+~~> [prefact (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+≡ [ (\f n. iszero n 1 (mul n (f (pred n)))) (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+~~> [\n. iszero n 1 (mul n ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred n)))] 2
+~~> iszero 2 1 (mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 2)))
+~~> mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 1)
+...
+~~> mul 2 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 0))
+≡ mul 2 (mul 1 (iszero 0 1 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 0)))))
+~~> mul 2 (mul 1 1)
+~~> mul 2 1
+~~> 2
+
The crucial step is the third from the last. We have our choice of
either evaluating the test `isZero 0 1 ...`, which evaluates to `1`,
+either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
no matter what the ... contains;
or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
produce another copy of `prefac`. If we postpone evaluting the
`isZero` test, we'll pump out copy after copy of `prefac`, and never
+or we can evaluate the `Y` pump, `(\x. prefact (x x)) (\x. prefact (x x))`, to
+produce another copy of `prefact`. If we postpone evaluting the
+`iszero` test, we'll pump out copy after copy of `prefact`, and never
realize that we've bottomed out in the recursion. But if we adopt a
leftmost/callbyname/normalorder evaluation strategy, we'll always
start with the `isZero` predicate, and only produce a fresh copy of
`prefac` if we are forced to.
+start with the `iszero` predicate, and only produce a fresh copy of
+`prefact` if we are forced to.
+
+
+#Q. You claimed that the Ackermann function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackermann function using full recursion.#
#Q. You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackerman function using full recursion.#
A. OK:
A(m,n) =
  when m == 0 > n + 1
  else when n == 0 > A(m1,1)
  else > A(m1, A(m,n1))

let A = Y (\A m n. isZero m (succ n) (isZero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in


For instance,

 A 1 2
 = A 0 (A 1 1)
 = A 0 (A 0 (A 1 0))
 = A 0 (A 0 (A 0 1))
 = A 0 (A 0 2)
 = A 0 3
 = 4

A 1 x is to A 0 x as addition is to the successor function;
A 2 x is to A 1 x as multiplication is to addition;
A 3 x is to A 2 x as exponentiation is to multiplication
so A 4 x is to A 3 x as hyperexponentiation is to exponentiation...
+ A(m,n) =
+  when m == 0 > n + 1
+  else when n == 0 > A(m1,1)
+  else > A(m1, A(m,n1))
+
+ let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n)))))
+
+So for instance:
+
+ A 1 2
+ ~~> A 0 (A 1 1)
+ ~~> A 0 (A 0 (A 1 0))
+ ~~> A 0 (A 0 (A 0 1))
+ ~~> A 0 (A 0 2)
+ ~~> A 0 3
+ ~~> 4
+
+`A 1 x` is to `A 0 x` as addition is to the successor function;
+`A 2 x` is to `A 1 x` as multiplication is to addition;
+`A 3 x` is to `A 2 x` as exponentiation is to multiplication
+so `A 4 x` is to `A 3 x` as hyperexponentiation is to exponentiation...
#Q. What other questions should I be asking?#
* What is it about the variant fixedpoint combinators that makes
them compatible with a callbyvalue evaluation strategy?
* How do you know that the Ackerman function can't be computed
+* How do you know that the Ackermann function can't be computed
using primitive recursion techniques?
* What *exactly* is primitive recursion?
@@ 242,7 +250,7 @@ So, if we were searching the list that implements some set to see if the number
we can stop. If we haven't found `5` already, we know it's not in the rest of the
list either.
This is an improvement, but it's still a "linear" search through the list.
+> *Comment*: This is an improvement, but it's still a "linear" search through the list.
There are even more efficient methods, which employ "binary" searching. They'd
represent the set in such a way that you could quickly determine whether some
element fell in one half, call it the left half, of the structure that
@@ 252,7 +260,7 @@ determination could be made for whichever half you were directed to. And then
for whichever quarter you were directed to next. And so on. Until you either
found the element or exhausted the structure and could then conclude that the
element in question was not part of the set. These sorts of structures are done
using **binary trees** (see below).
+using [binary trees](/implementing_trees).
#Aborting a search through a list#
@@ 268,10 +276,39 @@ can just deliver that answer, and not branch into any further recursion. If
you've got the right evaluation strategy in place, everything will work out
fine.
But what if you're using v3 lists? What options would you have then for
aborting a search?

Well, suppose we're searching through the list `[5;4;3;2;1]` to see if it
+But what if we wanted to use v3 lists instead?
+
+> Why would we want to do that? The advantage of the v3 lists and v3 (aka
+"Church") numerals is that they have their recursive capacity built into their
+very bones. So for many natural operations on them, you won't need to use a fixed
+point combinator.
+
+> Why is that an advantage? Well, if you use a fixed point combinator, then
+the terms you get won't be strongly normalizing: whether their reduction stops
+at a normal form will depend on what evaluation order you use. Our online
+[[lambda evaluator]] uses normalorder reduction, so it finds a normal form if
+there's one to be had. But if you want to build lambda terms in, say, Scheme,
+and you wanted to roll your own recursion as we've been doing, rather than
+relying on Scheme's native `let rec` or `define`, then you can't use the
+fixedpoint combinators `Y` or Θ
. Expressions using them
+will have nonterminating reductions, with Scheme's eager/callbyvalue
+strategy. There are other fixedpoint combinators you can use with Scheme (in
+the [week 3 notes](/week3/#index7h2) they were Y′
and
+Θ′
. But even with them, evaluation order still
+matters: for some (admittedly unusual) evaluation strategies, expressions using
+them will also be nonterminating.
+
+> The fixedpoint combinators may be the conceptual stars. They are cool and
+mathematically elegant. But for efficiency and implementation elegance, it's
+best to know how to do as much as you can without them. (Also, that knowledge
+could carry over to settings where the fixed point combinators are in principle
+unavailable.)
+
+
+So again, what if we're using v3 lists? What options would we have then for
+aborting a search or list traversal before it runs to completion?
+
+Suppose we're searching through the list `[5;4;3;2;1]` to see if it
contains the number `3`. The expression which represents this search would have
something like the following form:
@@ 291,7 +328,7 @@ more rightmost pieces of the list, too, regardless of what order the reduction
is computed by. Conceptually, it will be easiest if we think of the reduction
happening in the order displayed above.
Well, once we've found a match between our sought number `3` and some member of
+Once we've found a match between our sought number `3` and some member of
the list, we'd like to avoid any further unnecessary computations and just
deliver the answer `true` as "quickly" or directly as possible to the larger
computation in which the search was embedded.
@@ 416,64 +453,64 @@ Do you have the basic idea? Think about how you'd implement it. A good
understanding of the v2 lists will give you a helpful model.
In broad outline, a single stage of the search would look like before, except
now f would receive two extra, "handler" arguments.
+now `f` would receive two extra, "handler" arguments. We'll reserve the name `f` for the original fold function, and use `f2` for the function that accepts two additional handler arguments. To get the general idea, you can regard these as interchangeable. If the extra precision might help, then you can pay attention to when we're talking about the handlertaking `f2` or the original `f`. You'll only be *supplying* the `f2` function; the idea will be that the behavior of the original `f` will be implicitly encoded in `f2`'s behavior.
 f 3
+ f2 3
`f`'s job would be to check whether `3` matches the element we're searching for
+`f2`'s job would be to check whether `3` matches the element we're searching for
(here also `3`), and if it does, just evaluate to the result of passing `true` to
the abort handler. If it doesn't, then evaluate to the result of passing
`false` to the continueleftwards handler.
In this case, `f` wouldn't need to consult the result of folding `f` and `z` over `[2;
1]`, since if we had found the element `3` in more rightward positions of the
list, we'd have called the abort handler and this application of `f` to `3` etc
would never be needed. However, in other applications the result of folding `f`
and `z` over the more rightward parts of the list would be needed. Consider if
you were trying to multiply all the elements of the list, and were going to
abort (with the result `0`) if you came across any element in the list that was
zero. If you didn't abort, you'd need to know what the more rightward elements
of the list multiplied to, because that would affect the answer you passed
along to the continueleftwards handler.

A **version 5** list encodes the kind of fold operation we're envisaging here, in
the same way that v3 (and [v4](/advanced/#v4)) lists encoded the simpler fold operation.
Roughly, the list `[5;4;3;2;1]` would look like this:


 \f z continue_leftwards_handler abort_handler.

 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 continue_leftwards_handler abort_handler)
+In this case, `f2` wouldn't need to consult the result of folding `f` and `z`
+over `[2; 1]`, since if we had found the element `3` in more rightward
+positions of the list, we'd have called the abort handler and this application
+of `f2` to `3` etc would never be needed. However, in other applications the
+result of folding `f` and `z` over the more rightward parts of the list would
+be needed. Consider if you were trying to multiply all the elements of the
+list, and were going to abort (with the result `0`) if you came across any
+element in the list that was zero. If you didn't abort, you'd need to know what
+the more rightward elements of the list multiplied to, because that would
+affect the answer you passed along to the continueleftwards handler.
+
+A **version 5** list encodes the kind of fold operation we're envisaging here,
+in the same way that v3 (and [v4](/advanced_lambda/#index1h1)) lists encoded
+the simpler fold operation. Roughly, the list `[5;4;3;2;1]` would look like
+this:
+
+
+ \f2 z continue_leftwards_handler abort_handler.
+
+ (\result_of_folding_over_4321. f2 5 result_of_folding_over_4321 continue_leftwards_handler abort_handler)
abort_handler
; or, expanding the fold over [4;3;2;1]:
 \f z continue_leftwards_handler abort_handler.
+ \f2 z continue_leftwards_handler abort_handler.
(\continue_leftwards_handler abort_handler.

 (\result_of_fold_over_321. f 4 result_of_fold_over_321 continue_leftwards_handler abort_handler)
+
+ (\result_of_folding_over_321. f2 4 result_of_folding_over_321 continue_leftwards_handler abort_handler)
abort_handler
)
 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 continue_leftwards_handler abort_handler)
+ (\result_of_folding_over_4321. f2 5 result_of_folding_over_4321 continue_leftwards_handler abort_handler)
abort_handler
 ; and so on
+ ; and so on
Remarks: the `larger_computation` handler should be supplied as both the
`continue_leftwards_handler` and the `abort_handler` for the leftmost
application, where the head `5` is supplied to `f`; because the result of this
+application, where the head `5` is supplied to `f2`; because the result of this
application should be passed to the larger computation, whether it's a "fall
off the left end of the list" result or it's a "I'm finished, possibly early"
result. The `larger_computation` handler also then gets passed to the next
rightmost stage, where the head `4` is supplied to `f`, as the `abort_handler` to
+rightmost stage, where the head `4` is supplied to `f2`, as the `abort_handler` to
use if that stage decides it has an early answer.
Finally, notice that we don't have the result of applying `f` to `4` etc given as
an argument to the application of `f` to `5` etc. Instead, we pass
+Finally, notice that we're not supplying the application of `f2` to `4` etc as an argument to the application of `f2` to `5` etcat least, not directly. Instead, we pass
 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 )
+ (\result_of_folding_over_4321. f2 5 result_of_folding_over_4321 )
*to* the application of `f` to `4` as its "continue" handler. The application of `f`
+*to* the application of `f2` to `4` as its "continue" handler. The application of `f2`
to `4` can decide whether this handler, or the other, "abort" handler, should be
given an argument and constitute its result.
@@ 483,33 +520,32 @@ but really all we're in a position to mean by that are claims about the result
of the complex expression semantically depending only on this, not on that. A
demon evaluator who custompicked the evaluation order to make things maximally
bad for you could ensure that all the semantically unnecessary computations got
evaluated anyway. We don't have any way to prevent that. Later,
we'll see ways to *semantically guarantee* one evaluation order rather than
another. Though even then the demonic evaluationorderchooser could make it
take unnecessarily long to compute the semantically guaranteed result. Of
course, in any real computing environment you'll know you're dealing with a
fixed evaluation order and you'll be able to program efficiently around that.
+evaluated anyway. We don't yet know any way to prevent that. Later, we'll see
+ways to *guarantee* one evaluation order rather than another. Of
+course, in any real computing environment you'll know in advance that you're
+dealing with a fixed evaluation order and you'll be able to program efficiently
+around that.
In detail, then, here's what our v5 lists will look like:
 let empty = \f z continue_handler abort_handler. continue_handler z in
 let make_list = \h t. \f z continue_handler abort_handler.
 t f z (\sofar. f h sofar continue_handler abort_handler) abort_handler in
+ let empty = \f2 z continue_handler abort_handler. continue_handler z in
+ let make_list = \h t. \f2 z continue_handler abort_handler.
+ t f2 z (\sofar. f2 h sofar continue_handler abort_handler) abort_handler in
let isempty = \lst larger_computation. lst
 ; here's our f
+ ; here's our f2
(\hd sofar continue_handler abort_handler. abort_handler false)
; here's our z
true
 ; here's the continue_handler for the leftmost application of f
+ ; here's the continue_handler for the leftmost application of f2
larger_computation
; here's the abort_handler
larger_computation in
let extract_head = \lst larger_computation. lst
 ; here's our f
+ ; here's our f2
(\hd sofar continue_handler abort_handler. continue_handler hd)
; here's our z
junk
 ; here's the continue_handler for the leftmost application of f
+ ; here's the continue_handler for the leftmost application of f2
larger_computation
; here's the abort_handler
larger_computation in
@@ 533,39 +569,56 @@ Of course, like everything elegant and exciting in this seminar, [Oleg
discusses it in much more
detail](http://okmij.org/ftp/Streams.html#enumeratorstream).
*Comments*:

1. The technique deployed here, and in the v2 lists, and in our implementations
 of pairs and booleans, is known as **continuationpassing style** programming.

2. We're still building the list as a right fold, so in a sense the
 application of `f` to the leftmost element `5` is "outermost". However,
 this "outermost" application is getting lifted, and passed as a *handler*
 to the next right application. Which is in turn getting lifted, and
 passed to its next right application, and so on. So if you
 trace the evaluation of the `extract_head` function to the list `[5;4;3;2;1]`,
 you'll see `1` gets passed as a "this is the head sofar" answer to its
 `continue_handler`; then that answer is discarded and `2` is
 passed as a "this is the head sofar" answer to *its* `continue_handler`,
 and so on. All those steps have to be evaluated to finally get the result
 that `5` is the outer/leftmost head of the list. That's not an efficient way
 to get the leftmost head.

 We could improve this by building lists as left folds when implementing them
 as continuationpassing style folds. We'd just replace above:

 let make_list = \h t. \f z continue_handler abort_handler.
 f h z (\z. t f z continue_handler abort_handler) abort_handler

 now `extract_head` should return the leftmost head directly, using its `abort_handler`:

 let extract_head = \lst larger_computation. lst
 (\hd sofar continue_handler abort_handler. abort_handler hd)
 junk
 larger_computation
 larger_computation

3. To extract tails efficiently, too, it'd be nice to fuse the apparatus developed
 in these v5 lists with the ideas from [v4](/advanced/#v4) lists.
 But that also is left as an exercise.
+> *Comments*:
+
+> 1. The technique deployed here, and in the v2 lists, and in our
+> implementations of pairs and booleans, is known as
+> **continuationpassing style** programming.
+
+> 2. We're still building the list as a right fold, so in a sense the
+> application of `f2` to the leftmost element `5` is "outermost". However,
+> this "outermost" application is getting lifted, and passed as a *handler*
+> to the next right application. Which is in turn getting lifted, and
+> passed to its next right application, and so on. So if you
+> trace the evaluation of the `extract_head` function to the list `[5;4;3;2;1]`,
+> you'll see `1` gets passed as a "this is the head sofar" answer to its
+> `continue_handler`; then that answer is discarded and `2` is
+> passed as a "this is the head sofar" answer to *its* `continue_handler`,
+> and so on. All those steps have to be evaluated to finally get the result
+> that `5` is the outer/leftmost head of the list. That's not an efficient way
+> to get the leftmost head.
+>
+> We could improve this by building lists as **left folds**. What's that?
+>
+> Well, the right fold of `f` over a list `[a;b;c;d;e]`, using starting value z, is:
+>
+> f a (f b (f c (f d (f e z))))
+>
+> The left fold on the other hand starts combining `z` with elements from the left. `f z a` is then combined with `b`, and so on:
+>
+> f (f (f (f (f z a) b) c) d) e
+>
+> or, if we preferred the arguments to each `f` flipped:
+>
+> f e (f d (f c (f b (f a z))))
+>
+> Recall we implemented v3 lists as their own rightfold functions. We could
+> instead implement lists as their own leftfold functions. To do that with our
+> v5 lists, we'd replace above:
+>
+> let make_list = \h t. \f2 z continue_handler abort_handler.
+> f2 h z (\z. t f2 z continue_handler abort_handler) abort_handler
+>
+> Having done that, now `extract_head` can return the leftmost head
+> directly, using its `abort_handler`:
+>
+> let extract_head = \lst larger_computation. lst
+> (\hd sofar continue_handler abort_handler. abort_handler hd)
+> junk
+> larger_computation
+> larger_computation
+>
+> 3. To extract tails efficiently, too, it'd be nice to fuse the apparatus
+> developed in these v5 lists with the ideas from
+> [v4](/advanced_lambda/#index1h1) lists. But that is left as an exercise.