```
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
```

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`, `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 ≡ 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 (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: 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
number---had 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)...)
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 (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.
#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 fixed-point 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 `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.)
This is amazing, by the way: we're proving things about a term that
represents arithmetic infinity.
It's important to bear in mind the simplest term in question is not
infinite:
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 well-behaved base case, such as
the factorial function, evaluation order is crucial. In the following
computation, we will arrive at a normal form. Watch for the moment at
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 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 ([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`,
no matter what the ... contains;
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/call-by-name/normal-order evaluation strategy, we'll always
start with the `iszero` predicate, and only produce a fresh copy of
`prefact` if we are forced to.
#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(m-1,1)
| else -> A(m-1, A(m,n-1))
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 hyper-exponentiation is to exponentiation...
#Q. What other questions should I be asking?#
* What is it about the variant fixed-point combinators that makes
them compatible with a call-by-value evaluation strategy?
* How do you know that the Ackerman function can't be computed
using primitive recursion techniques?
* What *exactly* is primitive recursion?
* I hear that `Y` delivers the *least* fixed point. Least
according to what ordering? How do you know it's least?
Is leastness important?
#Sets#
You're now already in a position to implement sets: that is, collections with
no intrinsic order where elements can occur at most once. Like lists, we'll
understand the basic set structures to be *type-homogenous*. So you might have
a set of integers, or you might have a set of pairs of integers, but you
wouldn't have a set that mixed both types of elements. Something *like* the
last option is also achievable, but it's more difficult, and we won't pursue it
now. In fact, we won't talk about sets of pairs, either. We'll just talk about
sets of integers. The same techniques we discuss here could also be applied to
sets of pairs of integers, or sets of triples of booleans, or sets of pairs
whose first elements are booleans, and whose second elements are triples of
integers. And so on.
(You're also now in a position to implement *multi*sets: that is, collections
with no intrinsic order where elements can occur multiple times: the multiset
{a,a} is distinct from the multiset {a}. But we'll leave these as an exercise.)
The easiest way to implement sets of integers would just be to use lists. When
you "add" a member to a set, you'd get back a list that was either identical to
the original list, if the added member already was present in it, or consisted
of a new list with the added member prepended to the old list. That is:
let empty_set = empty in
; see the library for definitions of any and eq
let make_set = \new_member old_set. any (eq new_member) old_set
; if any element in old_set was eq new_member
old_set
; else
make_list new_member old_set
Think about how you'd implement operations like `set_union`,
`set_intersection`, and `set_difference` with this implementation of sets.
The implementation just described works, and it's the simplest to code.
However, it's pretty inefficient. If you had a 100-member set, and you wanted
to create a set which had all those 100-members and some possibly new element
`e`, you might need to check all 100 members to see if they're equal to `e`
before concluding they're not, and returning the new list. And comparing for
numeric equality is a moderately expensive operation, in the first place.
(You might say, well, what's the harm in just prepending `e` to the list even
if it already occurs later in the list. The answer is, if you don't keep track
of things like this, it will likely mess up your implementations of
`set_difference` and so on. You'll have to do the book-keeping for duplicates
at some point in your code. It goes much more smoothly if you plan this from
the very beginning.)
How might we make the implementation more efficient? Well, the *semantics* of
sets says that they have no intrinsic order. That means, there's no difference
between the set {a,b} and the set {b,a}; whereas there is a difference between
the *list* `[a;b]` and the list `[b;a]`. But this semantic point can be respected
even if we *implement* sets with something ordered, like list---as we're
already doing. And we might *exploit* the intrinsic order of lists to make our
implementation of sets more efficient.
What we could do is arrange it so that a list that implements a set always
keeps in elements in some specified order. To do this, there'd have *to be*
some way to order its elements. Since we're talking now about sets of numbers,
that's easy. (If we were talking about sets of pairs of numbers, we'd use
"lexicographic" ordering, where `(a,b) < (c,d)` iff `a < c or (a == c and b <
d)`.)
So, if we were searching the list that implements some set to see if the number
`5` belonged to it, once we get to elements in the list that are larger than `5`,
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.
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
implements the set, if it belonged to the set at all. Or that it fell in the
right half, it it belonged to the set at all. And then the same sort of
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).
#Aborting a search through a list#
We said that the sorted-list implementation of a set was more efficient than
the unsorted-list implementation, because as you were searching through the
list, you could come to a point where you knew the element wasn't going to be
found. So you wouldn't have to continue the search.
If your implementation of lists was, say v1 lists plus the Y-combinator, then
this is exactly right. When you get to a point where you know the answer, you
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
contains the number `3`. The expression which represents this search would have
something like the following form:
..................`extract_fst ≡ \pair. pair (\x y. x)`

but at a lower level, the pair is still accepting its handler as an argument,
rather than the handler taking the pair as an argument. (The handler gets *the
pair's elements*, not the pair itself, as arguments.)
> *Terminology*: we'll try to use names of the form `get_foo` for handlers, and
names of the form `extract_foo` for lifted versions of them, that accept the
lists (or whatever data structure we're working with) as arguments. But we may
sometimes forget.
The v2 implementation of lists followed a similar strategy:
v2list (\h t. do_something_with_h_and_t) result_if_empty
If the `v2list` here is not empty, then this will reduce to the result of
supplying the list's head and tail to the handler `(\h t.
do_something_with_h_and_t)`.
Now, what we've been imagining ourselves doing with the search through the v3
list is something like this:
larger_computation (search_through_the_list_for_3) other_arguments
That is, the result of our search is supplied as an argument (perhaps together
with other arguments) to the "larger computation". Without knowing the
evaluation order/reduction strategy, we can't say whether the search is
evaluated before or after it's substituted into the larger computation. But
semantically, the search is the argument and the larger computation is the
function to which it's supplied.
What if, instead, we did the same kind of thing we did with pairs and v2
lists? That is, what if we made the larger computation a "handler" that we
passed as an argument to the search?
the_search (\search_result. larger_computation search_result other_arguments)
What's the advantage of that, you say. Other than to show off how cleverly
you can lift.
Well, think about it. Think about the difficulty we were having aborting the
search. Does this switch-around offer us anything useful?
It could.
What if the way we implemented the search procedure looked something like this?
At a given stage in the search, we wouldn't just apply some function `f` to the
head at this stage and the result accumulated so far (from folding the same
function, and a base value, to the tail at this stage)...and then pass the result
of that application to the embedding, more leftward computation.
We'd *instead* give `f` a "handler" that expects the result of the current
stage *as an argument*, and then evaluates to what you'd get by passing that
result leftwards up the list, as before.
Why would we do that, you say? Just more flamboyant lifting?
Well, no, there's a real point here. If we give the function a "handler" that
encodes the normal continuation of the fold leftwards through the list, we can
also give it other "handlers" too. For example, we can also give it the underlined handler:
the_search (\search_result. larger_computation search_result other_arguments)
------------------------------------------------------------------
This "handler" encodes the search's having finished, and delivering a final
answer to whatever else you wanted your program to do with the result of the
search. If you like, at any stage in the search you might just give an argument
to *this* handler, instead of giving an argument to the handler that continues
the list traversal leftwards. Semantically, this would amount to *aborting* the
list traversal! (As we've said before, whether the rest of the list traversal
really gets evaluated will depend on what evaluation order is in place. But
semantically we'll have avoided it. Our larger computation won't depend on the
rest of the list traversal having been computed.)
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.
f 3