`x`_{n}

and `zero`, where `x`_{n}

is the last element of the list. This gives us `successor zero`, or `one`. That's the value we've accumuluted "so far." Then we go apply the function `\x sofar. successor sofar` to the two arguments `x`_{n-1}

and the value `one` that we've accumulated "so far." This gives us `two`. We continue until we get to the start of the list. The value we've then built up "so far" will be the length of the list.
We can use similar techniques to define many recursive operations on lists and numbers. The reason we can do this is that our "version 3," fold-based implementation of lists, and Church's implementations of numbers, have a internal structure that *mirrors* the common recursive operations we'd use lists and numbers for.
As we said before, it does take some ingenuity to define functions like `extract-tail` or `predecessor` for these implementations. However it can be done. (And it's not *that* difficult.) Given those functions, we can go on to define other functions like numeric equality, subtraction, and so on, just by exploiting the structure already present in our implementations of lists and numbers.
With sufficient ingenuity, a great many functions can be defined in the same way. For example, the factorial function is straightforward. The function which returns the nth term in the Fibonacci series is a bit more difficult, but also achievable.
##However...##
Some computable functions are just not definable in this way. We can't, for example, define a function that tells us, for whatever function `f` we supply it, what is the smallest integer `x` where `f x` is `true`.
Neither do the resources we've so far developed suffice to define the
[[!wikipedia Ackermann function]]:
A(m,n) =
| when m == 0 -> n + 1
| else when n == 0 -> A(m-1,1)
| else -> A(m-1, A(m,n-1))
A(0,y) = y+1
A(1,y) = 2+(y+3) - 3
A(2,y) = 2(y+3) - 3
A(3,y) = 2^(y+3) - 3
A(4,y) = 2^(2^(2^...2)) [where there are y+3 2s] - 3
...
Simpler functions always *could* be defined using the resources we've so far developed, although those definitions won't always be very efficient or easily intelligible.
But functions like the Ackermann function require us to develop a more general technique for doing recursion---and having developed it, it will often be easier to use it even in the cases where, in principle, we didn't have to.
##How to do recursion with lower-case omega##
Recall our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
\lst. (isempty lst) zero (add one (... (extract-tail lst)))
where this very same formula occupies the `...` position."
We are not going to exactly that, at least not yet. But we are going to do something close to it.
Consider a formula of the following form (don't worry yet about exactly how we'll fill the `...`s):
\h \lst. (isempty lst) zero (add one (... (extract-tail lst)))
Call that formula `H`. Now what would happen if we applied `H` to itself? Then we'd get back:
\lst. (isempty lst) zero (add one (... (extract-tail lst)))
where any occurrences of `h` inside the `...` were substituted with `H`. Call this `F`. `F` looks pretty close to what we're after: a function that takes a list and returns zero if it's empty, and so on. And `F` is the result of applying `H` to itself. But now inside `F`, the occurrences of `h` are substituted with the very formula `H` we started with. So if we want to get `F` again, all we have to do is apply `h` to itself---since as we said, the self-application of `H` is how we created `F` in the first place.
So, the way `F` should be completed is:
\lst. (isempty lst) zero (add one ((h h) (extract-tail lst)))
and our original `H` is:
\h \lst. (isempty lst) zero (add one ((h h) (extract-tail lst)))
The self-application of `H` will give us `F` with `H` substituted in for its free variable `h`.
Instead of writing out a long formula twice, we could write:
(\x. x x) LONG-FORMULA
and the initial `(\x. x x)` is just what we earlier called the `ω`

combinator (lower-case omega, not the non-terminating `Ω`

). So the self-application of `H` can be written:
```
ω (\h \lst. (isempty lst) zero (add one ((h h) (extract-tail lst))))
```

and this will indeed implement the recursive function we couldn't earlier figure out how to define.
In broad brush-strokes, `H` is half of the `get_length` function we're seeking, and `H` has the form:
\h other-arguments. ... (h h) ...
We get the whole `get_length` function by applying `H` to itself. Then `h` is replaced by the half `H`, and when we later apply `h` to itself, we re-create the whole `get_length` again.
##Neat! Can I make it easier to use?##
Suppose you wanted to wrap this up in a pretty interface, so that the programmer didn't need to write `(h h)` but could just write `g` for some function `g`. How could you do it?
Now the `F`-like expression we'd be aiming for---call it `F*`---would look like this:
\lst. (isempty lst) zero (add one (g (extract-tail lst)))
or, abbreviating:
\lst. ...g...
Here we have just a single `g` instead of `(h h)`. We'd want `F*` to be the result of self-applying some `H*`, and then binding to `g` that very self-application of `H*`. We'd get that if `H*` had the form:
\h. (\g lst. ...g...) (h h)
The self-application of `H*` would be:
(\h. (\g lst. ...g...) (h h)) (\h. (\g lst. ...g...) (h h))
or:
(\f. (\h. f (h h)) (\h. f (h h))) (\g lst. ...g...)
The left-hand side of this is known as **the Y-combinator** and so this could be written more compactly as:
Y (\g lst. ...g...)
or, replacing the abbreviated bits:
Y (\g lst. (isempty lst) zero (add one (g (extract-tail lst))))
So this is another way to implement the recursive function we couldn't earlier figure out how to define.
##Generalizing##
Let's step back and fill in some theory to help us understand why these tricks work.
In general, we call a **fixed point** of a function f any value *x* such that f `Ψ`

. That is, some function that returns, for any expression `f` we give it as argument, a fixed point for `f`. In other words:
`Ψ f <~~> f (Ψ f)`

Then applying `Ψ`

to the "starting formula" displayed above would give us our fixed point `X` for the starting formula:
`Ψ (\self (\lst. (isempty lst) zero (add one (self (extract-tail lst))) ))`

And this is the fully general strategy for
defining recursive functions in the lambda calculus. You begin with a "body formula":
...self...
containing free occurrences of `self` that you treat as being equivalent to the body formula itself. In the case we're considering, that was:
\lst. (isempty lst) zero (add one (self (extract-tail lst)))
You bind the free occurrence of `self` as: `\self. BODY`. And then you generate a fixed point for this larger expression:
`Ψ (\self. BODY)`

using some fixed-point combinator `Ψ`

.
Isn't that cool?
##Okay, then give me a fixed-point combinator, already!##
Many fixed-point combinators have been discovered. (And some fixed-point combinators give us models for building infinitely many more, non-equivalent fixed-point combinators.)
Two of the simplest:
```
Θ′ ≡ (\u f. f (\n. u u f n)) (\u f. f (\n. u u f n))
Y′ ≡ \f. (\u. f (\n. u u n)) (\u. f (\n. u u n))
```

`Θ′`

has the advantage that `f (Θ′ f)`

really *reduces to* `Θ′ f`

. Whereas `f (Y′ f)`

is only *convertible with* `Y′ f`

; that is, there's a common formula they both reduce to. For most purposes, though, either will do.
You may notice that both of these formulas have eta-redexes inside them: why can't we simplify the two `\n. u u f n` inside `Θ′`

to just `u u f`? And similarly for `Y′`

?
Indeed you can, getting the simpler:
```
Θ ≡ (\u f. f (u u f)) (\u f. f (u u f))
Y ≡ \f. (\u. f (u u)) (\u. f (u u))
```

I stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\self. BODY)`

and `Y (\self. BODY)` will in general not terminate. But evaluation of the eta-unreduced primed versions will.
Of course, if you define your `\self. BODY` stupidly, your formula will never terminate. For example, it doesn't matter what fixed point combinator you use for `Ψ`

in:
`Ψ (\self. \n. self n)`

When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
(\n. self n) M
where `self` is equivalent to the very formula `\n. self n` that contains it. So the evaluation will proceed:
(\n. self n) M ~~>
self M ~~>
(\n. self n) M ~~>
self M ~~>
...
You've written an infinite loop!
However, when we evaluate the application of our:
`Ψ (\self (\lst. (isempty lst) zero (add one (self (extract-tail lst))) ))`

to some list `L`, we're not going to go into an infinite evaluation loop of that sort. At each cycle, we're going to be evaluating the application of:
\lst. (isempty lst) zero (add one (self (extract-tail lst)))
to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the implementations we're using don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `zero`. So the recursion eventually bottoms out in a base value.
##Fixed-point Combinators Are a Bit Intoxicating##
![tatoo](/y-combinator.jpg)
There's a tendency for people to say "Y-combinator" to refer to fixed-point combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Y-combinator is only one of many fixed-point combinators.
I used `Ψ`

above to stand in for an arbitrary fixed-point combinator. I don't know of any broad conventions for this. But this seems a useful one.
As we said, there are many other fixed-point combinators as well. For example, Jan Willem Klop pointed out that if we define `L` to be:
\a b c d e f g h i j k l m n o p q s t u v w x y z r. (r (t h i s i s a f i x e d p o i n t c o m b i n a t o r))
then this is a fixed-point combinator:
L L L L L L L L L L L L L L L L L L L L L L L L L L
##Watching Y in action##
For those of you who like to watch ultra slow-mo movies of bullets
piercing apples, here's a stepwise computation of the application of a
recursive function. We'll use a function `sink`, which takes one
argument. If the argument is boolean true (i.e., `\x y.x`), it
returns itself (a copy of `sink`); if the argument is boolean false
(`\x y. y`), it returns `I`. That is, we want the following behavior:
sink false ~~> I
sink true false ~~> I
sink true true false ~~> I
sink true true true false ~~> I
So we make `sink = Y (\f b. b f I)`:
1. sink false
2. Y (\fb.bfI) false
3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) false
4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) false
5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) false
6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) false
7. false [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I
--------------------------------------------
8. I
So far so good. The crucial thing to note is that as long as we
always reduce the outermost redex first, we never have to get around
to computing the underlined redex: because `false` ignores its first
argument, we can throw it away unreduced.
Now we try the next most complex example:
1. sink true false
2. Y (\fb.bfI) true false
3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) true false
4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) true false
5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) true false
6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) true false
7. true [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I false
8. [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] false
We've now arrived at line (4) of the first computation, so the result
is again I.
You should be able to see that `sink` will consume as many `true`s as
we throw at it, then turn into the identity function after it
encounters the first `false`.
The key to the recursion is that, thanks to Y, the definition of
`sink` contains within it the ability to fully regenerate itself as
many times as is necessary. The key to *ending* the recursion is that
the behavior of `sink` is sensitive to the nature of the input: if the
input is the magic function `false`, the self-regeneration machinery
will be discarded, and the recursion will stop.
That's about as simple as recursion gets.
##Base cases, and their lack##
As any functional programmer quickly learns, writing a recursive
function divides into two tasks: figuring out how to handle the
recursive case, and remembering to insert a base case. The
interesting and enjoyable part is figuring out the recursive pattern,
but the base case cannot be ignored, since leaving out the base case
creates a program that runs forever. For instance, consider computing
a factorial: `n!` is `n * (n-1) * (n-2) * ... * 1`. The recursive
case says that the factorial of a number `n` is `n` times the
factorial of `n-1`. But if we leave out the base case, we get
3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * -1! ...
That's why it's crucial to declare that 0! = 1, in which case the
recursive rule does not apply. In our terms,
fac = Y (\fac n. iszero n 1 (fac (predecessor n)))
If `n` is 0, `fac` reduces to 1, without computing the recursive case.
There is a well-known problem in philosophy and natural language
semantics that has the flavor of a recursive function without a base
case: the truth-teller paradox (and related paradoxes).
(1) This sentence is true.
If we assume that the complex demonstrative "this sentence" can refer
to (1), then the proposition expressed by (1) will be true just in
case the thing referred to by *this sentence* is true. Thus (1) will
be true just in case (1) is true, and (1) is true just in case (1) is
true, and so on. If (1) is true, then (1) is true; but if (1) is not
true, then (1) is not true.
Without pretending to give a serious analysis of the paradox, let's
assume that sentences can have for their meaning boolean functions
like the ones we have been working with here. Then the sentence *John
is John* might denote the function `\x y. x`, our `true`.
Then (1) denotes a function from whatever the referent of *this
sentence* is to a boolean. So (1) denotes `\f. f true false`, where
the argument `f` is the referent of *this sentence*. Of course, if
`f` is a boolean, `f true false <~~> f`, so for our purposes, we can
assume that (1) denotes the identity function `I`.
If we use (1) in a context in which *this sentence* refers to the
sentence in which the demonstrative occurs, then we must find a
meaning `m` such that `I m = I`. But since in this context `m` is the
same as the meaning `I`, so we have `m = I m`. In other words, `m` is
a fixed point for the denotation of the sentence (when used in the
appropriate context).
That means that in a context in which *this sentence* refers to the
sentence in which it occurs, the sentence denotes a fixed point for
the identity function. Here's a fixed point for the identity
function:
```
Y I
(\f. (\h. f (h h)) (\h. f (h h))) I
(\h. I (h h)) (\h. I (h h)))
(\h. (h h)) (\h. (h h)))
ω ω
&Omega
```

Oh. Well! That feels right. The meaning of *This sentence is true*
in a context in which *this sentence* refers to the sentence in which
it occurs is `Ω`

, our prototypical infinite loop...
What about the liar paradox?
(2) This sentence is false.
Used in a context in which *this sentence* refers to the utterance of
(2) in which it occurs, (2) will denote a fixed point for `\f.neg f`,
or `\f l r. f r l`, which is the `C` combinator. So in such a
context, (2) might denote
Y C
(\f. (\h. f (h h)) (\h. f (h h))) I
(\h. C (h h)) (\h. C (h h)))
C ((\h. C (h h)) (\h. C (h h)))
C (C ((\h. C (h h))(\h. C (h h))))
C (C (C ((\h. C (h h))(\h. C (h h)))))
...
And infinite sequence of `C`s, each one negating the remainder of the
sequence. Yep, that feels like a reasonable representation of the
liar paradox.
See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
truth and circularity](http://tinyurl.com/2db62bk) for an approach
that is similar, but expressed in terms of non-well-founded sets
rather than recursive functions.
HOWEVER, you should be cautious about feeling too comfortable with
these results. Thinking again of the truth-teller paradox, yes,
`Ω`

is *a* fixed point for `I`, and perhaps it has
some a privileged status among all the fixed points for `I`, being the
one delivered by Y and all (though it is not obvious why Y should have
any special status).
But one could ask: look, literally every formula is a fixed point for
`I`, since
X <~~> I X
for any choice of X whatsoever.
So the Y combinator is only guaranteed to give us one fixed point out
of infinitely many---and not always the intuitively most useful
one. (For instance, the squaring function has zero as a fixed point,
since 0 * 0 = 0, and 1 as a fixed point, since 1 * 1 = 1, but `Y
(\x. mul x x)` doesn't give us 0 or 1.) So with respect to the
truth-teller paradox, why in the reasoning we've
just gone through should we be reaching for just this fixed point at
just this juncture?
One obstacle to thinking this through is the fact that a sentence
normally has only two truth values. We might consider instead a noun
phrase such as
(3) the entity that this noun phrase refers to
The reference of (3) depends on the reference of the embedded noun
phrase *this noun phrase*. It's easy to see that any object is a
fixed point for this referential function: if this pen cap is the
referent of *this noun phrase*, then it is the referent of (3), and so
for any object.
The chameleon nature of (3), by the way (a description that is equally
good at describing any object), makes it particularly well suited as a
gloss on pronouns such as *it*. In the system of
[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
pronouns denote (you guessed it!) identity functions...
Ultimately, in the context of this course, these paradoxes are more
useful as a way of gaining leverage on the concepts of fixed points
and recursion, rather than the other way around.