ωω (\h \lst. (isempty lst) zero (add one ((h h) (extracttail lst))))

and this will indeed implement the recursive function we couldn't earlier figure out how to define.
In broad brushstrokes, `H` is half of the `get_length` function we're seeking, and `H` has the form:
@@ 546,18 +545,17 @@ 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

+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...
+it occurs is Ω
, our prototypical infinite loop...
What about the liar paradox?
diff git a/week3a.mdwn b/week3a.mdwn
deleted file mode 100644
index 3ed45aef..00000000
 a/week3a.mdwn
+++ /dev/null
@@ 1,431 +0,0 @@
##Computing the length of a list##

How could we compute the length of a list? Without worrying yet about what lambdacalculus implementation we're using for the list, the basic idea would be to define this recursively:

> the empty list has length 0

> any nonempty list has length 1 + (the length of its tail)

In OCaml, you'd define that like this:

 let rec get_length = fun lst >
 if lst == [] then 0 else 1 + get_length (tail lst)
 in ... (* here you go on to use the function "get_length" *)

In Scheme you'd define it like this:

 (letrec [(get_length
 (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
 ... ; here you go on to use the function "get_length"
 )

Some comments on this:

1. `null?` is Scheme's way of saying `isempty`. That is, `(null? lst)` returns true (which Scheme writes as `#t`) iff `lst` is the empty list (which Scheme writes as `'()` or `(list)`).

2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of an ordered pair. It just turns out to return the tail of a list because of the particular way Scheme implements lists.)

3. I use `get_length` instead of the convention we've been following so far of hyphenated names, as in `makelist`, because we're discussing OCaml code here, too, and OCaml doesn't permit the hyphenated variable names. OCaml requires variables to always start with a lowercase letter (or `_`), and then continue with only letters, numbers, `_` or `'`. Most other programming languages are similar. Scheme is very relaxed, and permits you to use ``, `?`, `/`, and all sorts of other crazy characters in your variable names.

4. I alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.


The main question for us to dwell on here is: What are the `let rec` in the OCaml code and the `letrec` in the Scheme code?

Answer: These work like the `let` expressions we've already seen, except that they let you use the variable `get_length` *inside* the body of the function being bound to itwith the understanding that it will there refer to the same function that you're then in the process of binding to `get_length`. So our recursivelydefined function works the way we'd expect it to. In OCaml:

 let rec get_length = fun lst >
 if lst == [] then 0 else 1 + get_length (tail lst)
 in get_length [20; 30]
 (* this evaluates to 2 *)

In Scheme:

 (letrec [(get_length
 (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
 (get_length (list 20 30)))
 ; this evaluates to 2

If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:

 let get_length = fun lst >
 if lst == [] then 0 else 1 + get_length (tail lst)
 in get_length [20; 30]
 (* fails with error "Unbound value length" *)

Here's Scheme:

 (let* [(get_length
 (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )]
 (get_length (list 20 30)))
 ; fails with error "reference to undefined identifier: get_length"

Why? Because we said that constructions of this form:

 let get_length = A
 in B

really were just another way of saying:

 (\get_length. B) A

and so the occurrences of `get_length` in A *aren't bound by the `\get_length` that wraps B*. Those occurrences are free.

We can verify this by wrapping the whole expression in a more outer binding of `get_length` to some other function, say the constant function from any list to the integer 99:

 let get_length = fun lst > 99
 in let get_length = fun lst >
 if lst == [] then 0 else 1 + get_length (tail lst)
 in get_length [20; 30]
 (* evaluates to 1 + 99 *)

Here the use of `get_length` in `1 + get_length (tail lst)` can clearly be seen to be bound by the outermost `let`.

And indeed, if you tried to define `get_length` in the lambda calculus, how would you do it?

 \lst. (isempty lst) zero (add one (get_length (extracttail lst)))

We've defined all of `isempty`, `zero`, `add`, `one`, and `extracttail` in earlier discussion. But what about `get_length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.

What we really want to do is something like this:

 \lst. (isempty lst) zero (add one (... (extracttail lst)))

where this very same formula occupies the `...` position:

 \lst. (isempty lst) zero (add one (
 \lst. (isempty lst) zero (add one (... (extracttail lst)))
 (extracttail lst)))

but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.

[At this point, some of you will recall the discussion in the first
class concerning the conception of functions as sets of ordered pairs.
The problem, as you will recall, was that in the untyped lambda
calculus, we wanted a function to be capable of taking itself as an
argument. For instance, we wanted to be able to apply the identity
function to itself. And since the identity function always returns
its argument unchanged, the value it should return in that case is
itself:

 (\x.x)(\x.x) ~~> (\x.x)

If we conceive of a function as a set of ordered pairs, we would start
off like this:

 1 > 1
 2 > 2
 3 > 3
 ...
 [1 > 1, 2 > 2, 3 > 3, ..., [1 > 1, 2 > 2, 3 > 3, ...,

Eventually, we would get to the point where we want to say what the
identity function itself gets mapped to. But in order to say that, we
need to write down the identity function in the argument position as a
set of ordered pairs. The need to insert a copy of the entire
function definition inside of a copy of the entire function definition
inside of... is the same problem as the need to insert a complete
graph of the identity function inside of the graph for the identity function.]

So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?

1. OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.

2. If you tried this in Scheme:

 (define get_length
 (lambda (lst) (if (null? lst) 0 [+ 1 (get_length (cdr lst))] )) )

 (get_length (list 20 30))

 You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.

3. In fact, it *is* possible to define the `get_length` function in the lambda calculus despite these obstacles. This depends on using the "version 3" implementation of lists, and exploiting its internal structure: that it takes a function and a base value and returns the result of folding that function over the list, with that base value. So we could use this as a definition of `get_length`:

 \lst. lst (\x sofar. successor sofar) zero

 What's happening here? We start with the value zero, then we apply the function `\x sofar. successor sofar` to the two arguments 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_{n1}
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," foldbased 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 `extracttail` 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(m1,1)
  else > A(m1, A(m,n1))

 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 recursionand 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 lowercase 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 (... (extracttail 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 (... (extracttail 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 (... (extracttail 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 itselfsince as we said, the selfapplication 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) (extracttail lst)))

and our original `H` is:

 \h \lst. (isempty lst) zero (add one ((h h) (extracttail lst)))

The selfapplication 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) LONGFORMULA

and the initial `(\x. x x)` is just what we earlier called the ω
combinator (lowercase omega, not the nonterminating Ω
). So the selfapplication of `H` can be written:

ω (\h \lst. (isempty lst) zero (add one ((h h) (extracttail lst))))



and this will indeed implement the recursive function we couldn't earlier figure out how to define.

In broad brushstrokes, `H` is half of the `get_length` function we're seeking, and `H` has the form:

 \h otherarguments. ... (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 recreate 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 forcall it `F*`would look like this:

 \lst. (isempty lst) zero (add one (g (extracttail 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 selfapplying some `H*`, and then binding to `g` that very selfapplication of `H*`. We'd get that if `H*` had the form:

 \h. (\g lst. ...g...) (h h)

The selfapplication 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 lefthand side of this is known as **the Ycombinator** 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 (extracttail 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 x is equivalent to *x*. For example, what is a fixed point of the function from natural numbers to their squares? What is a fixed point of the successor function?

In the lambda calculus, we say a fixed point of an expression `f` is any formula `X` such that:

 X <~~> f X

What is a fixed point of the identity combinator I?

It's a theorem of the lambda calculus that every formula has a fixed point. In fact, it will have infinitely many, nonequivalent fixed points. And we don't just know that they exist: for any given formula, we can name many of them.

Yes, even the formula that you're using the define the successor function will have a fixed point. Isn't that weird? Think about how it might be true.

Well, you might think, only some of the formulas that we might give to the `successor` as arguments would really represent numbers. If we said something like:

 successor makepair

who knows what we'd get back? Perhaps there's some nonnumberrepresenting formula such that when we feed it to `successor` as an argument, we get the same formula back.

Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.

Moreover, the recipes that enable us to name fixed points for any given formula aren't *guaranteed* to give us *terminating* fixed points. They might give us formulas X such that neither `X` nor `f X` have normal forms. (Indeed, what they give us for the square function isn't any of the Church numerals, but is rather an expression with no normal form.) However, if we take care we can ensure that we *do* get terminating fixed points. And this gives us a principled, fully general strategy for doing recursion. It lets us define even functions like the Ackermann function, which were until now out of our reach. It would also let us define arithmetic and list functions on the "version 1" and "version 2" implementations, where it wasn't always clear how to force the computation to "keep going."

OK, so how do we make use of this?

Recall again 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 (... (extracttail lst)))

where this very same formula occupies the `...` position."

If we could somehow get ahold of this very formula, as an additional argument, then we could take the argument and plug it into the `...` position. Something like this:

 \self (\lst. (isempty lst) zero (add one (self (extracttail lst))) )

This is an abstract of the form:

 \self. BODY

where `BODY` is the expression:

 \lst. (isempty lst) zero (add one (self (extracttail lst)))

containing an occurrence of `self`.

Now consider what would be a fixed point of our expression `\self. BODY`? That would be some expression `X` such that:

 X <~~> (\self.BODY) X

Betareducing the righthand side, we get:

 X <~~> BODY [self := X]

Think about what this says. It says if you substitute `X` for `self` in our formula BODY:

 \lst. (isempty lst) zero (add one (X (extracttail lst)))

what you get is "equivalent" to (that is, convertible with) X itself. That is, the `X` inside the above expression is equivalent to the whole expression. So the expression *does*, in a sense, contain itself!

Let's go over that again. If we had a fixed point `X` for our expression `\self. ...self...`, then by the definition of a fixedpoint, this has to be true:

 X <~~> (\self. ...self...) X

but betareducing the righthand side, we get something of the form:

 X <~~> ...X...

So on the righthand side we have a complex expression, that contains some occurrences of whatever our fixedpoint `X` is, and `X` is convertible with *that very complex, righthand side expression.*

So we really *can* define `get_length` in the way we were initially attempting, in the bare lambda calculus, where Scheme and OCaml's soupedup `let rec` constructions aren't primitively available. (In fact, what we're doing here is the natural way to implement `let rec`.)

This all turns on having a way to generate a fixedpoint for our "starting formula":

 \self (\lst. (isempty lst) zero (add one (self (extracttail lst))) )

Where do we get it?

Suppose we have some **fixedpoint combinator**
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 (extracttail 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 (extracttail 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 fixedpoint combinator Ψ
.

Isn't that cool?

##Okay, then give me a fixedpoint combinator, already!##

Many fixedpoint combinators have been discovered. (And some fixedpoint combinators give us models for building infinitely many more, nonequivalent fixedpoint 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 etaredexes 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 *callbyvalue*, the evaluation of Θ (\self. BODY)
and `Y (\self. BODY)` will in general not terminate. But evaluation of the etaunreduced 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 (extracttail 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 (extracttail 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.

##Fixedpoint Combinators Are a Bit Intoxicating##

![tatoo](/ycombinator.jpg)

There's a tendency for people to say "Ycombinator" to refer to fixedpoint combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Ycombinator is only one of many fixedpoint combinators.

I used Ψ
above to stand in for an arbitrary fixedpoint combinator. I don't know of any broad conventions for this. But this seems a useful one.

As we said, there are many other fixedpoint 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 fixedpoint 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



2.11.0