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#Recursion: fixed points in the lambda calculus##

Sometimes when you type in a web search, Google will suggest
alternatives.  For instance, if you type in "Lingusitics", it will ask
you "Did you mean Linguistics?".  But the engineers at Google have
added some playfulness to the system.  For instance, if you search for
"anagram", Google asks you "Did you mean: nag a ram?"  And if you
search for "recursion", Google asks: "Did you mean: recursion?"

##What is the "rec" part of "letrec" doing?##

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

>       the empty list has length 0

>       any non-empty list has length 1 + (the length of its tail)

In OCaml, you'd define that like this:

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

In Scheme you'd define it like this:

        (letrec [(length
                                (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )]
                ... ; here you go on to use the function "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 `length` instead of the convention we've been following so far of hyphenated names, as in `make-list`, 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 lower-case 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 `length` *inside* the body of the function being bound to it---with the understanding that it will there refer to the same function that you're then in the process of binding to `length`. So our recursively-defined function works the way we'd expect it to. In OCaml:

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

In Scheme:

        (letrec [(length
                                (lambda (lst) (if (null? lst) 0 [+ 1 (length (cdr lst))] )) )]
                        (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 length = fun lst ->
                if lst == [] then 0 else 1 + length (tail lst)
        in length [20; 30]
        (* fails with error "Unbound value length" *)

Here's Scheme:

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

Why? Because we said that constructions of this form:

        let length = A
                in B

really were just another way of saying:

        (\length. B) A

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

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

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

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

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

        \lst. (isempty lst) zero (add one (length (extract-tail lst)))

We've defined all of `isempty`, `zero`, `add`, `one`, and `extract-tail` in earlier discussion. But what about `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 (... (extract-tail lst)))

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

        \lst. (isempty lst) zero (add one (
                \lst. (isempty lst) zero (add one (... (extract-tail lst)))
                        (extract-tail lst)))

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

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

                (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 `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 `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 <code>x<sub>n</sub></code> and `zero`, where <code>x<sub>n</sub></code> 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 <code>x<sub>n-1</sub></code> 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.  In a sense, the recursive
structure of the `length` operation is built into the data
structure we are using to represent the list.  The non-recursive
version of length exploits this embedding of the recursion into
the data type.

This is one of the themes of the course: using data structures to
encode the state of some recursive operation.  See discussions of the
[[zipper]] technique, and [[defunctionalization]].

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.

##Some functions require full-fledged recursive definitions##

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`.  (You may be thinking: but that
smallest-integer function is not a proper algorithm, since it is not
guaranteed to halt in any finite amount of time for every argument.
This is the famous [[!wikipedia Halting problem]].  But the fact that
an implementation may not terminate doesn't mean that such a function
isn't well-defined.  The point of interest here is that its definition
requires recursion in the function definition.)

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
        ...

Many 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.

##Using fixed-point combinators to define recursive functions##

###Fixed points###

In general, a **fixed point** of a function `f` is any value `x`
such that `f x` is equivalent to `x`. For example,
consider the squaring function `square` that maps natural numbers to their squares.
`square 2 = 4`, so `2` is not a fixed point.  But `square 1 = 1`, so `1` is a
fixed point of the squaring function.

There are many beautiful theorems guaranteeing the existence of a
fixed point for various classes of interesting functions.  For
instance, imainge that you are looking at a map of Manhattan, and you
are standing somewhere in Manhattan.  The the [[!wikipedia Brouwer
fixed-point theorem]] guarantees that there is a spot on the map that is
directly above the corresponding spot in Manhattan.  It's the spot
where the blue you-are-here dot should be.

Whether a function has a fixed point depends on the set of arguments
it is defined for.  For instance, consider the successor function `succ`
that maps each natural number to its successor.  If we limit our
attention to the natural numbers, then this function has no fixed
point.  (See the discussion below concerning a way of understanding
the successor function on which it does have a fixed point.)

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

        X <~~> f X

You should be able to immediately provide a fixed point of the
identity combinator I.  In fact, you should be able to provide a
whole bunch of distinct fixed points.

With a little thought, you should be able to provide a fixed point of
the false combinator, KI.  Here's how to find it: recall that KI
throws away its first argument, and always returns I.  Therefore, if
we give it I as an argument, it will throw away the argument, and
return I.  So KII ~~> I, which is all it takes for I to qualify as a
fixed point of KI.

What about K?  Does it have a fixed point?  You might not think so,
after trying on paper for a while.

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

Yes, as we've mentioned, 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.  We'll return to this point below.

###How fixed points help define recursive functions###

Recall our initial, abortive attempt above to define the `length` function in the lambda calculus. We said "What we really want to do is something like this:

        \list. if empty list then zero else add one (... (tail lst))

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

Imagine replacing the `...` with some function that computes the
length function.  Call that function `length`.  Then we have

        \list. if empty list then zero else add one (length (tail lst))

At this point, we have a definition of the length function, though
it's not complete, since we don't know what value to use for the
symbol `length`.  Technically, it has the status of an unbound
variable.

Imagine now binding the mysterious variable, and calling the resulting
function `h`:

        h := \length \list . if empty list then zero else add one (length (tail list))

Now we have no unbound variables, and we have complete non-recursive
definitions of each of the other symbols.

So `h` takes an argument, and returns a function that accurately
computes the length of a list---as long as the argument we supply is
already the length function we are trying to define.  (Dehydrated
water: to reconstitute, just add water!)

Here is where the discussion of fixed points becomes relevant.  Saying
that `h` is looking for an argument (call it `LEN`) that has the same
behavior as the result of applying `h` to `LEN` is just another way of
saying that we are looking for a fixed point for `h`.

    h LEN <~~> LEN

Replacing `h` with its definition, we have

    (\list . if empty list then zero else add one (LEN (tail list))) <~~> LEN

If we can find a value for `LEN` that satisfies this constraint, we'll
have a function we can use to compute the length of an arbitrary list.
All we have to do is find a fixed point for `h`.

The strategy we will present will turn out to be a general way of
finding a fixed point for any lambda term.

##Deriving Y, a fixed point combinator##

How shall we begin?  Well, we need to find an argument to supply to
`h`.  The argument has to be a function that computes the length of a
list.  The function `h` is *almost* a function that computes the
length of a list.  Let's try applying `h` to itself.  It won't quite
work, but examining the way in which it fails will lead to a solution.

    h h <~~> \list . if empty list then zero else 1 + h (tail list)

The problem is that in the subexpression `h (tail list)`, we've
applied `h` to a list, but `h` expects as its first argument the
length function.

So let's adjust h, calling the adjusted function H:

    H = \h \list . if empty list then zero else one plus ((h h) (tail list))

This is the key creative step.  Instead of applying `h` to a list, we
apply it first to itself.  After applying `h` to an argument, it's
ready to apply to a list, so we've solved the problem just noted.
We're not done yet, of course; we don't yet know what argument to give
to `H` that will behave in the desired way.

So let's reason about `H`.  What exactly is H expecting as its first
argument?  Based on the excerpt `(h h) (tail l)`, it appears that
`H`'s argument, `h`, should be a function that is ready to take itself
as an argument, and that returns a function that takes a list as an
argument.  `H` itself fits the bill:

    H H <~~> (\h \list . if empty list then zero else 1 + ((h h) (tail list))) H
        <~~> \list . if empty list then zero else 1 + ((H H) (tail list))
        == \list . if empty list then zero else 1 + ((\list . if empty list then zero else 1 + ((H H) (tail list))) (tail list))
        <~~> \list . if empty list then zero
                    else 1 + (if empty (tail list) then zero else 1 + ((H H) (tail (tail list))))

We're in business!

How does the recursion work?
We've defined `H` in such a way that `H H` turns out to be the length function.
In order to evaluate `H H`, we substitute `H` into the body of the
lambda term.  Inside the lambda term, once the substitution has
occurred, we are once again faced with evaluating `H H`.  And so on.

We've got the infinite regress we desired, defined in terms of a
finite lambda term with no undefined symbols.

Since `H H` turns out to be the length function, we can think of `H`
by itself as half of the length function (which is why we called it
`H`, of course).  Can you think up a recursion strategy that involves
"dividing" the recursive function into equal thirds `T`, such that the
length function <~~> T T T?

We've starting with a particular recursive definition, and arrived at
a fixed point for that definition.
What's the general recipe?

1.   Start with any recursive definition `h` that takes itself as an arg: `h := \fn ... fn ...`
2.   Next, define `H := \f . h (f f)` 
3.   Then compute `H H = ((\f . h (f f)) (\f . h (f f)))`
4.   That's the fixed point, the recursive function we're trying to define

So here is a general method for taking an arbitrary h-style recursive function
and returning a fixed point for that function:

     Y := \h. ((\f.h(ff))(\f.h(ff)))

Test:

    Yh == ((\f.h(ff))(\f.h(ff)))
       <~~> h((\f.h(ff))(\f.h(ff)))
       == h(Yh)

That is, Yh is a fixed point for h.

Works!

Let's do one more example to illustrate.  We'll do `K`, since we
wondered above whether it had a fixed point.

Before we begin, we can reason a bit about what the fixed point must
be like.  We're looking for a fixed point for `K`, i.e., `\xy.x`.  `K`
ignores its second argument.  That means that no matter what we give
`K` as its first argument, the result will ignore the next argument
(that is, `KX` ignores its first argument, no matter what `X` is).  So
if `KX <~~> X`, `X` had also better ignore its first argument.  But we
also have `KX == (\xy.x)X ~~> \y.X`.  This means that if `X` ignores
its first argument, then `\y.X` will ignore its first two arguments.
So once again, if `KX <~~> X`, `X` also had better ignore at least its
first two arguments.  Repeating this reasoning, we realize that `X`
must be a function that ignores an infinite series of arguments.  
Our expectation, then, is that our recipe for finding fixed points
will build us a function that somehow manages to ignore an infinite
series of arguments.

    h := \xy.x
    H := \f.h(ff) == \f.(\xy.x)(ff) ~~> \fy.ff
    H H := (\fy.ff)(\fy.ff) ~~> \y.(\fy.ff)(\fy.ff)

Let's check that it is in fact a fixed point:

    K(H H) == (\xy.x)((\fy.ff)(\fy.ff)
           ~~> \y.(\fy.ff)(\fy.ff)

Yep, `H H` and `K(H H)` both reduce to the same term.  

To see what this fixed point does, let's reduce it a bit more:

    H H == (\fy.ff)(\fy.ff)
        ~~> \y.(\fy.ff)(\fy.ff)
        ~~> \yy.(\fy.ff)(\fy.ff)
        ~~> \yyy.(\fy.ff)(\fy.ff)
    
Sure enough, this fixed point ignores an endless, infinite series of
arguments.  It's a write-only memory, a black hole.

Now that we have one fixed point, we can find others, for instance,

    (\fy.fff)(\fy.fff) 
    ~~> \y.(\fy.fff)(\fy.fff)(\fy.fff)
    ~~> \yy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)
    ~~> \yyy.(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)(\fy.fff)

Continuing in this way, you can now find an infinite number of fixed
points, all of which have the crucial property of ignoring an infinite
series of arguments.

##What is a fixed point for the successor function?##

As we've seen, the recipe just given for finding a fixed point worked
great for our `h`, which we wrote as a definition for the length
function.  But the recipe doesn't make any assumptions about the
internal structure of the function it works with.  That means it can
find a fixed point for literally any function whatsoever.

In particular, what could the fixed point for the
successor function possibly be like?

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 make-pair

who knows what we'd get back? Perhaps there's some non-number-representing 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.

Let's pick a way of defining the successor function and reason about it.
Here is one way that is compatible with the constraints given in
homework 2: `succ := \nfz.f(nfz)`.  This takes a Church
number, and returns the next Church number.  For instance, 

    succ 2 == succ (\fz.f(fz)) 
           == (\nfz.f(nfz)) (\fz.f(fz))
           ~~> \fz.f((\fz.f(fz))fz)
           ~~> \fz.f(f(fz))
           == 3

Using logic similar to the discussion above of the fixed point for K,
we can say that for any Church number argument to the successor
function, the result will be the next Church number.  Assume that
there is some Church number `n` that is a fixed point.  Then
`succ n <~~> n` (because `n` is a fixed point) and `succ n <~~> n + 1`
(since that's what the successor function does).  By the Church Rosser
theorem, `n <~~> n + 1`.  What kind of `n` could satisfy that
requirement?

Let's run the recipe:

    H := \f . succ (ff)
      == \f . (\nfz.f(nfz)) (ff)
      ~~> \h . (\nfz.f(nfz)) (hh)
      ~~> \hfz.f(hhfz)

    H H == (\hfz.f(hhfz)) (\hfz.f(hhfz))
        ~~> \fz.f((\hfz.f(hhfz))(\hfz.f(hhfz))fz)
        ~~> \fz.f(f((\hfz.f(hhfz))(\hfz.f(hhfz))fz))
        ~~> \fz.f(f(f((\hfz.f(hhfz))(\hfz.f(hhfz))fz))

We can see that the fixed point generates an endless series of `f`'s.
In terms of Church numbers, this is a way of representing infinity:
if the size of a Church number is the number `f`'s it contains, and
this Church number contains an unbounded number of `f`'s, then its
size is unbounded.  

We can also see how this candidate for infinity behaves with respect
to our other arithmetic operators.

    add 2 (HH) == (\mnfz.mf(nfz)) (\fz.f(fz)) (H H)
               ~~> \fz.(\fz.f(fz)) f ((HH)fz)
               ~~> \fz.\z.f(fz) ((HH)fz)
               ~~> \fz.f(f((HH)fz))
               ==  \fz.f(f(((\hfz.f(hhfz)) (\hfz.f(hhfz)))fz))
               ~~> \fz.f(f((\fz.f((\hfz.f(hhfz)) (\hfz.f(hhfz))))fz))
               ~~> \fz.f(f(f((\hfz.f(hhfz)) (\hfz.f(hhfz)))))

So `2 + (HH) <~~> (HH)`.  This is what we expect from arithmetic infinity.
You can check to see if `2 * (HH) <~~> (HH)`.

So our fixed point recipe has delivere a reasonable candidate for
arithmetic infinity. 

One (by now obvious) upshot is that 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."

###Varieties of fixed-point combinators###

OK, so how do we make use of this?

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:

<pre><code>&Theta;&prime; &equiv; (\u f. f (\n. u u f n)) (\u f. f (\n. u u f n))
Y&prime; &equiv; \f. (\u. f (\n. u u n)) (\u. f (\n. u u n))</code></pre>

<code>&Theta;&prime;</code> has the advantage that <code>f (&Theta;&prime; f)</code> really *reduces to* <code>&Theta;&prime; f</code>. Whereas <code>f (Y&prime; f)</code> is only *convertible with* <code>Y&prime; f</code>; 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 <code>&Theta;&prime;</code> to just `u u f`? And similarly for <code>Y&prime;</code>?

Indeed you can, getting the simpler:

<pre><code>&Theta; &equiv; (\u f. f (u u f)) (\u f. f (u u f))
Y &equiv; \f. (\u. f (u u)) (\u. f (u u))</code></pre>

I stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of <code>&Theta; (\self. BODY)</code> 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 <code>&Psi;</code> in:

<pre><code>&Psi; (\self. \n. self n)</code></pre>

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:

<pre><code>&Psi; (\self (\lst. (isempty lst) zero (add one (self (extract-tail lst))) ))</code></pre>

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-fixed.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 <code>&Psi;</code> 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.

##Application to the truth teller/liar paradoxes##

###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.

Curry originally called `Y` the paradoxical combinator, and discussed
it in connection with certain well-known paradoxes from the philosophy
literature.  The truth teller paradox 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:

<pre><code>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)))
&omega; &omega;
&Omega
</code></pre>

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 <code>&Omega;</code>, 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,
<code>&Omega;</code> 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.