`x`_{n}

and `0`, where `x`_{n}

is the last element of the list. This gives us `succ 0`, or `1`. That's the value we've accumulated "so far." Then we go apply the function `\x sofar. succ sofar` to the two arguments `x`_{n-1}

and the value `1` that we've accumulated "so far." This gives us `2`. 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
fold-based encoding of lists, and Church's encodings 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
definition of length, above, exploits this embedding of the recursion into
the data type.
This illustrates what will be one of the recurring themes of the course: using data structures to
encode the state of some recursive operation. See our discussions later this semester of the
[[zipper]] technique, and [[defunctionalization]].
As we've seen, it does take some ingenuity to define functions like `tail` or `pred` for our right-fold encoding of lists. 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 implementation 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 *n*th 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 natural number `x`
where `f x` is `true` (even if `f` itself is a function we do already know how to define).
Neither do the resources we've so far developed suffice to define the
[[!wikipedia Ackermann function]]. In OCaml:
let rec A = fun (m,n) ->
if m = 0 then n + 1
else if n = 0 then 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 mathematics, a **fixed point** of a function `f` is any value `ξ`
such that `f ξ` is equivalent to `ξ`. 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. (Can you think of another?)
There are many beautiful theorems guaranteeing the existence of a
fixed point for various classes of interesting functions. For
instance, imagine that you are looking at a map of Manhattan, and you
are standing somewhere in Manhattan. Then 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 on the map
where the blue you-are-here dot should go.
Whether a function has a fixed point depends on the domain 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* `ξ` such that:
ξ <~~> f ξ
This is a bit different than the general mathematical definition, in that here we're saying it is *terms* that are fixed points, not *values*. We like to think that some lambda terms represent values, such as our term `\f z. z` representing the numerical value zero (and also the truth-value false, and also the empty list... on the other hand, we never did explicitly agree that those three values are all the same thing, did we?). But some terms in the Lambda Calculus don't even have a normal form. We don't want to count them as *values*. Yet the way we're proposing to use the notion of a fixed point here, they too are allowed to be fixed points, and to have fixed points of their own.
Note that `M <~~> N` doesn't entail that `M` and `N` have a normal form (though if they do, they will have the same normal form). It just requires that there be some term that they both reduce to. It may be that *that* term itself never stops being reducible.
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* lambda term has
a fixed point. Even bare variables like `x`! In fact, they 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.
(As we mentioned, even the formula that you're using the define
the successor function will have a fixed point. Isn't that weird? There's some `ξ` such that it is equivalent to `succ ξ`?
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:
> \xs. (empty? xs) 0 (succ ((...) (tail xs)))
> where this very same formula occupies the `...` position...
Imagine replacing the `...` with some expression `LENGTH` that computes the
length function. Then we have
\xs. (empty? xs) 0 (succ (LENGTH (tail xs)))
(More generally, we might have some lambda term `Φ[...SELF...]` where we
want the contained `SELF` to refer to that very lambda term `Φ[...SELF...]`.)
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
term `h`:
h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
(More generally, convert `Φ[...SELF...]` to `\body. Φ[...body...]`, where
the variable `body` wants to be bound to the very lambda term that is that abstract's body.)
Now we have no unbound variables, and we have complete non-recursive
definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
So `h` takes a `length` 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 `LENGTH`) that has the same
behavior as the result of applying `h` to `LENGTH` is just another way of
saying that we are looking for a fixed point for `h`:
h LENGTH <~~> LENGTH
Replacing `h` with its definition, we have
(\xs. (empty? xs) 0 (succ (LENGTH (tail xs)))) <~~> LENGTH
If we can find a value for `LENGTH` 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 <~~> \xs. (empty? xs) 0 (succ (h (tail xs)))
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`. (We'll use `u` as the variable
that expects to be bound to the as-yet-*unknown* argument, rather than `length`. This will make it easier
to discuss generalizations of this strategy.)
h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
H ≡ \u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))
(We'll discuss the general case, when you're starting from `\body. Φ[...body...]` rather than this specific `h`, below.)
Shifting to `H` is the key creative step. Instead of applying `u` to a list, as happened
when we self-applied `h`, `H` applies its argument `u` first to *itself*: `u u`.
After `u` gets an argument, the *result* is ready to apply to a list, so we've solved the problem noted above with `h (tail list)`.
We're not done yet, of course; we don't yet know what argument `u` 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 `(u u) (tail xs)`, it appears that
`H`'s argument, `u`, 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 <~~> (\u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))) H
<~~> \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
<~~> \xs. (empty? xs) 0 (succ ((
\xs. (empty? xs) 0 (succ ((H H) (tail xs)))
) (tail xs)))
<~~> \xs. (empty? xs) 0 (succ (
(empty? (tail xs)) 0 (succ ((H H) (tail (tail xs))))
))
<~~> \xs. (empty? xs) 0 (succ (
(empty? (tail xs)) 0 (succ (
\xs. (empty? xs) 0 (succ ((H H) (tail xs)))
) (tail (tail xs))))
))
<~~> \xs. (empty? xs) 0 (succ (
(empty? (tail xs)) 0 (succ (
(empty? (tail (tail xs))) 0 (succ ((H H) (tail (tail (tail xs)))))
))
))
<~~> ...
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.
That is, `H H` is the `LENGTH` we were looking for.
In order to evaluate `H H`, we substitute `H` into the body of the
lambda term `H`. Inside that lambda term, once the substitution has
occurred, we are once again faced with evaluating `H H`. And so on.
We've got the (potentially) 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). (Thought exercise: 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 a formula `h` that takes the recursive function you're seeking as an argument: `h ≡ \length. ...length...` (This is what we also called `\body. Φ[...body...]`.)
2. Next, define `H ≡ \u. h (u u)`
3. Then compute `H H ≡ ((\u. h (u u)) (\u. h (u u)))`
4. That's the fixed point of `h`, the recursive function you're seeking.
Expressed in terms of a single formula, here is this method for taking an arbitrary `h`-style term and returning
that term's fixed point, which will be the recursive function that term expects as an argument:
Y ≡ \h. (\u. h (u u)) (\u. h (u u))
Let's test that `Y h` will indeed be `h`'s fixed point:
Y h ≡ (\h. (\u. h (u u)) (\u. h (u u))) h
~~> (\u. h (u u)) (\u. h (u u))
~~> h ((\u. h (u u)) (\u. h (u u)))
But the argument of `h` in the last line is just the same as the second line, which `<~~> Y h`. So the last line `<~~> h (Y h)`. In other words, `Y h <~~> h (Y h)`. So by definition, `Y h` is a fixed point for `h`.
Works!
## A fixed point for K? ##
Let's do one more example to illustrate. We'll do `K` (boolean true), 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., `\x y. x`. The term `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, `K ξ` ignores its first argument, no matter what `ξ` is). So
if `K ξ <~~> ξ`, `ξ` had also better ignore its first argument. But we
also have `K ξ ≡ (\x y. x) ξ ~~> \y. ξ`. This means that if `ξ` ignores
its first argument, then `K ξ <~~> \y. ξ` will ignore its first two arguments.
So once again, if `K ξ <~~> ξ`, `ξ` also had better ignore (at least) its
first two arguments. Repeating this reasoning, we realize that `ξ` here
must be a function that ignores as many arguments as you give it.
Our expectation, then, is that our recipe for finding fixed points
will build us a term that somehow manages to ignore arbitrarily many arguments.
h ≡ \x y. x
H ≡ \u. h (u u)
≡ \u. (\x y. x) (u u)
~~> \u. \y. u u
H H ~~> (\u y. u u) (\u y. u u)
~~> \y. (\u y. u u) (\u y. u u)
Let's check that it is in fact a fixed point for `K`:
K (H H) ~~> (\x y. x) ((\u y. u u) (\u y. u u))
~~> \y. (\u y. u u) (\u y. u u)
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 ~~> (\u y. u u) (\u y. u u)
~~> \y. (\u y. u u) (\u y. u u)
~~> \y. \y. (\u y. u u) (\u y. u u)
~~> \y. \y. \y. (\u y. u u) (\u y. u u)
Sure enough, this fixed point ignores an endless, arbitrarily-long 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,
(\uy.[uu]u) (\uy.uuu)
~~> \y. [(\uy.uuu) (\uy.uuu)] (\uy.uuu)
~~> \y. [\y. (\uy.uuu) (\uy.uuu) (\uy.uuu)] (\uy.uuu)
~~> \yyy. (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu)
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.
## A fixed point for succ? ##
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 term it works with. That means it can
find a fixed point for literally any lambda term whatsoever.
In particular, what could the fixed point for (our encoding of) the
successor function possibly be like?
Well, you might think, only some of the formulas that we might give to `succ` as arguments would really represent numbers. If we said something like:
succ pair
who knows what we'd get back? Perhaps there's some non-number-representing formula such that when we feed it to `succ` 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 encoded the successor function.
One (by now obvious) upshot is that the recipes that enable us to name
fixed points for any given formula `h` aren't *guaranteed* to give us
*terminating, normalizing* fixed points. They might give us formulas `ξ` such that
neither `ξ` nor `h ξ` 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 list
functions on [[the encodings we discussed last week|week3_lists#other-lists]], where it
wasn't always clear how to force the computation to "keep going."
## Varieties of fixed-point combinators ##
Many fixed-point combinators have been discovered. (And as we've seen, some
fixed-point combinators give us models for building infinitely many
more, non-equivalent fixed-point combinators.)
Two of the simplest:
Θ′ ≡ (\u h. h (\n. u u h n)) (\u h. h (\n. u u h n))
Y′ ≡ \h. (\u. h (\n. u u n)) (\u. h (\n. u u n))
Applying either of these to a term `h` gives a fixed point `ξ` for `h`, meaning that `h ξ` <~~> `ξ`. The combinator `Θ′` has the advantage that `h (Θ′ h)` really *reduces to* `Θ′ h`. Whereas `h (Y′ h)` is only *convertible with* `Y′ h`; 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 h n` inside `Θ′` to just `u u h`? And similarly for `Y′`?
Indeed you can, getting the simpler:
Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
Y ≡ \h. (\u. h (u u)) (\u. h (u u))
We stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\body. BODY)` and `Y (\body. BODY)` won't terminate. But evaluation of the eta-unreduced primed versions may.
Of course, if you define your `\body. BODY` stupidly, your formula won't terminate, no matter what fixed point combinator you use. For example, let `Ψ` be any fixed point combinator in:
Ψ (\body. \n. body n)
When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
(\n. BODY n) M
where `BODY` is equivalent to the very formula `\n. BODY n` that contains it. So the evaluation will proceed:
(\n. BODY n) M ~~>
BODY M <~~>
(\n. BODY n) M ~~>
BODY M <~~>
...
You've written an infinite loop!
However, when we evaluate the application of our:
Ψ (\body. (\xs. (empty? xs) 0 (succ (body (tail xs))) ))
to some list, 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:
\xs. (empty? xs) 0 (succ (body (tail xs)))
to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the encodings we've been using so far 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 `0`. So the recursion eventually bottoms out in a base value.
## Fixed-point Combinators Are a Bit Intoxicating ##
[[tatto|/images/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.
We used `Ψ` above to stand in for an arbitrary fixed-point combinator. We 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
## Sink: 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., `\y n. y`), it
returns itself (a copy of `sink`); if the argument is boolean false
(`\y n. n`), 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
Evidently, then, `sink true <~~> sink`. So we want `sink` to be the fixed point
of `\sink. \b. b sink I`. That is, `sink ≡ Y (\sb.bsI)`:
1. sink false
2. Y (\sb.bsI) false
3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) false
4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) false
5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
6. (\b. b [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] I) false
7. false [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] 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 (\sb.bsI) true false
3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) true false
4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) true false
5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] true false
6. (\b. b [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I) true false
7. true [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I false
8. [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] 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 when 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,
fact ≡ Y (\fact n. (zero? n) 1 (fact (pred n)))
If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.