-[[!toc levels=2]]
-
-~~~~
-
-**Chris:** I'll be working on this page heavily until 11--11:30 or so. Sorry not to do it last night, I crashed.
-
-ξ
-Θ′
-≡
-~~~~
+[[!toc levels=3]]
#Recursion: fixed points in the Lambda Calculus#
In OCaml, you'd define that like this:
let rec length = fun xs ->
- if xs = [] then 0 else 1 + length (tail xs)
+ if xs = [] then 0
+ else 1 + length (List.tl xs)
in ... (* here you go on to use the function "length" *)
In Scheme you'd define it like this:
- (letrec [(length (lambda (xs)
- (if (null? xs) 0
- (+ 1 (length (cdr xs))) )))]
+ (letrec [(length (lambda (xs)
+ (if (null? xs) 0
+ (+ 1 (length (cdr xs))) )))]
... ; here you go on to use the function "length"
)
1. `null?` is Scheme's way of saying `empty?`. That is, `(null? xs)` returns true (which Scheme writes as `#t`) iff `xs` 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 a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.)
+2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.) `List.tl` is the function that gets the tail of an OCaml list.
3. We 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 a lot like `let` expressions, 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 be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
+Answer: These work a lot like `let` expressions, 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 be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
let rec length = fun xs ->
- if xs = [] then 0 else 1 + length (tail xs)
+ if xs = [] then 0
+ else 1 + length (List.tl xs)
in length [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 xs ->
- if xs = [] then 0 else 1 + length (tail xs)
+ if xs = [] then 0
+ else 1 + length (List.tl xs)
in length [20; 30]
(* fails with error "Unbound value length" *)
let length = fun xs -> 99
in let length = fun xs ->
- if xs = [] then 0 else 1 + length (tail xs)
+ if xs = [] then 0
+ else 1 + length (List.tl xs)
in length [20; 30]
(* evaluates to 1 + 99 *)
-Here the use of `length` in `1 + length (tail xs)` can clearly be seen to be bound by the outermost `let`.
+Here the use of `length` in `1 + length (List.tl xs)` 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?
What we really want to do is something like this:
- \xs. (empty? xs) 0 (succ (... (tail xs)))
+ \xs. (empty? xs) 0 (succ ((...) (tail xs)))
where this very same formula occupies the `...` position:
- \xs. (empty? xs) 0 (succ (\xs. (empty? xs) 0 (succ (... (tail xs)))
- (tail xs)))
+ \xs. (empty? xs) 0 (succ (
+ \xs. (empty? xs) 0 (succ ((...) (tail xs)))
+ ) (tail xs)))
but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
\xs. xs (\x sofar. succ sofar) 0
- What's happening here? We start with the value `0`, then we apply the function `\x sofar. succ sofar` to the two arguments <code>x<sub>n</sub></code> and `0`, where <code>x<sub>n</sub></code> 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 <code>x<sub>n-1</sub></code> and the value `1` 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.
+ What's happening here? We start with the value `0`, then we apply the function `\x sofar. succ sofar` to the two arguments <code>x<sub>n</sub></code> and `0`, where <code>x<sub>n</sub></code> 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 <code>x<sub>n-1</sub></code> 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
+ 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
definition of length, above, exploits this embedding of the recursion into
the data type.
-This is one of the themes of the course: using data structures to
+ 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 these encodings. 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 encodings of lists and numbers.
+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 nth term in the Fibonacci series is a bit more difficult, but also achievable.
+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##
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.
+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##
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
-where the blue you-are-here dot should be.
+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`
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:
+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. But 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.
+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.
+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
fixed points. And we don't just know that they exist: for any given
formula, we can explicit define many of them.
-As we've mentioned, even the formula that you're using the define
+(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.
+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:
+Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said:
- \xs. if empty? xs then 0 else succ (... (tail xs))
+> What we really want to do is something like this:
-where this very same formula occupies the `...` position."
+> \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. if empty? xs then 0 else succ (LENGTH (tail xs))
+ \xs. (empty? xs) 0 (succ (LENGTH (tail xs)))
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
variable.
Imagine now binding the mysterious variable, and calling the resulting
-function `h`:
+term `h`:
- h ≡ \length \xs. if empty? xs then 0 else succ (length (tail xs))
+ h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
Now we have no unbound variables, and we have complete non-recursive
-definitions of each of the other symbols.
+definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
-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
+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`.
+saying that we are looking for a fixed point for `h`:
h LENGTH <~~> LENGTH
Replacing `h` with its definition, we have
- (\xs. if empty? xs then 0 else succ (LENGTH (tail xs))) <~~> LENGTH
+ (\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.
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. if empty? xs then 0 else 1 + h (tail xs)
+ 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:
+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 = \h \xs. if empty? xs then 0 else 1 + ((h h) (tail xs))
+ h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
+ H ≡ \u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))
-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
+This 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 `(h h) (tail l)`, it appears that
-`H`'s argument, `h`, should be a function that is ready to take itself
+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 <~~> (\h \xs. if empty? xs then 0 else 1 + ((h h) (tail xs))) H
- <~~> \xs. if empty? xs then 0 else 1 + ((H H) (tail xs))
- ≡ \xs. if empty? xs then 0 else 1 + ((\xs. if empty? xs then 0 else 1 + ((H H) (tail xs))) (tail xs))
- <~~> \xs. if empty? xs then 0
- else 1 + (if empty? (tail xs) then 0 else 1 + ((H H) (tail (tail xs))))
+ 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)))) ))
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. Inside the lambda term, once the substitution has
+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
+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). Can you think up a recursion strategy that involves
+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?
+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
+1. Start with a formula `h` that takes the recursive function you're seeking as an argument: `h ≡ \length. ... length ...`
+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.
-So here is a general method for taking an arbitrary h-style recursive function
-and returning a fixed point for that function:
+Expressed in terms of a single formula, here is this method for taking an arbitrary `h`-style term and returning
+the recursive function that term expects as an argument, which as we've seen will be the `h`-term's fixed point:
- Y ≡ \h. ((\f.h(ff))(\f.h(ff)))
+ Y ≡ \h. (\u. h (u u)) (\u. h (u u))
-Test:
+Let's test that `Y h` will indeed be `h`'s fixed point:
- Yh ≡ ((\f.h(ff))(\f.h(ff)))
- <~~> h((\f.h(ff))(\f.h(ff)))
- ≡ h(Yh)
+ 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)))
-That is, Yh is a fixed point for h.
+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!
+###Coming at it another way###
+
+TODO
+
+
+##A fixed point for K?##
+
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`
+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, `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
+also have `KX ≡ (\x y. 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
+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.
+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 function that somehow manages to ignore an infinite
-series of arguments.
+will build us a term that somehow manages to ignore arbitrarily many 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)
+ H ≡ \u.h(uu) ≡ \u.(\xy.x)(uu) ~~> \uy.uu
+ H H ≡ (\uy.uu)(\uy.uu) ~~> \y.(\uy.uu)(\uy.uu)
-Let's check that it is in fact a fixed point:
+Let's check that it is in fact a fixed point for `K`:
- K(H H) ≡ (\xy.x)((\fy.ff)(\fy.ff)
- ~~> \y.(\fy.ff)(\fy.ff)
+ K(H H) ≡ (\xy.x)((\uy.uu)(\uy.uu))
+ ~~> \y.(\uy.uu)(\uy.uu)
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)
+ H H ≡ (\uy.uu)(\uy.uu)
+ ~~> \y.(\uy.uu)(\uy.uu)
+ ~~> \yy.(\uy.uu)(\uy.uu)
+ ~~> \yyy.(\uy.uu)(\uy.uu)
-Sure enough, this fixed point ignores an endless, infinite series of
+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,
- (\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)
+ (\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
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.
+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 the
+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 the `succ` as arguments would really represent numbers. If we said something 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 implemented the succ function.
+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 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
+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 arithmetic and list
-functions on the "version 1" and "version 2" encodings, where it
+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###
OK, so how do we make use of this?
-Many fixed-point combinators have been discovered. (And some
+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 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′ ≡ (\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 `f` gives a fixed point `ξ` for `f`, meaning that `f ξ` <~~> `ξ`. `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.
+Applying either of these to a term `h` gives a fixed point `ξ` for `h`, meaning that `h ξ` <~~> `ξ`. `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 f n` inside `Θ′` to just `u u f`? And similarly for `Y′`?
+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 f. f (u u f)) (\u f. f (u u f))
- Y ≡ \f. (\u. f (u u)) (\u. f (u u))
+ Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
+ Y ≡ \h. (\u. h (u u)) (\u. h (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.
+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)` 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:
+Of course, if you define your `\body. 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)
+ 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. self n) M
+ (\n. body n) M
-where `self` is equivalent to the very formula `\n. self n` that contains it. So the evaluation will proceed:
+where `body` is equivalent to the very formula `\n. body n` that contains it. So the evaluation will proceed:
- (\n. self n) M ~~>
- self M <~~>
- (\n. self n) M ~~>
- self M <~~>
+ (\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 (\self (\xs. (empty? xs) 0 (succ (self (tail xs))) ))
+ Ψ (\body. (\xs. (empty? xs) 0 (succ (body (tail xs))) ))
-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:
+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 (self (tail xs)))
+ \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'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 `0`. So the recursion eventually bottoms out in a base value.
+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.
-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.
+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:
sink true true false ~~> I
sink true true true false ~~> I
-So we make `sink = Y (\f b. b f I)`:
+So we make `sink = Y (\s b. b s 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
+ 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
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
+ 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.
+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
+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
That's about as simple as recursion gets.
+<!-- TODO Perhaps move rest to new document? -->
+
##Application to the truth teller/liar paradoxes##
###Base cases, and their lack###
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 (predecessor n)))
+ fact ≡ Y (\fact n. (zero? n) 1 (fact (pred n)))
If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.
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))) ≡
+ (\h. (\u. h (u u)) (\u. h (u u))) I ~~>
+ (\u. I (u u)) (\u. I (u u))) ~~>
+ (\u. (u u)) (\u. (u u))) ≡
ω ω
Ω
(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
+(2) in which that noun phrase occurs, (2) will denote a fixed point for `\f. neg f`,
+or `\f y n. f n y`, 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)))))
+ (\h. (\u. h (u u)) (\u. h (u u))) C
+ (\u. C (u u)) (\u. C (u u)))
+ C ((\u. C (u u)) (\u. C (u u)))
+ C (C ((\u. C (u u)) (\u. C (u u))))
+ C (C (C ((\u. C (u u)) (\u. C (u u)))))
...
And infinite sequence of `C`s, each one negating the remainder of the
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 `0` as a fixed point,
-since `0 * 0 = 0`, and `1` as a fixed point, since `1 * 1 = 1`, but `Y
+of infinitely many --- and not always the intuitively most useful
+one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,
+since `square 0 <~~> 0`, and `1` as a fixed point, since `square 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
[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
pronouns denote (you guessed it!) identity functions...
+<!-- Jim says: haven't made clear how we got from the self-referential (3) to I. -->
+
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.