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"
)
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?
\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:
- \xs. if empty? xs then 0 else succ (... (tail xs))
+ \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
Imagine now binding the mysterious variable, and calling the resulting
function `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.
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
+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!)
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
So let's adjust h, calling the adjusted function H:
- H = \h \xs. if empty? xs then 0 else 1 + ((h h) (tail xs))
+ H = \h \xs. (empty? xs) 0 (succ ((h h) (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
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!
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)))`
+1. Start with any recursive definition `h` that takes itself as an arg: `h ≡ \self ... self ...`
+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, 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)))
+ Y ≡ \h. (\u. h (u u)) (\u. h (u u))
Test:
- 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., `\xy.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
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)
+ 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:
- 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
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.uuu)(\uy.uuu)
+ ~~> \y.(\uy.uuu)(\uy.uuu)(\uy.uuu)
+ ~~> \yy.(\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
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.
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
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 (predecessor 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`,
+(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)))))
+ (\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
+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
(\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the