`x`_{n}

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

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

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 `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 `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,"
+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
-version of length exploits this embedding of the recursion into
+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
-encode the state of some recursive operation. See discussions of the
+encode the state of some recursive operation. See our discussions later this semester of the
[[zipper]] technique, and [[defunctionalization]].
-As we said before, it does take some ingenuity to define functions like `tail` or `predecessor` 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 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.
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.
@@ -159,28 +162,22 @@ With sufficient ingenuity, a great many functions can be defined in the same way
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.)
+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]]:
+[[!wikipedia Ackermann function]]. In OCaml:
- A(m,n) =
- | when m == 0 -> n + 1
- | else when n == 0 -> A(m-1,1)
- | else -> A(m-1, A(m,n-1))
+ 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
+ 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.
@@ -191,54 +188,59 @@ But functions like the Ackermann function require us to develop a more general t
###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,
+In mathematics, a **fixed point** of a function `f` is any value `Î¾`
+such that `f Î¾` is equivalent to `Î¾`. 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.
+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, imainge that you are looking at a map of Manhattan, and you
-are standing somewhere in Manhattan. The the [[!wikipedia Brouwer
+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.
-Whether a function has a fixed point depends on the set of arguments
+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.)
+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:
+In the Lambda Calculus, we say a fixed point of a term `f` is any term `Î¾` such that:
- X <~~> f X
+ Î¾ <~~> f Î¾
+
+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.
+
+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
+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.
+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,
+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
+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.
-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?
+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? There's some `Î¾` such that it is equivalent to `succ Î¾`?
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:
@@ -247,20 +249,20 @@ Recall our initial, abortive attempt above to define the `length` function in th
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
+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. if empty? xs then 0 else 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
-symbol `length`. Technically, it has the status of an unbound
+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 \xs. if empty? xs then 0 else succ (length (tail xs))
+ h â¡ \length \xs. if empty? xs then 0 else succ (length (tail xs))
Now we have no unbound variables, and we have complete non-recursive
definitions of each of the other symbols.
@@ -271,23 +273,24 @@ 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
+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 LEN <~~> LEN
+ h LENGTH <~~> LENGTH
Replacing `h` with its definition, we have
- (\xs. if empty? xs then 0 else succ (LEN (tail xs))) <~~> LEN
+ (\xs. if empty? xs then 0 else succ (LENGTH (tail xs))) <~~> LENGTH
-If we can find a value for `LEN` that satisfies this constraint, we'll
+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
@@ -320,7 +323,7 @@ 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))
+ â¡ \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))))
@@ -332,7 +335,7 @@ 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
+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`
@@ -345,21 +348,21 @@ 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)))`
+1. Start with any recursive definition `h` that takes itself as an arg: `h â¡ \fn ... fn ...`
+2. Next, define `H â¡ \f . h (f f)`
+3. Then compute `H H â¡ ((\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)))
+ Y â¡ \h. ((\f.h(ff))(\f.h(ff)))
Test:
- Yh == ((\f.h(ff))(\f.h(ff)))
+ Yh â¡ ((\f.h(ff))(\f.h(ff)))
<~~> h((\f.h(ff))(\f.h(ff)))
- == h(Yh)
+ â¡ h(Yh)
That is, Yh is a fixed point for h.
@@ -374,7 +377,7 @@ 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 â¡ (\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`
@@ -383,20 +386,20 @@ 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)
+ h â¡ \xy.x
+ H â¡ \f.h(ff) â¡ \f.(\xy.x)(ff) ~~> \fy.ff
+ H H â¡ (\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)
+ K(H H) â¡ (\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)
+ H H â¡ (\fy.ff)(\fy.ff)
~~> \y.(\fy.ff)(\fy.ff)
~~> \yy.(\fy.ff)(\fy.ff)
~~> \yyy.(\fy.ff)(\fy.ff)
@@ -426,13 +429,13 @@ 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:
+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:
- successor make-pair
+ succ 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.
+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 successor function.
+Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the succ 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
@@ -460,7 +463,7 @@ Two of the simplest:
Îâ² â¡ (\u f. f (\n. u u f n)) (\u f. f (\n. u u f n))
Yâ² â¡ \f. (\u. f (\n. u u n)) (\u. f (\n. u u n))
-`Îâ²` has the advantage that `f (Îâ² f)` really *reduces to* `Îâ² f`. Whereas `f (Yâ² f)` is only *convertible with* `Yâ² 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 `f` gives a fixed point `Î¾` for `f`, meaning that `f Î¾` <~~> `Î¾`. `Îâ²` has the advantage that `f (Îâ² f)` really *reduces to* `Îâ² f`. Whereas `f (Yâ² f)` is only *convertible with* `Yâ² f`; that is, there's a common formula they both reduce to. For most purposes, though, either will do.
You may notice that both of these formulas have eta-redexes inside them: why can't we simplify the two `\n. u u f n` inside `Îâ²` to just `u u f`? And similarly for `Yâ²`?
@@ -482,9 +485,9 @@ When you try to evaluate the application of that to some argument `M`, it's goin
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 ~~>
+ self M <~~>
(\n. self n) M ~~>
- self M ~~>
+ self M <~~>
...
You've written an infinite loop!
@@ -565,7 +568,7 @@ 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
+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
@@ -590,18 +593,17 @@ 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
+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. zero? n 1 (fac (predecessor n)))
+ fact = Y (\fact n. zero? n 1 (fact (predecessor n)))
-If `n` is 0, `fac` reduces to 1, without computing the recursive case.
+If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.
-Curry originally called `Y` the paradoxical combinator, and discussed
+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).
+literature. The truth-teller paradox has the flavor of a recursive
+function without a base case:
(1) This sentence is true.
@@ -617,6 +619,8 @@ 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
@@ -635,10 +639,10 @@ sentence in which it occurs, the sentence denotes a fixed point for
the identity function. Here's a fixed point for the identity
function:
- Y I
- (\f. (\h. f (h h)) (\h. f (h h))) I
- (\h. I (h h)) (\h. I (h h)))
- (\h. (h h)) (\h. (h h)))
+ Y I â¡
+ (\f. (\h. f (h h)) (\h. f (h h))) I ~~>
+ (\h. I (h h)) (\h. I (h h))) ~~>
+ (\h. (h h)) (\h. (h h))) â¡
Ï Ï
Î©
@@ -677,18 +681,18 @@ rather than recursive functions.
You should be cautious about feeling too comfortable with
these results. Thinking again of the truth-teller paradox, yes,
`Î©` 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).
+some 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, versus other fixed point combinators).
But one could ask: look, literally every formula is a fixed point for
`I`, since
X <~~> I X
-for any choice of X whatsoever.
+for any choice of `X` whatsoever.
-So the Y combinator is only guaranteed to give us one fixed point out
+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