XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=week3.mdwn;h=39e472bf9a644c1bdac774790ed134f26ab7cf31;hp=c9e26757dfac9cac8e94e7b8391f2045f8ed0cb0;hb=d9ef81bf6980969f16aebfed4b6fda9c3c5463bf;hpb=410a9889c779d960e4570c056833f7b3ba12a94a
diff git a/week3.mdwn b/week3.mdwn
index c9e26757..39e472bf 100644
 a/week3.mdwn
+++ b/week3.mdwn
@@ 1,3 +1,12 @@
+[[!toc]]
+
+##More on evaluation strategies##
+
+Here are notes on [[evaluation order]] that make the choice of which
+lambda to reduce next the selection of a route through a network of
+links.
+
+
##Computing the length of a list##
How could we compute the length of a list? Without worrying yet about what lambdacalculus implementation we're using for the list, the basic idea would be to define this recursively:
@@ 99,6 +108,34 @@ where this very same formula occupies the `...` position:
but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
+[At this point, some of you will recall the discussion in the first
+class concerning the conception of functions as sets of ordered pairs.
+The problem, as you will recall, was that in the untyped lambda
+calculus, we wanted a function to be capable of taking itself as an
+argument. For instance, we wanted to be able to apply the identity
+function to itself. And since the identity function always returns
+its argument unchanged, the value it should return in that case is
+itself:
+
+ (\x.x)(\x.x) ~~> (\x.x)
+
+If we conceive of a function as a set of ordered pairs, we would start
+off like this:
+
+ 1 > 1
+ 2 > 2
+ 3 > 3
+ ...
+ [1 > 1, 2 > 2, 3 > 3, ..., [1 > 1, 2 > 2, 3 > 3, ...,
+
+Eventually, we would get to the point where we want to say what the
+identity function itself gets mapped to. But in order to say that, we
+need to write down the identity function in the argument position as a
+set of ordered pairs. The need to insert a copy of the entire
+function definition inside of a copy of the entire function definition
+inside of... is the same problem as the need to insert a complete
+graph of the identity function inside of the graph for the identity function.]
+
So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?
1. OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.
@@ 126,7 +163,10 @@ 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. The simplest function that *simply cannot* be defined using the resources we've so far developed is the Ackermann function:
+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`.
+
+Neither do the resources we've so far developed suffice to define the
+[[!wikipedia Ackermann function]]:
A(m,n) =
 when m == 0 > n + 1
@@ 134,9 +174,9 @@ Some computable functions are just not definable in this way. The simplest funct
 else > A(m1, A(m,n1))
A(0,y) = y+1
 A(1,y) = y+2
 A(2,y) = 2y + 3
 A(3,y) = 2^(y+3) 3
+ 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
...
@@ 146,11 +186,91 @@ But functions like the Ackermann function require us to develop a more general t
##How to do recursion with lowercase omega##
[TODO]
+Recall our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
+
+ \lst. (isempty lst) zero (add one (... (extracttail lst)))
+
+where this very same formula occupies the `...` position."
+
+We are not going to exactly that, at least not yet. But we are going to do something close to it.
+
+Consider a formula of the following form (don't worry yet about exactly how we'll fill the `...`s):
+
+ \h \lst. (isempty lst) zero (add one (... (extracttail lst)))
+
+Call that formula `H`. Now what would happen if we applied `H` to itself? Then we'd get back:
+
+ \lst. (isempty lst) zero (add one (... (extracttail lst)))
+
+where any occurrences of `h` inside the `...` were substituted with `H`. Call this `F`. `F` looks pretty close to what we're after: a function that takes a list and returns zero if it's empty, and so on. And `F` is the result of applying `H` to itself. But now inside `F`, the occurrences of `h` are substituted with the very formula `H` we started with. So if we want to get `F` again, all we have to do is apply `h` to itselfsince as we said, the selfapplication of `H` is how we created `F` in the first place.
+
+So, the way `F` should be completed is:
+
+ \lst. (isempty lst) zero (add one ((h h) (extracttail lst)))
+
+and our original `H` is:
+
+ \h \lst. (isempty lst) zero (add one ((h h) (extracttail lst)))
+
+The selfapplication of `H` will give us `F` with `H` substituted in for its free variable `h`.
+
+Instead of writing out a long formula twice, we could write:
+
+ (\x. x x) LONGFORMULA
+
+and the initial `(\x. x x)` is just what we earlier called the ω
combinator (lowercase omega, not the nonterminating Ω
). So the selfapplication of `H` can be written:
+
+ω (\h \lst. (isempty lst) zero (add one ((h h) (extracttail lst))))
+
+
+and this will indeed implement the recursive function we couldn't earlier figure out how to define.
+
+In broad brushstrokes, `H` is half of the `get_length` function we're seeking, and `H` has the form:
+
+ \h otherarguments. ... (h h) ...
+
+We get the whole `get_length` function by applying `H` to itself. Then `h` is replaced by the half `H`, and when we later apply `h` to itself, we recreate the whole `get_length` again.
+
+##Neat! Can I make it easier to use?##
+
+Suppose you wanted to wrap this up in a pretty interface, so that the programmer didn't need to write `(h h)` but could just write `g` for some function `g`. How could you do it?
+
+Now the `F`like expression we'd be aiming forcall it `F*`would look like this:
+
+ \lst. (isempty lst) zero (add one (g (extracttail lst)))
+
+or, abbreviating:
+
+ \lst. ...g...
+
+Here we have just a single `g` instead of `(h h)`. We'd want `F*` to be the result of selfapplying some `H*`, and then binding to `g` that very selfapplication of `H*`. We'd get that if `H*` had the form:
+
+ \h. (\g lst. ...g...) (h h)
+
+The selfapplication of `H*` would be:
+
+ (\h. (\g lst. ...g...) (h h)) (\h. (\g lst. ...g...) (h h))
+
+or:
+
+ (\f. (\h. f (h h)) (\h. f (h h))) (\g lst. ...g...)
+
+The lefthand side of this is known as **the Ycombinator** and so this could be written more compactly as:
+
+ Y (\g lst. ...g...)
+
+or, replacing the abbreviated bits:
+
+ Y (\g lst. (isempty lst) zero (add one (g (extracttail lst))))
+
+So this is another way to implement the recursive function we couldn't earlier figure out how to define.
+
##Generalizing##
In general, a **fixed point** of a function f is a value *x* such that fx is equivalent to *x*. For example, what is a fixed point of the function from natural numbers to their squares? What is a fixed point of the successor function?
+Let's step back and fill in some theory to help us understand why these tricks work.
+
+In general, we call a **fixed point** of a function f any value *x* such that f x is equivalent to *x*. For example, what is a fixed point of the function from natural numbers to their squares? What is a fixed point of the successor function?
In the lambda calculus, we say a fixed point of an expression `f` is any formula `X` such that:
@@ 158,6 +278,8 @@ In the lambda calculus, we say a fixed point of an expression `f` is any formula
What is a fixed point of the identity combinator I?
+What is a fixed point of the false combinator, KI?
+
It's a theorem of the lambda calculus that every formula has a fixed point. In fact, it will have infinitely many, nonequivalent fixed points. And we don't just know that they exist: for any given formula, we can name many of them.
Yes, even the formula that you're using the define the successor function will have a fixed point. Isn't that weird? Think about how it might be true.
@@ 170,39 +292,39 @@ who knows what we'd get back? Perhaps there's some nonnumberrepresenting formu
Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the successor function.
Moreover, the recipes that enable us to name fixed points for any given formula aren't *guaranteed* to give us *terminating* fixed points. They might give us formulas X such that neither `X` nor `f X` have normal forms. (Indeed, what they give us for the square function isn't any of the Church numerals, but is rather an expression with no normal form.) However, if we take care we can ensure that we *do* get terminating fixed points. And this gives us a principled, fully general strategy for doing recursion. It lets us define even functions like the Ackermann function, which were until now out of our reach. It would let us define arithmetic and list functions on the "version 1" and "version 2" implementations, where it wasn't always clear how to force the computation to "keep going."
+Moreover, the recipes that enable us to name fixed points for any given formula aren't *guaranteed* to give us *terminating* fixed points. They might give us formulas X such that neither `X` nor `f X` have normal forms. (Indeed, what they give us for the square function isn't any of the Church numerals, but is rather an expression with no normal form.) However, if we take care we can ensure that we *do* get terminating fixed points. And this gives us a principled, fully general strategy for doing recursion. It lets us define even functions like the Ackermann function, which were until now out of our reach. It would also let us define arithmetic and list functions on the "version 1" and "version 2" implementations, where it wasn't always clear how to force the computation to "keep going."
OK, so how do we make use of this?
Recall our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
+Recall again our initial, abortive attempt above to define the `get_length` function in the lambda calculus. We said "What we really want to do is something like this:
\lst. (isempty lst) zero (add one (... (extracttail lst)))
where this very same formula occupies the `...` position."
Now, what if we *were* somehow able to get ahold of this formula, as an additional argument? We could take that argument and plug it into the `...` position. Something like this:
+If we could somehow get ahold of this very formula, as an additional argument, then we could take the argument and plug it into the `...` position. Something like this:
\self (\lst. (isempty lst) zero (add one (self (extracttail lst))) )
This is an abstract of the form:
 \self. body
+ \self. BODY
where `body` is the expression:
+where `BODY` is the expression:
\lst. (isempty lst) zero (add one (self (extracttail lst)))
containing an occurrence of `self`.
Now consider what would be a fixed point of our expression `\self. body`? That would be some expression `X` such that:
+Now consider what would be a fixed point of our expression `\self. BODY`? That would be some expression `X` such that:
 X <~~> (\self.body) X
+ X <~~> (\self.BODY) X
Betareducing the righthand side, we get:
 X <~~> body [self := X]
+ X <~~> BODY [self := X]
Think about what this says. It says if you substitute `X` for `self` in our formula body:
+Think about what this says. It says if you substitute `X` for `self` in our formula BODY:
\lst. (isempty lst) zero (add one (X (extracttail lst)))
@@ 244,9 +366,9 @@ containing free occurrences of `self` that you treat as being equivalent to the
\lst. (isempty lst) zero (add one (self (extracttail lst)))
You bind the free occurrence of `self` as: `\self. body`. And then you generate a fixed point for this larger expression:
+You bind the free occurrence of `self` as: `\self. BODY`. And then you generate a fixed point for this larger expression:
Ψ (\self. body)
+Ψ (\self. BODY)
using some fixedpoint combinator Ψ
.
@@ 254,16 +376,14 @@ Isn't that cool?
##Okay, then give me a fixedpoint combinator, already!##
Many fixedpoint combinators have been discovered. (And given a fixedpoint combinators, there are ways to use it as a model to build infinitely many more, nonequivalent fixedpoint combinators.)
+Many fixedpoint combinators have been discovered. (And some fixedpoint combinators give us models for building infinitely many more, nonequivalent fixedpoint 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′ has the advantage that f (Θ′ f)
really *reduces to* Θ′ f
.

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.
+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.
You may notice that both of these formulas have etaredexes 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′
?
@@ 272,9 +392,9 @@ 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))
I stated the more complex formulas for the following reason: in a language whose evaluation order is *callbyvalue*, the evaluation of Θ (\self. body)
and `Y (\self. body)` will in general not terminate. But evaluation of the etaunreduced primed versions will.
+I stated the more complex formulas for the following reason: in a language whose evaluation order is *callbyvalue*, the evaluation of Θ (\self. BODY)
and `Y (\self. BODY)` will in general not terminate. But evaluation of the etaunreduced 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 `\self. 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)
@@ 304,7 +424,7 @@ to *the tail* of the list we were evaluating its application to at the previous
##Fixedpoint Combinators Are a Bit Intoxicating##
![tatoo](/ycombinator.jpg)
+![tatoo](/ycombinatorfixed.jpg)
There's a tendency for people to say "Ycombinator" to refer to fixedpoint combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Ycombinator is only one of many fixedpoint combinators.
@@ 319,7 +439,202 @@ then this is a fixedpoint combinator:
L L L L L L L L L L L L L L L L L L L L L L L L L L
+##Watching Y in action##
+
+For those of you who like to watch ultra slowmo movies of bullets
+piercing apples, here's a stepwise computation of the application of a
+recursive function. We'll use a function `sink`, which takes one
+argument. If the argument is boolean true (i.e., `\x y.x`), it
+returns itself (a copy of `sink`); if the argument is boolean false
+(`\x y. y`), it returns `I`. That is, we want the following behavior:
+
+ sink false ~~> I
+ sink true false ~~> I
+ sink true true false ~~> I
+ sink true true true false ~~> I
+
+So we make `sink = Y (\f b. b f I)`:
+
+ 1. sink false
+ 2. Y (\fb.bfI) false
+ 3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) false
+ 4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) false
+ 5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) false
+ 6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) false
+ 7. false [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I
+ 
+ 8. I
+
+So far so good. The crucial thing to note is that as long as we
+always reduce the outermost redex first, we never have to get around
+to computing the underlined redex: because `false` ignores its first
+argument, we can throw it away unreduced.
+
+Now we try the next most complex example:
+
+ 1. sink true false
+ 2. Y (\fb.bfI) true false
+ 3. (\f. (\h. f (h h)) (\h. f (h h))) (\fb.bfI) true false
+ 4. (\h. [\fb.bfI] (h h)) (\h. [\fb.bfI] (h h)) true false
+ 5. [\fb.bfI] ((\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))) true false
+ 6. (\b.b[(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))]I) true false
+ 7. true [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] I false
+ 8. [(\h. [\fb.bsI] (h h))(\h. [\fb.bsI] (h h))] false
+
+We've now arrived at line (4) of the first computation, so the result
+is again I.
+
+You should be able to see that `sink` will consume as many `true`s as
+we throw at it, then turn into the identity function after it
+encounters the first `false`.
+
+The key to the recursion is that, thanks to Y, the definition of
+`sink` contains within it the ability to fully regenerate itself as
+many times as is necessary. The key to *ending* the recursion is that
+the behavior of `sink` is sensitive to the nature of the input: if the
+input is the magic function `false`, the selfregeneration 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 * (n1) * (n2) * ... * 1`. The recursive
+case says that the factorial of a number `n` is `n` times the
+factorial of `n1`. But if we leave out the base case, we get
+
+ 3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * 1! ...
+
+That's why it's crucial to declare that 0! = 1, in which case the
+recursive rule does not apply. In our terms,
+
+ fac = Y (\fac n. iszero n 1 (fac (predecessor n)))
+
+If `n` is 0, `fac` reduces to 1, without computing the recursive case.
+
+There is a wellknown problem in philosophy and natural language
+semantics that has the flavor of a recursive function without a base
+case: the truthteller paradox (and related paradoxes).
+
+(1) This sentence is true.
+
+If we assume that the complex demonstrative "this sentence" can refer
+to (1), then the proposition expressed by (1) will be true just in
+case the thing referred to by *this sentence* is true. Thus (1) will
+be true just in case (1) is true, and (1) is true just in case (1) is
+true, and so on. If (1) is true, then (1) is true; but if (1) is not
+true, then (1) is not true.
+
+Without pretending to give a serious analysis of the paradox, let's
+assume that sentences can have for their meaning boolean functions
+like the ones we have been working with here. Then the sentence *John
+is John* might denote the function `\x y. x`, our `true`.
+
+Then (1) denotes a function from whatever the referent of *this
+sentence* is to a boolean. So (1) denotes `\f. f true false`, where
+the argument `f` is the referent of *this sentence*. Of course, if
+`f` is a boolean, `f true false <~~> f`, so for our purposes, we can
+assume that (1) denotes the identity function `I`.
+
+If we use (1) in a context in which *this sentence* refers to the
+sentence in which the demonstrative occurs, then we must find a
+meaning `m` such that `I m = I`. But since in this context `m` is the
+same as the meaning `I`, so we have `m = I m`. In other words, `m` is
+a fixed point for the denotation of the sentence (when used in the
+appropriate context).
+
+That means that in a context in which *this sentence* refers to the
+sentence in which it occurs, the sentence denotes a fixed point for
+the identity function. Here's a fixed point for the identity
+function:
+
+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)))
+ω ω
+&Omega
+
+
+Oh. Well! That feels right. The meaning of *This sentence is true*
+in a context in which *this sentence* refers to the sentence in which
+it occurs is Ω
, our prototypical infinite loop...
+
+What about the liar paradox?
+
+(2) This sentence is false.
+
+Used in a context in which *this sentence* refers to the utterance of
+(2) in which it occurs, (2) will denote a fixed point for `\f.neg f`,
+or `\f l r. f r l`, which is the `C` combinator. So in such a
+context, (2) might denote
+
+ Y C
+ (\f. (\h. f (h h)) (\h. f (h h))) I
+ (\h. C (h h)) (\h. C (h h)))
+ C ((\h. C (h h)) (\h. C (h h)))
+ C (C ((\h. C (h h))(\h. C (h h))))
+ C (C (C ((\h. C (h h))(\h. C (h h)))))
+ ...
+
+And infinite sequence of `C`s, each one negating the remainder of the
+sequence. Yep, that feels like a reasonable representation of the
+liar paradox.
+
+See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
+truth and circularity](http://tinyurl.com/2db62bk) for an approach
+that is similar, but expressed in terms of nonwellfounded sets
+rather than recursive functions.
[TODO: Explain how what we've done relates to the version using lowercase ω.]

+##However...##
+You should be cautious about feeling too comfortable with
+these results. Thinking again of the truthteller paradox, yes,
+Ω
is *a* fixed point for `I`, and perhaps it has
+some a privileged status among all the fixed points for `I`, being the
+one delivered by Y and all (though it is not obvious why Y should have
+any special status).
+
+But one could ask: look, literally every formula is a fixed point for
+`I`, since
+
+ X <~~> I X
+
+for any choice of X whatsoever.
+
+So the Y combinator is only guaranteed to give us one fixed point out
+of infinitely manyand not always the intuitively most useful
+one. (For instance, the squaring function has zero as a fixed point,
+since 0 * 0 = 0, and 1 as a fixed point, since 1 * 1 = 1, but `Y
+(\x. mul x x)` doesn't give us 0 or 1.) So with respect to the
+truthteller paradox, why in the reasoning we've
+just gone through should we be reaching for just this fixed point at
+just this juncture?
+
+One obstacle to thinking this through is the fact that a sentence
+normally has only two truth values. We might consider instead a noun
+phrase such as
+
+(3) the entity that this noun phrase refers to
+
+The reference of (3) depends on the reference of the embedded noun
+phrase *this noun phrase*. It's easy to see that any object is a
+fixed point for this referential function: if this pen cap is the
+referent of *this noun phrase*, then it is the referent of (3), and so
+for any object.
+
+The chameleon nature of (3), by the way (a description that is equally
+good at describing any object), makes it particularly well suited as a
+gloss on pronouns such as *it*. In the system of
+[Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
+pronouns denote (you guessed it!) identity functions...
+
+Ultimately, in the context of this course, these paradoxes are more
+useful as a way of gaining leverage on the concepts of fixed points
+and recursion, rather than the other way around.