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[[!toc]]
#These notes return to the topic of fixed point combiantors for one more return to the topic of fixed point combinators#

#Q: How do you know that every term in the untyped lambda calculus has
a fixed point?#
+#Q: How do you know that every term in the untyped lambda calculus has a fixed point?#
A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
`T` has a fixed point, then there exists some `X` such that `X <~~>
TX` (that's what it means to *have* a fixed point).
let W = \x.T(xx) in
let X = WW in
X = WW = (\x.T(xx))W = T(WW) = TX

+let L = \x. T (x x) in
+let X = L L in
+X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
+
Please slow down and make sure that you understand what justified each
of the equalities in the last line.
#Q: How do you know that for any term `T`, `YT` is a fixed point of `T`?#
+#Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?#
A: Note that in the proof given in the previous answer, we chose `T`
and then set `X = WW = (\x.T(xx))(\x.T(xx))`. If we abstract over
`T`, we get the Y combinator, `\T.(\x.T(xx))(\x.T(xx))`. No matter
what argument `T` we feed Y, it returns some `X` that is a fixed point
+and then set `X = L L = (\x. T (x x)) (\x. T (x x))`. If we abstract over
+`T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`. No matter
+what argument `T` we feed `Y`, it returns some `X` that is a fixed point
of `T`, by the reasoning in the previous answer.
#Q: So if every term has a fixed point, even `Y` has fixed point.#
A: Right:
 let Y = \T.(\x.T(xx))(\x.T(xx)) in
 Y Y = \T.(\x.T(xx))(\x.T(xx)) Y
 = (\x.Y(xx))(\x.Y(xx))
 = Y((\x.Y(xx))(\x.Y(xx)))
 = Y(Y((\x.Y(xx))(\x.Y(xx))))
 = Y(Y(Y(...(Y(YY))...)))
+let Y = \T. (\x. T (x x)) (\x. T (x x)) in
+Y Y ≡ \T. (\x. T (x x)) (\x. T (x x)) Y
+~~> (\x. Y (x x)) (\x. Y (x x))
+~~> Y ((\x. Y (x x)) (\x. Y (x x)))
+~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
+~~> Y (Y (Y (...(Y (Y Y))...)))
+
#Q: Ouch! Stop hurting my brain.#
A: Let's come at it from the direction of arithmetic. Recall that we
+A: Is that a question?
+
+Let's come at it from the direction of arithmetic. Recall that we
claimed that even `succ`the function that added one to any
numberhad a fixed point. How could there be an X such that X = X+1?
That would imply that
 X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
+ X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...)
In other words, the fixed point of `succ` is a term that is its own
successor. Let's just check that `X = succ X`:
 let succ = \n s z. s (n s z) in
 let X = (\x.succ(xx))(\x.succ(xx)) in
 succ X
 = succ ((\x.succ(xx))(\x.succ(xx)))
 = succ (succ ((\x.succ(xx))(\x.succ(xx))))
 = succ (succ X)

You should see the close similarity with YY here.

#Q. So `Y` applied to `succ` returns a number that is not finite!#

A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
numeral:

 [same definitions]
 succ X
 = (\n s z. s (n s z)) X
 = \s z. s (X s z)
 = succ (\s z. s (X s z)) ; using fixedpoint reasoning
 = \s z. s ([succ (\s z. s (X s z))] s z)
 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
 = \s z. s (s (succ (\s z. s (X s z))))

So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
and returns a sequence of nested applications of `s`...

You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
succ)(Y succ)`? What would you expect infinity minus infinity to be?
(Hint: choose your evaluation strategy so that you add two `s`s to the
first number for every `s` that you add to the second number.)

This is amazing, by the way: we're proving things about a term that
represents arithmetic infinity.

It's important to bear in mind the simplest term in question is not
infinite:

 Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))

The way that infinity enters into the picture is that this term has
no normal form: no matter how many times we perform beta reduction,
there will always be an opportunity for more beta reduction. (Lather,
rinse, repeat!)

#Q. That reminds me, what about [[evaluation order]]?#

A. For a recursive function that has a wellbehaved base case, such as
the factorial function, evaluation order is crucial. In the following
computation, we will arrive at a normal form. Watch for the moment at
which we have to make a choice about which beta reduction to perform
next: one choice leads to a normal form, the other choice leads to
endless reduction:

 let prefac = \f n. isZero n 1 (mult n (f (pred n))) in
 let fac = Y prefac in
 fac 2
 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
 = [(\x.prefac(xx))(\x.prefac(xx))] 2
 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
 = [(\f n. isZero n 1 (mult n (f (pred n))))
 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
 = [\n. isZero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
 = isZero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
 ...
 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
 = mult 2 (mult 1 (isZero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
 = mult 2 (mult 1 1)
 = mult 2 1
 = 2

The crucial step is the third from the last. We have our choice of
either evaluating the test `isZero 0 1 ...`, which evaluates to `1`,
no matter what the ... contains;
or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
produce another copy of `prefac`. If we postpone evaluting the
`isZero` test, we'll pump out copy after copy of `prefac`, and never
realize that we've bottomed out in the recursion. But if we adopt a
leftmost/callbyname/normalorder evaluation strategy, we'll always
start with the `isZero` predicate, and only produce a fresh copy of
`prefac` if we are forced to.

#Q. You claimed that the Ackerman function couldn't be implemented using our primitive recursion techniques (such as the techniques that allow us to define addition and multiplication). But you haven't shown that it is possible to define the Ackerman function using full recursion.#

A. OK:

A(m,n) =
  when m == 0 > n + 1
  else when n == 0 > A(m1,1)
  else > A(m1, A(m,n1))

let A = Y (\A m n. isZero m (succ n) (isZero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in


For instance,

 A 1 2
 = A 0 (A 1 1)
 = A 0 (A 0 (A 1 0))
 = A 0 (A 0 (A 0 1))
 = A 0 (A 0 2)
 = A 0 3
 = 4

A 1 x is to A 0 x as addition is to the successor function;
A 2 x is to A 1 x as multiplication is to addition;
A 3 x is to A 2 x as exponentiation is to multiplication
so A 4 x is to A 3 x as hyperexponentiation is to exponentiation...

#Q. What other questions should I be asking?#

* What is it about the variant fixedpoint combinators that makes
 them compatible with a callbyvalue evaluation strategy?

* How do you know that the Ackerman function can't be computed
 using primitive recursion techniques?

* What *exactly* is primitive recursion?
+let succ = \n s z. s (n s z) in
+let X = (\x. succ (x x)) (\x. succ (x x)) in
+succ X
+≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
+~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
+≡ succ (succ X)
+
+
+You should see the close similarity with `Y Y` here.
+