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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?
-
-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
-
-
-Q: How do you know that for any term T, YT 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
-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))...)))
-
-Q: Ouch! Stop hurting my brain.
-
-A: Let's come at it from the direction of arithmetic. Recall that we
-claimed that even `succ`---the function that added one to any
-number---had a fixed point. How could there be an X such that X = X+1?
-Then
-
- 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 infinity!
-
-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 fixed-point 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 well-behaved 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`,
-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/call by name/normal order 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(m-1,1)
- | else -> A(m-1, A(m,n-1))
-
-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 super-exponentiation is to exponentiation...
-