[[!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 <~~>
X = WW = (\x.T(xx))W = T(WW) = TX
</pre>
-Q: How do you know that for any term T, YT is a fixed point of T?
+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`?#
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
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.
+#Q: So if every term has a fixed point, even `Y` has fixed point.#
A: Right:
= Y(Y((\x.Y(xx))(\x.Y(xx))))
= Y(Y(Y(...(Y(YY))...)))
-Q: Ouch! Stop hurting my brain.
+#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
+That would imply that
X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
You should see the close similarity with YY here.
-Q. So `Y` applied to `succ` returns infinity!
+#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:
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:
+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))
there will always be an opportunity for more beta reduction. (Lather,
rinse, repeat!)
-Q. That reminds me, what about [[evaluation order]]?
+#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
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/call by name/normal order evaluation strategy, we'll always
-start with the isZero predicate, and only produce a fresh copy of
+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.
+#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 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...
+so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...
+
+#Q. What other questions should I be asking?#
+
+* What is it about the variant fixed-point combinators that makes
+ them compatible with a call-by-value evaluation strategy?
+
+* How do you know that the Ackerman function can't be computed
+ using primitive recursion techniques?
+
+* What *exactly* is primitive recursion?
+
+* I hear that `Y` delivers the *least* fixed point. Least
+ according to what ordering? How do you know it's least?
+ Is leastness important?
+