3 #These notes return to the topic of fixed point combiantors for one more return to the topic of fixed point combinators#
5 #Q: How do you know that every term in the untyped lambda calculus has
8 A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
9 `T` has a fixed point, then there exists some `X` such that `X <~~>
10 TX` (that's what it means to *have* a fixed point).
15 X = WW = (\x.T(xx))W = T(WW) = TX
18 Please slow down and make sure that you understand what justified each
19 of the equalities in the last line.
21 #Q: How do you know that for any term `T`, `YT` is a fixed point of `T`?#
23 A: Note that in the proof given in the previous answer, we chose `T`
24 and then set `X = WW = (\x.T(xx))(\x.T(xx))`. If we abstract over
25 `T`, we get the Y combinator, `\T.(\x.T(xx))(\x.T(xx))`. No matter
26 what argument `T` we feed Y, it returns some `X` that is a fixed point
27 of `T`, by the reasoning in the previous answer.
29 #Q: So if every term has a fixed point, even `Y` has fixed point.#
33 let Y = \T.(\x.T(xx))(\x.T(xx)) in
34 Y Y = \T.(\x.T(xx))(\x.T(xx)) Y
35 = (\x.Y(xx))(\x.Y(xx))
36 = Y((\x.Y(xx))(\x.Y(xx)))
37 = Y(Y((\x.Y(xx))(\x.Y(xx))))
38 = Y(Y(Y(...(Y(YY))...)))
40 #Q: Ouch! Stop hurting my brain.#
42 A: Let's come at it from the direction of arithmetic. Recall that we
43 claimed that even `succ`---the function that added one to any
44 number---had a fixed point. How could there be an X such that X = X+1?
47 X = succ X = succ (succ X) = succ (succ (succ (X))) = succ (... (succ X)...)
49 In other words, the fixed point of `succ` is a term that is its own
50 successor. Let's just check that `X = succ X`:
52 let succ = \n s z. s (n s z) in
53 let X = (\x.succ(xx))(\x.succ(xx)) in
55 = succ ((\x.succ(xx))(\x.succ(xx)))
56 = succ (succ ((\x.succ(xx))(\x.succ(xx))))
59 You should see the close similarity with YY here.
61 #Q. So `Y` applied to `succ` returns a number that is not finite!#
63 A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
68 = (\n s z. s (n s z)) X
70 = succ (\s z. s (X s z)) ; using fixed-point reasoning
71 = \s z. s ([succ (\s z. s (X s z))] s z)
72 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
73 = \s z. s (s (succ (\s z. s (X s z))))
75 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
76 and returns a sequence of nested applications of `s`...
78 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
79 likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
80 succ)(Y succ)`? What would you expect infinity minus infinity to be?
81 (Hint: choose your evaluation strategy so that you add two `s`s to the
82 first number for every `s` that you add to the second number.)
84 This is amazing, by the way: we're proving things about a term that
85 represents arithmetic infinity.
87 It's important to bear in mind the simplest term in question is not
90 Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
92 The way that infinity enters into the picture is that this term has
93 no normal form: no matter how many times we perform beta reduction,
94 there will always be an opportunity for more beta reduction. (Lather,
97 #Q. That reminds me, what about [[evaluation order]]?#
99 A. For a recursive function that has a well-behaved base case, such as
100 the factorial function, evaluation order is crucial. In the following
101 computation, we will arrive at a normal form. Watch for the moment at
102 which we have to make a choice about which beta reduction to perform
103 next: one choice leads to a normal form, the other choice leads to
106 let prefac = \f n. isZero n 1 (mult n (f (pred n))) in
107 let fac = Y prefac in
109 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
110 = [(\x.prefac(xx))(\x.prefac(xx))] 2
111 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
112 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
113 = [(\f n. isZero n 1 (mult n (f (pred n))))
114 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
115 = [\n. isZero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
116 = isZero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
117 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
119 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
120 = mult 2 (mult 1 (isZero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
125 The crucial step is the third from the last. We have our choice of
126 either evaluating the test `isZero 0 1 ...`, which evaluates to `1`,
127 no matter what the ... contains;
128 or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
129 produce another copy of `prefac`. If we postpone evaluting the
130 `isZero` test, we'll pump out copy after copy of `prefac`, and never
131 realize that we've bottomed out in the recursion. But if we adopt a
132 leftmost/call-by-name/normal-order evaluation strategy, we'll always
133 start with the `isZero` predicate, and only produce a fresh copy of
134 `prefac` if we are forced to.
136 #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.#
142 | when m == 0 -> n + 1
143 | else when n == 0 -> A(m-1,1)
144 | else -> A(m-1, A(m,n-1))
146 let A = Y (\A m n. isZero m (succ n) (isZero n (A (pred m) 1) (A (pred m) (A m (pred n))))) in
159 A 1 x is to A 0 x as addition is to the successor function;
160 A 2 x is to A 1 x as multiplication is to addition;
161 A 3 x is to A 2 x as exponentiation is to multiplication---
162 so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...
164 #Q. What other questions should I be asking?#
166 * What is it about the variant fixed-point combinators that makes
167 them compatible with a call-by-value evaluation strategy?
169 * How do you know that the Ackerman function can't be computed
170 using primitive recursion techniques?
172 * What *exactly* is primitive recursion?