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