<pre><code>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</code></pre>
+X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
+</code></pre>
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`, `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 = L L = (\x. T (x x)) (\x. T (x x))`. If we abstract over
+and then set <code>X ≡ L L ≡ (\x. T (x x)) (\x. T (x x))</code>. 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.
A: Right:
<pre><code>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))...)))</code></pre>
+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))...)))
+</code></pre>
+
#Q: Ouch! Stop hurting my brain.#
number---had 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`:
<pre><code>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)</code></pre>
+≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
+~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x)) ))
+≡ succ (succ X)
+</code></pre>
You should see the close similarity with `Y Y` 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 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))))
+<pre><code>[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
+≡ (\n s z. s (n s z)) (\s z. s (X s z))
+~~> \s z. s ((\s z. s (X s z)) s z)
+~~> \s z. s (s (X s z))
+</code></pre>
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?
+likewise for `mul`, `sub`, `pow`. What happens if we try `sub (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.)
It's important to bear in mind the simplest term in question is not
infinite:
- Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))
+ Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\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
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
+<pre><code>let prefact = \f n. iszero n 1 (mul n (f (pred n))) in
+let fact = Y prefact in
+fact 2
+≡ [(\f. (\x. f (x x)) (\x. f (x x))) prefact] 2
+~~> [(\x. prefact (x x)) (\x. prefact (x x))] 2
+~~> [prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 2
+~~> [prefact (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+≡ [ (\f n. iszero n 1 (mul n (f (pred n)))) (prefact ((\x. prefact (x x)) (\x. prefact (x x))))] 2
+~~> [\n. iszero n 1 (mul n ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred n)))] 2
+~~> iszero 2 1 (mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 2)))
+~~> mul 2 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 1)
+...
+~~> mul 2 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] 0))
+≡ mul 2 (mul 1 (iszero 0 1 (mul 1 ([prefact ((\x. prefact (x x)) (\x. prefact (x x)))] (pred 0)))))
+~~> mul 2 (mul 1 1)
+~~> mul 2 1
+~~> 2
+</code></pre>
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
+or we can evaluate the `Y` pump, `(\x. prefact (x x)) (\x. prefact (x x))`, to
+produce another copy of `prefact`. If we postpone evaluting the
+`iszero` test, we'll pump out copy after copy of `prefact`, 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.
+`prefact` 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:
-<pre>
-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
-</pre>
-
-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 hyper-exponentiation is to exponentiation...
+ 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)))))
+
+So 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 hyper-exponentiation is to exponentiation...
#Q. What other questions should I be asking?#
we can stop. If we haven't found `5` already, we know it's not in the rest of the
list either.
-This is an improvement, but it's still a "linear" search through the list.
+*Comment*: This is an improvement, but it's still a "linear" search through the list.
There are even more efficient methods, which employ "binary" searching. They'd
represent the set in such a way that you could quickly determine whether some
element fell in one half, call it the left half, of the structure that
for whichever quarter you were directed to next. And so on. Until you either
found the element or exhausted the structure and could then conclude that the
element in question was not part of the set. These sorts of structures are done
-using **binary trees** (see below).
+using [binary trees](/implementing_trees).
#Aborting a search through a list#
you've got the right evaluation strategy in place, everything will work out
fine.
+--
+An advantage of the v3 lists and v3 (aka "Church") numerals is that they
+have a recursive capacity built into their skeleton. So for many natural
+operations on them, you won't need to use a fixed point combinator. Why is
+that an advantage? Well, if you use a fixed point combinator, then the terms
+you get
+won't be strongly normalizing: whether their reduction stops at a normal form
+will depend on what evaluation order you use. Our online [[lambda evaluator]]
+uses normal-order reduction, so it finds a normal form if there's one to be
+had. But if you want to build lambda terms in, say, Scheme, and you wanted to
+roll your own recursion as we've been doing, rather than relying on Scheme's
+native `let rec` or `define`, then you can't use the fixed-point combinators
+`Y` or <code>Θ</code>. Expressions using them will have non-terminating
+reductions, with Scheme's eager/call-by-value strategy. There are other
+fixed-point combinators you can use with Scheme (in the [week 3 notes](/week3/#index7h2) they
+were <code>Y′</code> and <code>Θ′</code>. But even with
+them, evaluation order still matters: for some (admittedly unusual)
+evaluation strategies, expressions using them will also be non-terminating.
+
+The fixed-point combinators may be the conceptual stars. They are cool and
+mathematically elegant. But for efficiency and implementation elegance, it's
+best to know how to do as much as you can without them. (Also, that knowledge
+could carry over to settings where the fixed point combinators are in
+principle unavailable.)
+
+This is why the v3 lists and numbers are so lovely..
+
+--
+
But what if you're using v3 lists? What options would you have then for
aborting a search?
3. To extract tails efficiently, too, it'd be nice to fuse the apparatus developed
in these v5 lists with the ideas from [v4](/advanced/#index1h1) lists.
But that also is left as an exercise.
-