XGitUrl: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=blobdiff_plain;f=week4.mdwn;h=150e88354cd7f8b56884ffe201e3afd1255693ba;hp=e3f0fb956068b0e6c33b07ce9a8ef0e247b98d66;hb=ca367b32e54d5816fcea12dd590a76306901d1ae;hpb=792521cd3b2e337d425aadc7d8f09a2eab2eb391
diff git a/week4.mdwn b/week4.mdwn
index e3f0fb95..150e8835 100644
 a/week4.mdwn
+++ b/week4.mdwn
@@ 6,8 +6,7 @@ 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 L = \x. T (x x) in
+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
@@ 15,66 +14,74 @@ X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
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`?#
+#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 = 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
+and then set X ≡ L L ≡ (\x. T (x x)) (\x. T (x x))
. 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.
#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))...)))
+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))...)))
+
+
#Q: Ouch! Stop hurting my brain.#
A: Let's come at it from the direction of arithmetic. Recall that we
+A: Is that a question?
+
+Let's come at it from the direction of arithmetic. Recall that we
claimed that even `succ`the function that added one to any
numberhad 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`:
 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)
+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)
+
+
+You should see the close similarity with `Y Y` here.
You should see the close similarity with YY 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 fixedpoint 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))))
+[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 fixedpoint 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))
+
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.)
@@ 84,13 +91,14 @@ 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))
+ 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 wellbehaved base case, such as
@@ 100,63 +108,63 @@ 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
+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
+
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`,
+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/callbyname/normalorder evaluation strategy, we'll always
start with the `isZero` predicate, and only produce a fresh copy of
`prefac` if we are forced to.
+start with the `iszero` predicate, and only produce a fresh copy of
+`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:
A(m,n) =
  when m == 0 > n + 1
  else when n == 0 > A(m1,1)
  else > A(m1, A(m,n1))

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 hyperexponentiation is to exponentiation...
+ A(m,n) =
+  when m == 0 > n + 1
+  else when n == 0 > A(m1,1)
+  else > A(m1, A(m,n1))
+
+ 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 hyperexponentiation is to exponentiation...
#Q. What other questions should I be asking?#
@@ 242,7 +250,7 @@ So, if we were searching the list that implements some set to see if the number
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
@@ 252,7 +260,7 @@ determination could be made for whichever half you were directed to. And then
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#
@@ 268,6 +276,35 @@ can just deliver that answer, and not branch into any further recursion. If
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 normalorder 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 fixedpoint combinators
+`Y` or Θ
. Expressions using them will have nonterminating
+reductions, with Scheme's eager/callbyvalue strategy. There are other
+fixedpoint combinators you can use with Scheme (in the [week 3 notes](/week3/#index7h2) they
+were Y′
and Θ′
. But even with
+them, evaluation order still matters: for some (admittedly unusual)
+evaluation strategies, expressions using them will also be nonterminating.
+
+The fixedpoint 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?
@@ 568,4 +605,3 @@ detail](http://okmij.org/ftp/Streams.html#enumeratorstream).
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.
