`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.
@@ -27,11 +27,14 @@ of `T`, by the reasoning in the previous answer.
A: Right:
```
let Y = \T. (\x. T (x x)) (\x. T (x x)) in
-Y Y ≡ \T. (\x. T (x x)) (\x. T (x x)) Y
+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))...)))
```

+~~> Y (Y (Y (...(Y (Y Y))...)))
+
+
#Q: Ouch! Stop hurting my brain.#
@@ -42,7 +45,7 @@ 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?
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`:
@@ -50,32 +53,35 @@ successor. Let's just check that `X = 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)
```

+≡ 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.
+
#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))))
+```
[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))
+
```

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.)
@@ -85,13 +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 (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
@@ -101,70 +108,70 @@ 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`,
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 Ackermann 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 Ackermann 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(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 -- -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?# * 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 +* How do you know that the Ackermann function can't be computed using primitive recursion techniques? * What *exactly* is primitive recursion? @@ -243,7 +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 @@ -253,7 +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# @@ -269,10 +276,39 @@ 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. -But what if you're using v3 lists? What options would you have then for -aborting a search? - -Well, suppose we're searching through the list `[5;4;3;2;1]` to see if it +But what if we wanted to use v3 lists instead? + +> Why would we want to do that? The advantage of the v3 lists and v3 (aka +"Church") numerals is that they have their recursive capacity built into their +very bones. 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

`Θ`

. 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 `Y′`

and
+`Θ′`

. 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.)
+
+
+So again, what if we're using v3 lists? What options would we have then for
+aborting a search or list traversal before it runs to completion?
+
+Suppose we're searching through the list `[5;4;3;2;1]` to see if it
contains the number `3`. The expression which represents this search would have
something like the following form:
@@ -292,7 +328,7 @@ more rightmost pieces of the list, too, regardless of what order the reduction
is computed by. Conceptually, it will be easiest if we think of the reduction
happening in the order displayed above.
-Well, once we've found a match between our sought number `3` and some member of
+Once we've found a match between our sought number `3` and some member of
the list, we'd like to avoid any further unnecessary computations and just
deliver the answer `true` as "quickly" or directly as possible to the larger
computation in which the search was embedded.
@@ -417,64 +453,64 @@ Do you have the basic idea? Think about how you'd implement it. A good
understanding of the v2 lists will give you a helpful model.
In broad outline, a single stage of the search would look like before, except
-now f would receive two extra, "handler" arguments.
+now `f` would receive two extra, "handler" arguments. We'll reserve the name `f` for the original fold function, and use `f2` for the function that accepts two additional handler arguments. To get the general idea, you can regard these as interchangeable. If the extra precision might help, then you can pay attention to when we're talking about the handler-taking `f2` or the original `f`. You'll only be *supplying* the `f2` function; the idea will be that the behavior of the original `f` will be implicitly encoded in `f2`'s behavior.
- f 3