-let W = \x.T(xx) in -let X = WW in -X = WW = (\x.T(xx))W = T(WW) = TX -+

```
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
+
```

+
+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.
+#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.
+#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
number---had a fixed point. How could there be an X such that X = X+1?
-Then
+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 infinity!
+#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.)
This is amazing, by the way: we're proving things about a term that
-represents arithmetic infinity. It's important to bear in mind the
-simplest term in question is not infinite:
+represents arithmetic infinity.
- Y succ = (\f.(\x.f(xx))(\x.f(xx)))(\n s z. s (n s z))
+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))
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]]?
+
+#Q. That reminds me, what about [[evaluation order]]?#
A. For a recursive function that has a well-behaved base case, such as
the factorial function, evaluation order is crucial. In the following
@@ -98,64 +108,517 @@ 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`,
-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
+either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
+no matter what the ... contains;
+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.
+leftmost/call-by-name/normal-order evaluation strategy, we'll always
+start with the `iszero` predicate, and only produce a fresh copy of
+`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 super-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 Ackermann function can't be computed + using primitive recursion techniques? + +* What *exactly* is primitive recursion? + +* I hear that `Y` delivers the *least* fixed point. Least + according to what ordering? How do you know it's least? + Is leastness important? + + + +#Sets# + +You're now already in a position to implement sets: that is, collections with +no intrinsic order where elements can occur at most once. Like lists, we'll +understand the basic set structures to be *type-homogenous*. So you might have +a set of integers, or you might have a set of pairs of integers, but you +wouldn't have a set that mixed both types of elements. Something *like* the +last option is also achievable, but it's more difficult, and we won't pursue it +now. In fact, we won't talk about sets of pairs, either. We'll just talk about +sets of integers. The same techniques we discuss here could also be applied to +sets of pairs of integers, or sets of triples of booleans, or sets of pairs +whose first elements are booleans, and whose second elements are triples of +integers. And so on. + +(You're also now in a position to implement *multi*sets: that is, collections +with no intrinsic order where elements can occur multiple times: the multiset +{a,a} is distinct from the multiset {a}. But we'll leave these as an exercise.) + +The easiest way to implement sets of integers would just be to use lists. When +you "add" a member to a set, you'd get back a list that was either identical to +the original list, if the added member already was present in it, or consisted +of a new list with the added member prepended to the old list. That is: + + let empty_set = empty in + ; see the library for definitions of any and eq + let make_set = \new_member old_set. any (eq new_member) old_set + ; if any element in old_set was eq new_member + old_set + ; else + make_list new_member old_set + +Think about how you'd implement operations like `set_union`, +`set_intersection`, and `set_difference` with this implementation of sets. + +The implementation just described works, and it's the simplest to code. +However, it's pretty inefficient. If you had a 100-member set, and you wanted +to create a set which had all those 100-members and some possibly new element +`e`, you might need to check all 100 members to see if they're equal to `e` +before concluding they're not, and returning the new list. And comparing for +numeric equality is a moderately expensive operation, in the first place. + +(You might say, well, what's the harm in just prepending `e` to the list even +if it already occurs later in the list. The answer is, if you don't keep track +of things like this, it will likely mess up your implementations of +`set_difference` and so on. You'll have to do the book-keeping for duplicates +at some point in your code. It goes much more smoothly if you plan this from +the very beginning.) + +How might we make the implementation more efficient? Well, the *semantics* of +sets says that they have no intrinsic order. That means, there's no difference +between the set {a,b} and the set {b,a}; whereas there is a difference between +the *list* `[a;b]` and the list `[b;a]`. But this semantic point can be respected +even if we *implement* sets with something ordered, like list---as we're +already doing. And we might *exploit* the intrinsic order of lists to make our +implementation of sets more efficient. + +What we could do is arrange it so that a list that implements a set always +keeps in elements in some specified order. To do this, there'd have *to be* +some way to order its elements. Since we're talking now about sets of numbers, +that's easy. (If we were talking about sets of pairs of numbers, we'd use +"lexicographic" ordering, where `(a,b) < (c,d)` iff `a < c or (a == c and b < +d)`.) + +So, if we were searching the list that implements some set to see if the number +`5` belonged to it, once we get to elements in the list that are larger than `5`, +we can stop. If we haven't found `5` already, we know it's not in the rest of the +list either. + +> *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 +implements the set, if it belonged to the set at all. Or that it fell in the +right half, it it belonged to the set at all. And then the same sort of +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](/implementing_trees). + + +#Aborting a search through a list# + +We said that the sorted-list implementation of a set was more efficient than +the unsorted-list implementation, because as you were searching through the +list, you could come to a point where you knew the element wasn't going to be +found. So you wouldn't have to continue the search. + +If your implementation of lists was, say v1 lists plus the Y-combinator, then +this is exactly right. When you get to a point where you know the answer, you +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 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:
+
+ ..................`extract_fst ≡ \pair. pair (\x y. x)`

+
+but at a lower level, the pair is still accepting its handler as an argument,
+rather than the handler taking the pair as an argument. (The handler gets *the
+pair's elements*, not the pair itself, as arguments.)
+
+> *Terminology*: we'll try to use names of the form `get_foo` for handlers, and
+names of the form `extract_foo` for lifted versions of them, that accept the
+lists (or whatever data structure we're working with) as arguments. But we may
+sometimes forget.
+
+The v2 implementation of lists followed a similar strategy:
+
+ v2list (\h t. do_something_with_h_and_t) result_if_empty
+
+If the `v2list` here is not empty, then this will reduce to the result of
+supplying the list's head and tail to the handler `(\h t.
+do_something_with_h_and_t)`.
+
+Now, what we've been imagining ourselves doing with the search through the v3
+list is something like this:
+
+
+ larger_computation (search_through_the_list_for_3) other_arguments
+
+That is, the result of our search is supplied as an argument (perhaps together
+with other arguments) to the "larger computation". Without knowing the
+evaluation order/reduction strategy, we can't say whether the search is
+evaluated before or after it's substituted into the larger computation. But
+semantically, the search is the argument and the larger computation is the
+function to which it's supplied.
+
+What if, instead, we did the same kind of thing we did with pairs and v2
+lists? That is, what if we made the larger computation a "handler" that we
+passed as an argument to the search?
+
+ the_search (\search_result. larger_computation search_result other_arguments)
+
+What's the advantage of that, you say. Other than to show off how cleverly
+you can lift.
+
+Well, think about it. Think about the difficulty we were having aborting the
+search. Does this switch-around offer us anything useful?
+
+It could.
+
+What if the way we implemented the search procedure looked something like this?
+
+At a given stage in the search, we wouldn't just apply some function `f` to the
+head at this stage and the result accumulated so far (from folding the same
+function, and a base value, to the tail at this stage)...and then pass the result
+of that application to the embedding, more leftward computation.
+
+We'd *instead* give `f` a "handler" that expects the result of the current
+stage *as an argument*, and then evaluates to what you'd get by passing that
+result leftwards up the list, as before.
+
+Why would we do that, you say? Just more flamboyant lifting?
+
+Well, no, there's a real point here. If we give the function a "handler" that
+encodes the normal continuation of the fold leftwards through the list, we can
+also give it other "handlers" too. For example, we can also give it the underlined handler:
+
+
+ the_search (\search_result. larger_computation search_result other_arguments)
+ ------------------------------------------------------------------
+
+This "handler" encodes the search's having finished, and delivering a final
+answer to whatever else you wanted your program to do with the result of the
+search. If you like, at any stage in the search you might just give an argument
+to *this* handler, instead of giving an argument to the handler that continues
+the list traversal leftwards. Semantically, this would amount to *aborting* the
+list traversal! (As we've said before, whether the rest of the list traversal
+really gets evaluated will depend on what evaluation order is in place. But
+semantically we'll have avoided it. Our larger computation won't depend on the
+rest of the list traversal having been computed.)
+
+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. 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.
+
+ f2 3