`T` has a fixed point, then there exists some `X` such that `X <~~>
TX` (that's what it means to *have* a fixed point).
-<pre>
-let W = \x.T(xx) in
-let X = WW in
-X = WW = (\x.T(xx))W = T(WW) = TX
-</pre>
+<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>
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 <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.
#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))...)))
+<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>
+
#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?
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)
+<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>
+
+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 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(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 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`,
+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.
+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:
-<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.
-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 <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.)
+
+
+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:
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.
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.
f 3 <result of folding f and z over [2; 1]> <handler to continue folding leftwards> <handler to abort the traversal>
of the list multiplied to, because that would affect the answer you passed
along to the continue-leftwards handler.
-A **version 5** list encodes the kind of fold operation we're envisaging here, in
-the same way that v3 (and [v4](/advanced/#v4)) lists encoded the simpler fold operation.
-Roughly, the list `[5;4;3;2;1]` would look like this:
+A **version 5** list encodes the kind of fold operation we're envisaging here,
+in the same way that v3 (and [v4](/advanced_lambda/#index1h1)) lists encoded
+the simpler fold operation. Roughly, the list `[5;4;3;2;1]` would look like
+this:
\f z continue_leftwards_handler abort_handler.
discusses it in much more
detail](http://okmij.org/ftp/Streams.html#enumerator-stream).
-*Comments*:
-
-1. The technique deployed here, and in the v2 lists, and in our implementations
- of pairs and booleans, is known as **continuation-passing style** programming.
-
-2. We're still building the list as a right fold, so in a sense the
- application of `f` to the leftmost element `5` is "outermost". However,
- this "outermost" application is getting lifted, and passed as a *handler*
- to the next right application. Which is in turn getting lifted, and
- passed to its next right application, and so on. So if you
- trace the evaluation of the `extract_head` function to the list `[5;4;3;2;1]`,
- you'll see `1` gets passed as a "this is the head sofar" answer to its
- `continue_handler`; then that answer is discarded and `2` is
- passed as a "this is the head sofar" answer to *its* `continue_handler`,
- and so on. All those steps have to be evaluated to finally get the result
- that `5` is the outer/leftmost head of the list. That's not an efficient way
- to get the leftmost head.
-
- We could improve this by building lists as left folds when implementing them
- as continuation-passing style folds. We'd just replace above:
-
- let make_list = \h t. \f z continue_handler abort_handler.
- f h z (\z. t f z continue_handler abort_handler) abort_handler
-
- now `extract_head` should return the leftmost head directly, using its `abort_handler`:
-
- let extract_head = \lst larger_computation. lst
- (\hd sofar continue_handler abort_handler. abort_handler hd)
- junk
- larger_computation
- larger_computation
-
-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/#v4) lists.
- But that also is left as an exercise.
+> *Comments*:
+
+> 1. The technique deployed here, and in the v2 lists, and in our
+> implementations of pairs and booleans, is known as
+> **continuation-passing style** programming.
+
+> 2. We're still building the list as a right fold, so in a sense the
+> application of `f` to the leftmost element `5` is "outermost". However,
+> this "outermost" application is getting lifted, and passed as a *handler*
+> to the next right application. Which is in turn getting lifted, and
+> passed to its next right application, and so on. So if you
+> trace the evaluation of the `extract_head` function to the list `[5;4;3;2;1]`,
+> you'll see `1` gets passed as a "this is the head sofar" answer to its
+> `continue_handler`; then that answer is discarded and `2` is
+> passed as a "this is the head sofar" answer to *its* `continue_handler`,
+> and so on. All those steps have to be evaluated to finally get the result
+> that `5` is the outer/leftmost head of the list. That's not an efficient way
+> to get the leftmost head.
+>
+> We could improve this by building lists as left folds when implementing them
+> as continuation-passing style folds. We'd just replace above:
+>
+> let make_list = \h t. \f z continue_handler abort_handler.
+> f h z (\z. t f z continue_handler abort_handler) abort_handler
+>
+> now `extract_head` should return the leftmost head directly, using its
+> `abort_handler`:
+>
+> let extract_head = \lst larger_computation. lst
+> (\hd sofar continue_handler abort_handler. abort_handler hd)
+> junk
+> larger_computation
+> larger_computation
+>
+> 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_lambda/#index1h1) lists. But that is left as an exercise.