3 #Q: How do you know that every term in the untyped lambda calculus has a fixed point?#
5 A: That's easy: let `T` be an arbitrary term in the lambda calculus. If
6 `T` has a fixed point, then there exists some `X` such that `X <~~>
7 TX` (that's what it means to *have* a fixed point).
9 <pre><code>let L = \x. T (x x) in
11 X ≡ L L ≡ (\x. T (x x)) L ~~> T (L L) ≡ T X
14 Please slow down and make sure that you understand what justified each
15 of the equalities in the last line.
17 #Q: How do you know that for any term `T`, `Y T` is a fixed point of `T`?#
19 A: Note that in the proof given in the previous answer, we chose `T`
20 and then set `X = L L = (\x. T (x x)) (\x. T (x x))`. If we abstract over
21 `T`, we get the Y combinator, `\T. (\x. T (x x)) (\x. T (x x))`. No matter
22 what argument `T` we feed `Y`, it returns some `X` that is a fixed point
23 of `T`, by the reasoning in the previous answer.
25 #Q: So if every term has a fixed point, even `Y` has fixed point.#
29 <pre><code>let Y = \T. (\x. T (x x)) (\x. T (x x)) in
30 Y Y ≡ \T. (\x. T (x x)) (\x. T (x x)) Y
31 ~~> (\x. Y (x x)) (\x. Y (x x))
32 ~~> Y ((\x. Y (x x)) (\x. Y (x x)))
33 ~~> Y (Y ((\x. Y (x x)) (\x. Y (x x))))
34 ~~> Y (Y (Y (...(Y (Y Y))...)))</code></pre>
37 #Q: Ouch! Stop hurting my brain.#
39 A: Is that a question?
41 Let's come at it from the direction of arithmetic. Recall that we
42 claimed that even `succ`---the function that added one to any
43 number---had a fixed point. How could there be an X such that X = X+1?
46 X <~~> succ X <~~> succ (succ X) <~~> succ (succ (succ (X))) <~~> succ (... (succ X)...)
48 In other words, the fixed point of `succ` is a term that is its own
49 successor. Let's just check that `X = succ X`:
51 <pre><code>let succ = \n s z. s (n s z) in
52 let X = (\x. succ (x x)) (\x. succ (x x)) in
54 ≡ succ ( (\x. succ (x x)) (\x. succ (x x)) )
55 ~~> succ (succ ( (\x. succ (x x)) (\x. succ (x x))))
59 You should see the close similarity with `Y Y` here.
62 #Q. So `Y` applied to `succ` returns a number that is not finite!#
64 A. Yes! Let's see why it makes sense to think of `Y succ` as a Church
69 = (\n s z. s (n s z)) X
71 = succ (\s z. s (X s z)) ; using fixed-point reasoning
72 = \s z. s ([succ (\s z. s (X s z))] s z)
73 = \s z. s ([\s z. s ([succ (\s z. s (X s z))] s z)] s z)
74 = \s z. s (s (succ (\s z. s (X s z))))
76 So `succ X` looks like a numeral: it takes two arguments, `s` and `z`,
77 and returns a sequence of nested applications of `s`...
79 You should be able to prove that `add 2 (Y succ) <~~> Y succ`,
80 likewise for `mult`, `minus`, `pow`. What happens if we try `minus (Y
81 succ)(Y succ)`? What would you expect infinity minus infinity to be?
82 (Hint: choose your evaluation strategy so that you add two `s`s to the
83 first number for every `s` that you add to the second number.)
85 This is amazing, by the way: we're proving things about a term that
86 represents arithmetic infinity.
88 It's important to bear in mind the simplest term in question is not
91 Y succ = (\f. (\x. f (x x)) (\x. f (x x))) (\n s z. s (n s z))
93 The way that infinity enters into the picture is that this term has
94 no normal form: no matter how many times we perform beta reduction,
95 there will always be an opportunity for more beta reduction. (Lather,
99 #Q. That reminds me, what about [[evaluation order]]?#
101 A. For a recursive function that has a well-behaved base case, such as
102 the factorial function, evaluation order is crucial. In the following
103 computation, we will arrive at a normal form. Watch for the moment at
104 which we have to make a choice about which beta reduction to perform
105 next: one choice leads to a normal form, the other choice leads to
108 let prefac = \f n. iszero n 1 (mult n (f (pred n))) in
109 let fac = Y prefac in
111 = [(\f.(\x.f(xx))(\x.f(xx))) prefac] 2
112 = [(\x.prefac(xx))(\x.prefac(xx))] 2
113 = [prefac((\x.prefac(xx))(\x.prefac(xx)))] 2
114 = [prefac(prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
115 = [(\f n. iszero n 1 (mult n (f (pred n))))
116 (prefac((\x.prefac(xx))(\x.prefac(xx))))] 2
117 = [\n. iszero n 1 (mult n ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred n)))] 2
118 = iszero 2 1 (mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 2)))
119 = mult 2 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 1)
121 = mult 2 (mult 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] 0))
122 = mult 2 (mult 1 (iszero 0 1 ([prefac((\x.prefac(xx))(\x.prefac(xx)))] (pred 0))))
127 The crucial step is the third from the last. We have our choice of
128 either evaluating the test `iszero 0 1 ...`, which evaluates to `1`,
129 no matter what the ... contains;
130 or we can evaluate the `Y` pump, `(\x.prefac(xx))(\x.prefac(xx))`, to
131 produce another copy of `prefac`. If we postpone evaluting the
132 `iszero` test, we'll pump out copy after copy of `prefac`, and never
133 realize that we've bottomed out in the recursion. But if we adopt a
134 leftmost/call-by-name/normal-order evaluation strategy, we'll always
135 start with the `iszero` predicate, and only produce a fresh copy of
136 `prefac` if we are forced to.
139 #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.#
145 | when m == 0 -> n + 1
146 | else when n == 0 -> A(m-1,1)
147 | else -> A(m-1, A(m,n-1))
149 let A = Y (\A m n. iszero m (succ n) (iszero n (A (pred m) 1) (A (pred m) (A m (pred n)))))
161 A 1 x is to A 0 x as addition is to the successor function;
162 A 2 x is to A 1 x as multiplication is to addition;
163 A 3 x is to A 2 x as exponentiation is to multiplication---
164 so A 4 x is to A 3 x as hyper-exponentiation is to exponentiation...
166 #Q. What other questions should I be asking?#
168 * What is it about the variant fixed-point combinators that makes
169 them compatible with a call-by-value evaluation strategy?
171 * How do you know that the Ackerman function can't be computed
172 using primitive recursion techniques?
174 * What *exactly* is primitive recursion?
176 * I hear that `Y` delivers the *least* fixed point. Least
177 according to what ordering? How do you know it's least?
178 Is leastness important?
184 You're now already in a position to implement sets: that is, collections with
185 no intrinsic order where elements can occur at most once. Like lists, we'll
186 understand the basic set structures to be *type-homogenous*. So you might have
187 a set of integers, or you might have a set of pairs of integers, but you
188 wouldn't have a set that mixed both types of elements. Something *like* the
189 last option is also achievable, but it's more difficult, and we won't pursue it
190 now. In fact, we won't talk about sets of pairs, either. We'll just talk about
191 sets of integers. The same techniques we discuss here could also be applied to
192 sets of pairs of integers, or sets of triples of booleans, or sets of pairs
193 whose first elements are booleans, and whose second elements are triples of
196 (You're also now in a position to implement *multi*sets: that is, collections
197 with no intrinsic order where elements can occur multiple times: the multiset
198 {a,a} is distinct from the multiset {a}. But we'll leave these as an exercise.)
200 The easiest way to implement sets of integers would just be to use lists. When
201 you "add" a member to a set, you'd get back a list that was either identical to
202 the original list, if the added member already was present in it, or consisted
203 of a new list with the added member prepended to the old list. That is:
205 let empty_set = empty in
206 ; see the library for definitions of any and eq
207 let make_set = \new_member old_set. any (eq new_member) old_set
208 ; if any element in old_set was eq new_member
211 make_list new_member old_set
213 Think about how you'd implement operations like `set_union`,
214 `set_intersection`, and `set_difference` with this implementation of sets.
216 The implementation just described works, and it's the simplest to code.
217 However, it's pretty inefficient. If you had a 100-member set, and you wanted
218 to create a set which had all those 100-members and some possibly new element
219 `e`, you might need to check all 100 members to see if they're equal to `e`
220 before concluding they're not, and returning the new list. And comparing for
221 numeric equality is a moderately expensive operation, in the first place.
223 (You might say, well, what's the harm in just prepending `e` to the list even
224 if it already occurs later in the list. The answer is, if you don't keep track
225 of things like this, it will likely mess up your implementations of
226 `set_difference` and so on. You'll have to do the book-keeping for duplicates
227 at some point in your code. It goes much more smoothly if you plan this from
230 How might we make the implementation more efficient? Well, the *semantics* of
231 sets says that they have no intrinsic order. That means, there's no difference
232 between the set {a,b} and the set {b,a}; whereas there is a difference between
233 the *list* `[a;b]` and the list `[b;a]`. But this semantic point can be respected
234 even if we *implement* sets with something ordered, like list---as we're
235 already doing. And we might *exploit* the intrinsic order of lists to make our
236 implementation of sets more efficient.
238 What we could do is arrange it so that a list that implements a set always
239 keeps in elements in some specified order. To do this, there'd have *to be*
240 some way to order its elements. Since we're talking now about sets of numbers,
241 that's easy. (If we were talking about sets of pairs of numbers, we'd use
242 "lexicographic" ordering, where `(a,b) < (c,d)` iff `a < c or (a == c and b <
245 So, if we were searching the list that implements some set to see if the number
246 `5` belonged to it, once we get to elements in the list that are larger than `5`,
247 we can stop. If we haven't found `5` already, we know it's not in the rest of the
250 This is an improvement, but it's still a "linear" search through the list.
251 There are even more efficient methods, which employ "binary" searching. They'd
252 represent the set in such a way that you could quickly determine whether some
253 element fell in one half, call it the left half, of the structure that
254 implements the set, if it belonged to the set at all. Or that it fell in the
255 right half, it it belonged to the set at all. And then the same sort of
256 determination could be made for whichever half you were directed to. And then
257 for whichever quarter you were directed to next. And so on. Until you either
258 found the element or exhausted the structure and could then conclude that the
259 element in question was not part of the set. These sorts of structures are done
260 using **binary trees** (see below).
263 #Aborting a search through a list#
265 We said that the sorted-list implementation of a set was more efficient than
266 the unsorted-list implementation, because as you were searching through the
267 list, you could come to a point where you knew the element wasn't going to be
268 found. So you wouldn't have to continue the search.
270 If your implementation of lists was, say v1 lists plus the Y-combinator, then
271 this is exactly right. When you get to a point where you know the answer, you
272 can just deliver that answer, and not branch into any further recursion. If
273 you've got the right evaluation strategy in place, everything will work out
276 But what if you're using v3 lists? What options would you have then for
279 Well, suppose we're searching through the list `[5;4;3;2;1]` to see if it
280 contains the number `3`. The expression which represents this search would have
281 something like the following form:
283 ..................<eq? 1 3> ~~>
284 .................. false ~~>
285 .............<eq? 2 3> ~~>
286 ............. false ~~>
287 .........<eq? 3 3> ~~>
291 Of course, whether those reductions actually followed in that order would
292 depend on what reduction strategy was in place. But the result of folding the
293 search function over the part of the list whose head is `3` and whose tail is `[2;
294 1]` will *semantically* depend on the result of applying that function to the
295 more rightmost pieces of the list, too, regardless of what order the reduction
296 is computed by. Conceptually, it will be easiest if we think of the reduction
297 happening in the order displayed above.
299 Well, once we've found a match between our sought number `3` and some member of
300 the list, we'd like to avoid any further unnecessary computations and just
301 deliver the answer `true` as "quickly" or directly as possible to the larger
302 computation in which the search was embedded.
304 With a Y-combinator based search, as we said, we could do this by just not
305 following a recursion branch.
307 But with the v3 lists, the fold is "pre-programmed" to continue over the whole
308 list. There is no way for us to bail out of applying the search function to the
309 parts of the list that have head `4` and head `5`, too.
311 We *can* avoid *some* unneccessary computation. The search function can detect
312 that the result we've accumulated so far during the fold is now `true`, so we
313 don't need to bother comparing `4` or `5` to `3` for equality. That will simplify the
314 computation to some degree, since as we said, numerical comparison in the
315 system we're working in is moderately expensive.
317 However, we're still going to have to traverse the remainder of the list. That
318 `true` result will have to be passed along all the way to the leftmost head of
319 the list. Only then can we deliver it to the larger computation in which the
322 It would be better if there were some way to "abort" the list traversal. If,
323 having found the element we're looking for (or having determined that the
324 element isn't going to be found), we could just immediately stop traversing the
325 list with our answer. **Continuations** will turn out to let us do that.
327 We won't try yet to fully exploit the terrible power of continuations. But
328 there's a way that we can gain their benefits here locally, without yet having
329 a fully general machinery or understanding of what's going on.
331 The key is to recall how our implementations of booleans and pairs worked.
332 Remember that with pairs, we supply the pair "handler" to the pair as *an
333 argument*, rather than the other way around:
341 to get the first element of the pair. Of course you can lift that if you want:
343 <pre><code>extract_fst ≡ \pair. pair (\x y. x)</code></pre>
345 but at a lower level, the pair is still accepting its handler as an argument,
346 rather than the handler taking the pair as an argument. (The handler gets *the
347 pair's elements*, not the pair itself, as arguments.)
349 > *Terminology*: we'll try to use names of the form `get_foo` for handlers, and
350 names of the form `extract_foo` for lifted versions of them, that accept the
351 lists (or whatever data structure we're working with) as arguments. But we may
354 The v2 implementation of lists followed a similar strategy:
356 v2list (\h t. do_something_with_h_and_t) result_if_empty
358 If the `v2list` here is not empty, then this will reduce to the result of
359 supplying the list's head and tail to the handler `(\h t.
360 do_something_with_h_and_t)`.
362 Now, what we've been imagining ourselves doing with the search through the v3
363 list is something like this:
366 larger_computation (search_through_the_list_for_3) other_arguments
368 That is, the result of our search is supplied as an argument (perhaps together
369 with other arguments) to the "larger computation". Without knowing the
370 evaluation order/reduction strategy, we can't say whether the search is
371 evaluated before or after it's substituted into the larger computation. But
372 semantically, the search is the argument and the larger computation is the
373 function to which it's supplied.
375 What if, instead, we did the same kind of thing we did with pairs and v2
376 lists? That is, what if we made the larger computation a "handler" that we
377 passed as an argument to the search?
379 the_search (\search_result. larger_computation search_result other_arguments)
381 What's the advantage of that, you say. Other than to show off how cleverly
384 Well, think about it. Think about the difficulty we were having aborting the
385 search. Does this switch-around offer us anything useful?
389 What if the way we implemented the search procedure looked something like this?
391 At a given stage in the search, we wouldn't just apply some function `f` to the
392 head at this stage and the result accumulated so far (from folding the same
393 function, and a base value, to the tail at this stage)...and then pass the result
394 of that application to the embedding, more leftward computation.
396 We'd *instead* give `f` a "handler" that expects the result of the current
397 stage *as an argument*, and then evaluates to what you'd get by passing that
398 result leftwards up the list, as before.
400 Why would we do that, you say? Just more flamboyant lifting?
402 Well, no, there's a real point here. If we give the function a "handler" that
403 encodes the normal continuation of the fold leftwards through the list, we can
404 also give it other "handlers" too. For example, we can also give it the underlined handler:
407 the_search (\search_result. larger_computation search_result other_arguments)
408 ------------------------------------------------------------------
410 This "handler" encodes the search's having finished, and delivering a final
411 answer to whatever else you wanted your program to do with the result of the
412 search. If you like, at any stage in the search you might just give an argument
413 to *this* handler, instead of giving an argument to the handler that continues
414 the list traversal leftwards. Semantically, this would amount to *aborting* the
415 list traversal! (As we've said before, whether the rest of the list traversal
416 really gets evaluated will depend on what evaluation order is in place. But
417 semantically we'll have avoided it. Our larger computation won't depend on the
418 rest of the list traversal having been computed.)
420 Do you have the basic idea? Think about how you'd implement it. A good
421 understanding of the v2 lists will give you a helpful model.
423 In broad outline, a single stage of the search would look like before, except
424 now f would receive two extra, "handler" arguments.
426 f 3 <result of folding f and z over [2; 1]> <handler to continue folding leftwards> <handler to abort the traversal>
428 `f`'s job would be to check whether `3` matches the element we're searching for
429 (here also `3`), and if it does, just evaluate to the result of passing `true` to
430 the abort handler. If it doesn't, then evaluate to the result of passing
431 `false` to the continue-leftwards handler.
433 In this case, `f` wouldn't need to consult the result of folding `f` and `z` over `[2;
434 1]`, since if we had found the element `3` in more rightward positions of the
435 list, we'd have called the abort handler and this application of `f` to `3` etc
436 would never be needed. However, in other applications the result of folding `f`
437 and `z` over the more rightward parts of the list would be needed. Consider if
438 you were trying to multiply all the elements of the list, and were going to
439 abort (with the result `0`) if you came across any element in the list that was
440 zero. If you didn't abort, you'd need to know what the more rightward elements
441 of the list multiplied to, because that would affect the answer you passed
442 along to the continue-leftwards handler.
444 A **version 5** list encodes the kind of fold operation we're envisaging here, in
445 the same way that v3 (and [v4](/advanced/#index1h1)) lists encoded the simpler fold operation.
446 Roughly, the list `[5;4;3;2;1]` would look like this:
449 \f z continue_leftwards_handler abort_handler.
450 <fold f and z over [4;3;2;1]>
451 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 continue_leftwards_handler abort_handler)
454 ; or, expanding the fold over [4;3;2;1]:
456 \f z continue_leftwards_handler abort_handler.
457 (\continue_leftwards_handler abort_handler.
458 <fold f and z over [3;2;1]>
459 (\result_of_fold_over_321. f 4 result_of_fold_over_321 continue_leftwards_handler abort_handler)
462 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 continue_leftwards_handler abort_handler)
467 Remarks: the `larger_computation` handler should be supplied as both the
468 `continue_leftwards_handler` and the `abort_handler` for the leftmost
469 application, where the head `5` is supplied to `f`; because the result of this
470 application should be passed to the larger computation, whether it's a "fall
471 off the left end of the list" result or it's a "I'm finished, possibly early"
472 result. The `larger_computation` handler also then gets passed to the next
473 rightmost stage, where the head `4` is supplied to `f`, as the `abort_handler` to
474 use if that stage decides it has an early answer.
476 Finally, notice that we don't have the result of applying `f` to `4` etc given as
477 an argument to the application of `f` to `5` etc. Instead, we pass
479 (\result_of_fold_over_4321. f 5 result_of_fold_over_4321 <one_handler> <another_handler>)
481 *to* the application of `f` to `4` as its "continue" handler. The application of `f`
482 to `4` can decide whether this handler, or the other, "abort" handler, should be
483 given an argument and constitute its result.
486 I'll say once again: we're using temporally-loaded vocabulary throughout this,
487 but really all we're in a position to mean by that are claims about the result
488 of the complex expression semantically depending only on this, not on that. A
489 demon evaluator who custom-picked the evaluation order to make things maximally
490 bad for you could ensure that all the semantically unnecessary computations got
491 evaluated anyway. We don't have any way to prevent that. Later,
492 we'll see ways to *semantically guarantee* one evaluation order rather than
493 another. Though even then the demonic evaluation-order-chooser could make it
494 take unnecessarily long to compute the semantically guaranteed result. Of
495 course, in any real computing environment you'll know you're dealing with a
496 fixed evaluation order and you'll be able to program efficiently around that.
498 In detail, then, here's what our v5 lists will look like:
500 let empty = \f z continue_handler abort_handler. continue_handler z in
501 let make_list = \h t. \f z continue_handler abort_handler.
502 t f z (\sofar. f h sofar continue_handler abort_handler) abort_handler in
503 let isempty = \lst larger_computation. lst
505 (\hd sofar continue_handler abort_handler. abort_handler false)
508 ; here's the continue_handler for the leftmost application of f
510 ; here's the abort_handler
511 larger_computation in
512 let extract_head = \lst larger_computation. lst
514 (\hd sofar continue_handler abort_handler. continue_handler hd)
517 ; here's the continue_handler for the leftmost application of f
519 ; here's the abort_handler
520 larger_computation in
521 let extract_tail = ; left as exercise
523 These functions are used like this:
525 let my_list = make_list a (make_list b (make_list c empty) in
526 extract_head my_list larger_computation
528 If you just want to see `my_list`'s head, the use `I` as the
529 `larger_computation`.
531 What we've done here does take some work to follow. But it should be within
532 your reach. And once you have followed it, you'll be well on your way to
533 appreciating the full terrible power of continuations.
535 <!-- (Silly [cultural reference](http://www.newgrounds.com/portal/view/33440).) -->
537 Of course, like everything elegant and exciting in this seminar, [Oleg
538 discusses it in much more
539 detail](http://okmij.org/ftp/Streams.html#enumerator-stream).
543 1. The technique deployed here, and in the v2 lists, and in our implementations
544 of pairs and booleans, is known as **continuation-passing style** programming.
546 2. We're still building the list as a right fold, so in a sense the
547 application of `f` to the leftmost element `5` is "outermost". However,
548 this "outermost" application is getting lifted, and passed as a *handler*
549 to the next right application. Which is in turn getting lifted, and
550 passed to its next right application, and so on. So if you
551 trace the evaluation of the `extract_head` function to the list `[5;4;3;2;1]`,
552 you'll see `1` gets passed as a "this is the head sofar" answer to its
553 `continue_handler`; then that answer is discarded and `2` is
554 passed as a "this is the head sofar" answer to *its* `continue_handler`,
555 and so on. All those steps have to be evaluated to finally get the result
556 that `5` is the outer/leftmost head of the list. That's not an efficient way
557 to get the leftmost head.
559 We could improve this by building lists as left folds when implementing them
560 as continuation-passing style folds. We'd just replace above:
562 let make_list = \h t. \f z continue_handler abort_handler.
563 f h z (\z. t f z continue_handler abort_handler) abort_handler
565 now `extract_head` should return the leftmost head directly, using its `abort_handler`:
567 let extract_head = \lst larger_computation. lst
568 (\hd sofar continue_handler abort_handler. abort_handler hd)
573 3. To extract tails efficiently, too, it'd be nice to fuse the apparatus developed
574 in these v5 lists with the ideas from [v4](/advanced/#index1h1) lists.
575 But that also is left as an exercise.