5 **Chris:** I'll be working on this page heavily until 11--11:30 or so. Sorry not to do it last night, I crashed.
12 #Recursion: fixed points in the Lambda Calculus#
14 Sometimes when you type in a web search, Google will suggest
15 alternatives. For instance, if you type in "Lingusitics", it will ask
16 you "Did you mean Linguistics?". But the engineers at Google have
17 added some playfulness to the system. For instance, if you search for
18 "anagram", Google asks you "Did you mean: nag a ram?" And if you
19 [search for "recursion"](http://www.google.com/search?q=recursion), Google asks: "Did you mean: recursion?"
21 ##What is the "rec" part of "letrec" doing?##
23 How could we compute the length of a list? Without worrying yet about what Lambda Calculus encoding we're using for the list, the basic idea is to define this recursively:
25 > the empty list has length 0
27 > any non-empty list has length 1 + (the length of its tail)
29 In OCaml, you'd define that like this:
31 let rec length = fun xs ->
32 if xs = [] then 0 else 1 + length (List.tl xs)
33 in ... (* here you go on to use the function "length" *)
35 In Scheme you'd define it like this:
37 (letrec [(length (lambda (xs)
39 (+ 1 (length (cdr xs))) )))]
40 ... ; here you go on to use the function "length"
43 Some comments on this:
45 1. `null?` is Scheme's way of saying `empty?`. That is, `(null? xs)` returns true (which Scheme writes as `#t`) iff `xs` is the empty list (which Scheme writes as `'()` or `(list)`).
47 2. `cdr` is function that gets the tail of a Scheme list. (By definition, it's the function for getting the second member of a [[dotted pair|week3_unit#imp]]. As we discussed in notes for last week, it just turns out to return the tail of a list because of the particular way Scheme implements lists.)
49 3. We alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
52 The main question for us to dwell on here is: What are the `let rec` in the OCaml code and the `letrec` in the Scheme code?
54 Answer: These work a lot like `let` expressions, except that they let you use the variable `length` *inside* the body of the function being bound to it --- with the understanding that it will there be bound to *the same function* that you're *then* in the process of binding `length` to. So our recursively-defined function works the way we'd expect it to. Here is OCaml:
56 let rec length = fun xs ->
57 if xs = [] then 0 else 1 + length (List.tl xs)
59 (* this evaluates to 2 *)
63 (letrec [(length (lambda (xs)
65 (+ 1 (length (cdr xs))) )))]
66 (length (list 20 30)))
69 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
71 let length = fun xs ->
72 if xs = [] then 0 else 1 + length (List.tl xs)
74 (* fails with error "Unbound value length" *)
78 (let* [(length (lambda (xs)
80 (+ 1 (length (cdr xs))) )))]
81 (length (list 20 30)))
82 ; fails with error "reference to undefined identifier: length"
84 Why? Because we said that constructions of this form:
90 really were just another way of saying:
94 and so the occurrences of `length` in A *aren't bound by the `\length` that wraps B*. Those occurrences are free.
96 We can verify this by wrapping the whole expression in a more outer binding of `length` to some other function, say the constant function from any list to the integer `99`:
98 let length = fun xs -> 99
99 in let length = fun xs ->
100 if xs = [] then 0 else 1 + length (List.tl xs)
102 (* evaluates to 1 + 99 *)
104 Here the use of `length` in `1 + length (List.tl xs)` can clearly be seen to be bound by the outermost `let`.
106 And indeed, if you tried to define `length` in the Lambda Calculus, how would you do it?
108 \xs. (empty? xs) 0 (succ (length (tail xs)))
110 We've defined all of `empty?`, `0`, `succ`, and `tail` in earlier discussion. But what about `length`? That's not yet defined! In fact, that's the very formula we're trying here to specify.
112 What we really want to do is something like this:
114 \xs. (empty? xs) 0 (succ (... (tail xs)))
116 where this very same formula occupies the `...` position:
118 \xs. (empty? xs) 0 (succ (\xs. (empty? xs) 0 (succ (... (tail xs)))
121 but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
123 So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?
125 1. OCaml and Scheme do it using a trick. Well, not a trick. Actually an impressive, conceptually deep technique, which we haven't yet developed. Since we want to build up all the techniques we're using by hand, then, we shouldn't permit ourselves to rely on `let rec` or `letrec` until we thoroughly understand what's going on under the hood.
127 2. If you tried this in Scheme:
129 (define length (lambda (xs)
131 (+ 1 (length (cdr xs))) )))
133 (length (list 20 30))
135 You'd find that it works! This is because `define` in Scheme is really shorthand for `letrec`, not for plain `let` or `let*`. So we should regard this as cheating, too.
137 3. In fact, it *is* possible to define the `length` function in the Lambda Calculus despite these obstacles, without yet knowing how to implement `letrec` in general. We've already seen how to do it, using our right-fold (or left-fold) encoding for lists, and exploiting their internal structure. Those encodings take a function and a seed value and returns the result of folding that function over the list, with that seed value. So we could use this as a definition of `length`:
139 \xs. xs (\x sofar. succ sofar) 0
141 What's happening here? We start with the value `0`, then we apply the function `\x sofar. succ sofar` to the two arguments <code>x<sub>n</sub></code> and `0`, where <code>x<sub>n</sub></code> is the last element of the list. This gives us `succ 0`, or `1`. That's the value we've accumulated "so far." Then we go apply the function `\x sofar. succ sofar` to the two arguments <code>x<sub>n-1</sub></code> and the value `1` that we've accumulated "so far." This gives us `2`. We continue until we get to the start of the list. The value we've then built up "so far" will be the length of the list.
143 We can use similar techniques to define many recursive operations on
144 lists and numbers. The reason we can do this is that our
145 fold-based encoding of lists, and Church's encodings of
146 numbers, have a internal structure that *mirrors* the common recursive
147 operations we'd use lists and numbers for. In a sense, the recursive
148 structure of the `length` operation is built into the data
149 structure we are using to represent the list. The non-recursive
150 definition of length, above, exploits this embedding of the recursion into
153 This illustrates what will be one of the recurring themes of the course: using data structures to
154 encode the state of some recursive operation. See our discussions later this semester of the
155 [[zipper]] technique, and [[defunctionalization]].
157 As we've seen, it does take some ingenuity to define functions like `tail` or `pred` for our right-fold encoding of lists. However it can be done. (And it's not *that* difficult.) Given those functions, we can go on to define other functions like numeric equality, subtraction, and so on, just by exploiting the structure already present in our implementation of lists and numbers.
159 With sufficient ingenuity, a great many functions can be defined in the same way. For example, the factorial function is straightforward. The function which returns the *n*th term in the Fibonacci series is a bit more difficult, but also achievable.
161 ##Some functions require full-fledged recursive definitions##
163 However, some computable functions are just not definable in this
164 way. We can't, for example, define a function that tells us, for
165 whatever function `f` we supply it, what is the smallest natural number `x`
166 where `f x` is `true` (even if `f` itself is a function we do already know how to define).
168 Neither do the resources we've so far developed suffice to define the
169 [[!wikipedia Ackermann function]]. In OCaml:
171 let rec A = fun (m,n) ->
173 else if n = 0 then A(m-1,1)
174 else A(m-1, A(m,n-1));;
180 A(4,y) = 2^(2^(2^...2)) (* where there are y+3 2s *) - 3
183 Many simpler functions always *could* be defined using the resources we've so far developed, although those definitions won't always be very efficient or easily intelligible.
185 But functions like the Ackermann function require us to develop a more general technique for doing recursion --- and having developed it, it will often be easier to use it even in the cases where, in principle, we didn't have to.
187 ##Using fixed-point combinators to define recursive functions##
191 In mathematics, a **fixed point** of a function `f` is any value `ξ`
192 such that `f ξ` is equivalent to `ξ`. For example,
193 consider the squaring function `square` that maps natural numbers to their squares.
194 `square 2 = 4`, so `2` is not a fixed point. But `square 1 = 1`, so `1` is a
195 fixed point of the squaring function. (Can you think of another?)
197 There are many beautiful theorems guaranteeing the existence of a
198 fixed point for various classes of interesting functions. For
199 instance, imagine that you are looking at a map of Manhattan, and you
200 are standing somewhere in Manhattan. Then the [[!wikipedia Brouwer
201 fixed-point theorem]] guarantees that there is a spot on the map that is
202 directly above the corresponding spot in Manhattan. It's the spot on the map
203 where the blue you-are-here dot should go.
205 Whether a function has a fixed point depends on the domain of arguments
206 it is defined for. For instance, consider the successor function `succ`
207 that maps each natural number to its successor. If we limit our
208 attention to the natural numbers, then this function has no fixed
209 point. (See the discussion below concerning a way of understanding
210 the successor function on which it *does* have a fixed point.)
212 In the Lambda Calculus, we say a fixed point of a term `f` is any *term* `ξ` such that:
216 This is a bit different than the general mathematical definition, in that here we're saying it is *terms* that are fixed points, not *values*. We like to think that some lambda terms represent values, such as our term `\f z. z` representing the numerical value zero (and also the truth-value false, and also the empty list... on the other hand, we never did explicitly agree that those three values are all the same thing, did we?). But some terms in the Lambda Calculus don't even have a normal form. We don't want to count them as *values*. Yet the way we're proposing to use the notion of a fixed point here, they too are allowed to be fixed points, and to have fixed points of their own.
218 Note that `M <~~> N` doesn't entail that `M` and `N` have a normal form (though if they do, they will have the same normal form). It just requires that there be some term that they both reduce to. It may be that *that* term itself never stops being reducible.
220 You should be able to immediately provide a fixed point of the
221 identity combinator `I`. In fact, you should be able to provide a
222 whole bunch of distinct fixed points.
224 With a little thought, you should be able to provide a fixed point of
225 the false combinator, `KI`. Here's how to find it: recall that `KI`
226 throws away its first argument, and always returns `I`. Therefore, if
227 we give it `I` as an argument, it will throw away the argument, and
228 return `I`. So `KII` ~~> `I`, which is all it takes for `I` to qualify as a
231 What about `K`? Does it have a fixed point? You might not think so,
232 after trying on paper for a while.
234 However, it's a theorem of the Lambda Calculus that *every* lambda term has
235 a fixed point. Even bare variables like `x`! In fact, they will have infinitely many, non-equivalent
236 fixed points. And we don't just know that they exist: for any given
237 formula, we can explicit define many of them.
239 (As we mentioned, even the formula that you're using the define
240 the successor function will have a fixed point. Isn't that weird? There's some `ξ` such that it is equivalent to `succ ξ`?
241 Think about how it might be true. We'll return to this point below.)
244 ###How fixed points help define recursive functions###
246 Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said "What we really want to do is something like this:
248 \xs. (empty? xs) 0 (succ (... (tail xs)))
250 where this very same formula occupies the `...` position."
252 Imagine replacing the `...` with some expression `LENGTH` that computes the
253 length function. Then we have
255 \xs. (empty? xs) 0 (succ (LENGTH (tail xs)))
257 At this point, we have a definition of the length function, though
258 it's not complete, since we don't know what value to use for the
259 symbol `LENGTH`. Technically, it has the status of an unbound
262 Imagine now binding the mysterious variable, and calling the resulting
265 h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
267 Now we have no unbound variables, and we have complete non-recursive
268 definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
270 So `h` takes an argument, and returns a function that accurately
271 computes the length of a list --- as long as the argument we supply is
272 already the length function we are trying to define. (Dehydrated
273 water: to reconstitute, just add water!)
275 Here is where the discussion of fixed points becomes relevant. Saying
276 that `h` is looking for an argument (call it `LENGTH`) that has the same
277 behavior as the result of applying `h` to `LENGTH` is just another way of
278 saying that we are looking for a fixed point for `h`.
282 Replacing `h` with its definition, we have
284 (\xs. (empty? xs) 0 (succ (LENGTH (tail xs)))) <~~> LENGTH
286 If we can find a value for `LENGTH` that satisfies this constraint, we'll
287 have a function we can use to compute the length of an arbitrary list.
288 All we have to do is find a fixed point for `h`.
290 The strategy we will present will turn out to be a general way of
291 finding a fixed point for any lambda term.
294 ##Deriving Y, a fixed point combinator##
296 How shall we begin? Well, we need to find an argument to supply to
297 `h`. The argument has to be a function that computes the length of a
298 list. The function `h` is *almost* a function that computes the
299 length of a list. Let's try applying `h` to itself. It won't quite
300 work, but examining the way in which it fails will lead to a solution.
302 h h <~~> \xs. (empty? xs) 0 (succ (h (tail xs)))
304 The problem is that in the subexpression `h (tail list)`, we've
305 applied `h` to a list, but `h` expects as its first argument the
308 So let's adjust `h`, calling the adjusted function `H`. (We'll use `u` as the variable
309 that expects to be bound to `H`'s fixed point, rather than `length`. This will make it easier
310 to discuss generalizations of this strategy.)
312 H ≡ \u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))
314 This is the key creative step. Instead of applying `u` to a list, we
315 apply it first to itself. After applying `u` to an argument, it's
316 ready to apply to a list, so we've solved the problem noted above with `h (tail list)`.
317 We're not done yet, of course; we don't yet know what argument to give
318 to `H` that will behave in the desired way.
320 So let's reason about `H`. What exactly is H expecting as its first
321 argument? Based on the excerpt `(u u) (tail xs)`, it appears that
322 `H`'s argument, `u`, should be a function that is ready to take itself
323 as an argument, and that returns a function that takes a list as an
324 argument. `H` itself fits the bill:
326 H H <~~> (\u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))) H
327 <~~> \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
328 ≡ \xs. (empty? xs) 0 (succ (
329 (\xs. (empty? xs) 0 (succ ((H H) (tail xs))))
331 <~~> \xs. (empty? xs) 0 (succ (
332 (empty? (tail xs)) 0 (succ ((H H) (tail (tail xs)))) ))
336 How does the recursion work?
337 We've defined `H` in such a way that `H H` turns out to be the length function.
338 In order to evaluate `H H`, we substitute `H` into the body of the
339 lambda term. Inside the lambda term, once the substitution has
340 occurred, we are once again faced with evaluating `H H`. And so on.
342 We've got the potentially infinite regress we desired, defined in terms of a
343 finite lambda term with no undefined symbols.
345 Since `H H` turns out to be the length function, we can think of `H`
346 by itself as half of the length function (which is why we called it
347 `H`, of course). Can you think up a recursion strategy that involves
348 "dividing" the recursive function into equal thirds `T`, such that the
349 length function <~~> `T T T`?
351 We've starting with a particular recursive definition, and arrived at
352 a fixed point for that definition.
353 What's the general recipe?
355 1. Start with any recursive definition `h` that takes itself as an arg: `h ≡ \self ... self ...`
356 2. Next, define `H ≡ \u . h (u u)`
357 3. Then compute `H H ≡ ((\u . h (u u)) (\u . h (u u)))`
358 4. That's the fixed point, the recursive function we're trying to define
360 So here is a general method for taking an arbitrary `h`-style recursive function
361 and returning a fixed point for that function:
363 Y ≡ \h. (\u. h (u u)) (\u. h (u u))
367 Y h ≡ (\h. (\u. h (u u)) (\u. h (u u))) h
368 ~~> (\u. h (u u)) (\u. h (u u))
369 ~~> h ((\u. h (u u)) (\u. h (u u)))
371 But the argument of `h` in the last line is just the same as the second line, which `<~~> Y h`. So the last line `<~~> h (Y h)`. In other words, `Y h <~~> h (Y h)`. So by definition, `Y h` is a fixed point for `h`.
375 ##Coming at it another way##
380 ##A fixed point for K?##
382 Let's do one more example to illustrate. We'll do `K`, since we
383 wondered above whether it had a fixed point.
385 Before we begin, we can reason a bit about what the fixed point must
386 be like. We're looking for a fixed point for `K`, i.e., `\xy.x`. The term `K`
387 ignores its second argument. That means that no matter what we give
388 `K` as its first argument, the result will ignore the next argument
389 (that is, `KX` ignores its first argument, no matter what `X` is). So
390 if `KX <~~> X`, `X` had also better ignore its first argument. But we
391 also have `KX ≡ (\xy.x)X ~~> \y.X`. This means that if `X` ignores
392 its first argument, then `\y.X` will ignore its first two arguments.
393 So once again, if `KX <~~> X`, `X` also had better ignore (at least) its
394 first two arguments. Repeating this reasoning, we realize that `X`
395 must be a function that ignores as many arguments as you give it.
396 Our expectation, then, is that our recipe for finding fixed points
397 will build us a term that somehow manages to ignore arbitrarily many arguments.
400 H ≡ \u.h(uu) ≡ \u.(\xy.x)(uu) ~~> \uy.uu
401 H H ≡ (\uy.uu)(\uy.uu) ~~> \y.(\uy.uu)(\uy.uu)
403 Let's check that it is in fact a fixed point:
405 K(H H) ≡ (\xy.x)((\uy.uu)(\uy.uu)
406 ~~> \y.(\uy.uu)(\uy.uu)
408 Yep, `H H` and `K(H H)` both reduce to the same term.
410 To see what this fixed point does, let's reduce it a bit more:
412 H H ≡ (\uy.uu)(\uy.uu)
413 ~~> \y.(\uy.uu)(\uy.uu)
414 ~~> \yy.(\uy.uu)(\uy.uu)
415 ~~> \yyy.(\uy.uu)(\uy.uu)
417 Sure enough, this fixed point ignores an endless, infinite series of
418 arguments. It's a write-only memory, a black hole.
420 Now that we have one fixed point, we can find others, for instance,
423 ~~> \y.(\uy.uuu)(\uy.uuu)(\uy.uuu)
424 ~~> \yy.(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)
425 ~~> \yyy.(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)(\uy.uuu)
427 Continuing in this way, you can now find an infinite number of fixed
428 points, all of which have the crucial property of ignoring an infinite
431 ##What is a fixed point for the successor function?##
433 As we've seen, the recipe just given for finding a fixed point worked
434 great for our `h`, which we wrote as a definition for the length
435 function. But the recipe doesn't make any assumptions about the
436 internal structure of the function it works with. That means it can
437 find a fixed point for literally any function whatsoever.
439 In particular, what could the fixed point for the
440 successor function possibly be like?
442 Well, you might think, only some of the formulas that we might give to the `succ` as arguments would really represent numbers. If we said something like:
446 who knows what we'd get back? Perhaps there's some non-number-representing formula such that when we feed it to `succ` as an argument, we get the same formula back.
448 Yes! That's exactly right. And which formula this is will depend on the particular way you've implemented the succ function.
450 One (by now obvious) upshot is that the recipes that enable us to name
451 fixed points for any given formula aren't *guaranteed* to give us
452 *terminating* fixed points. They might give us formulas `ξ` such that
453 neither `ξ` nor `f ξ` have normal forms. (Indeed, what they give us
454 for the `square` function isn't any of the Church numerals, but is
455 rather an expression with no normal form.) However, if we take care we
456 can ensure that we *do* get terminating fixed points. And this gives
457 us a principled, fully general strategy for doing recursion. It lets
458 us define even functions like the Ackermann function, which were until
459 now out of our reach. It would also let us define list
460 functions on [[the encodings we discussed last week|week3_lists#other-lists]], where it
461 wasn't always clear how to force the computation to "keep going."
463 ###Varieties of fixed-point combinators###
465 OK, so how do we make use of this?
467 Many fixed-point combinators have been discovered. (And some
468 fixed-point combinators give us models for building infinitely many
469 more, non-equivalent fixed-point combinators.)
473 Θ′ ≡ (\u h. h (\n. u u h n)) (\u h. h (\n. u u h n))
474 Y′ ≡ \h. (\u. h (\n. u u n)) (\u. h (\n. u u n))
476 Applying either of these to a term `h` gives a fixed point `ξ` for `h`, meaning that `h ξ` <~~> `ξ`. `Θ′` has the advantage that `h (Θ′ h)` really *reduces to* `Θ′ h`. Whereas `h (Y′ h)` is only *convertible with* `Y′ h`; that is, there's a common formula they both reduce to. For most purposes, though, either will do.
478 You may notice that both of these formulas have eta-redexes inside them: why can't we simplify the two `\n. u u h n` inside `Θ′` to just `u u h`? And similarly for `Y′`?
480 Indeed you can, getting the simpler:
482 Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
483 Y ≡ \h. (\u. h (u u)) (\u. h (u u))
485 I stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\self. BODY)` and `Y (\self. BODY)` will in general not terminate. But evaluation of the eta-unreduced primed versions will.
487 Of course, if you define your `\self. BODY` stupidly, your formula will never terminate. For example, it doesn't matter what fixed point combinator you use for `Ψ` in:
489 Ψ (\self. \n. self n)
491 When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
495 where `self` is equivalent to the very formula `\n. self n` that contains it. So the evaluation will proceed:
503 You've written an infinite loop!
505 However, when we evaluate the application of our:
507 Ψ (\self (\xs. (empty? xs) 0 (succ (self (tail xs))) ))
509 to some list, we're not going to go into an infinite evaluation loop of that sort. At each cycle, we're going to be evaluating the application of:
511 \xs. (empty? xs) 0 (succ (self (tail xs)))
513 to *the tail* of the list we were evaluating its application to at the previous stage. Assuming our lists are finite (and the encodings we've been using so far don't permit otherwise), at some point one will get a list whose tail is empty, and then the evaluation of that formula to that tail will return `0`. So the recursion eventually bottoms out in a base value.
515 ##Fixed-point Combinators Are a Bit Intoxicating##
517 [[tatto|/images/y-combinator-fixed.jpg]]
519 There's a tendency for people to say "Y-combinator" to refer to fixed-point combinators generally. We'll probably fall into that usage ourselves. Speaking correctly, though, the Y-combinator is only one of many fixed-point combinators.
521 We used `Ψ` above to stand in for an arbitrary fixed-point combinator. We don't know of any broad conventions for this. But this seems a useful one.
523 As we said, there are many other fixed-point combinators as well. For example, Jan Willem Klop pointed out that if we define `L` to be:
525 \a b c d e f g h i j k l m n o p q s t u v w x y z r. (r (t h i s i s a f i x e d p o i n t c o m b i n a t o r))
527 then this is a fixed-point combinator:
529 L L L L L L L L L L L L L L L L L L L L L L L L L L
532 ##Watching Y in action##
534 For those of you who like to watch ultra slow-mo movies of bullets
535 piercing apples, here's a stepwise computation of the application of a
536 recursive function. We'll use a function `sink`, which takes one
537 argument. If the argument is boolean true (i.e., `\x y.x`), it
538 returns itself (a copy of `sink`); if the argument is boolean false
539 (`\x y. y`), it returns `I`. That is, we want the following behavior:
542 sink true false ~~> I
543 sink true true false ~~> I
544 sink true true true false ~~> I
546 So we make `sink = Y (\s b. b s I)`:
550 3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) false
551 4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) false
552 5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
553 6. (\b. b [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] I) false
554 7. false [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] I
555 --------------------------------------------
558 So far so good. The crucial thing to note is that as long as we
559 always reduce the outermost redex first, we never have to get around
560 to computing the underlined redex: because `false` ignores its first
561 argument, we can throw it away unreduced.
563 Now we try the next most complex example:
566 2. Y (\sb.bsI) true false
567 3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) true false
568 4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) true false
569 5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] true false
570 6. (\b.b [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I) true false
571 7. true [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I false
572 8. [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
574 We've now arrived at line (4) of the first computation, so the result
577 You should be able to see that `sink` will consume as many `true`s as
578 we throw at it, then turn into the identity function after it
579 encounters the first `false`.
581 The key to the recursion is that, thanks to `Y`, the definition of
582 `sink` contains within it the ability to fully regenerate itself as
583 many times as is necessary. The key to *ending* the recursion is that
584 the behavior of `sink` is sensitive to the nature of the input: if the
585 input is the magic function `false`, the self-regeneration machinery
586 will be discarded, and the recursion will stop.
588 That's about as simple as recursion gets.
590 ##Application to the truth teller/liar paradoxes##
592 ###Base cases, and their lack###
594 As any functional programmer quickly learns, writing a recursive
595 function divides into two tasks: figuring out how to handle the
596 recursive case, and remembering to insert a base case. The
597 interesting and enjoyable part is figuring out the recursive pattern,
598 but the base case cannot be ignored, since leaving out the base case
599 creates a program that runs forever. For instance, consider computing
600 a factorial: `n!` is `n * (n-1) * (n-2) * ... * 1`. The recursive
601 case says that the factorial of a number `n` is `n` times the
602 factorial of `n-1`. But if we leave out the base case, we get
604 3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * -1! ...
606 That's why it's crucial to declare that `0!` = `1`, in which case the
607 recursive rule does not apply. In our terms,
609 fact ≡ Y (\fact n. zero? n 1 (fact (predecessor n)))
611 If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.
613 Curry originally called `Y` the "paradoxical" combinator, and discussed
614 it in connection with certain well-known paradoxes from the philosophy
615 literature. The truth-teller paradox has the flavor of a recursive
616 function without a base case:
618 (1) This sentence is true.
620 If we assume that the complex demonstrative "this sentence" can refer
621 to (1), then the proposition expressed by (1) will be true just in
622 case the thing referred to by *this sentence* is true. Thus (1) will
623 be true just in case (1) is true, and (1) is true just in case (1) is
624 true, and so on. If (1) is true, then (1) is true; but if (1) is not
625 true, then (1) is not true.
627 Without pretending to give a serious analysis of the paradox, let's
628 assume that sentences can have for their meaning boolean functions
629 like the ones we have been working with here. Then the sentence *John
630 is John* might denote the function `\x y. x`, our `true`.
632 <!-- Jim says: I haven't yet followed the next chunk to my satisfaction -->
634 Then (1) denotes a function from whatever the referent of *this
635 sentence* is to a boolean. So (1) denotes `\f. f true false`, where
636 the argument `f` is the referent of *this sentence*. Of course, if
637 `f` is a boolean, `f true false <~~> f`, so for our purposes, we can
638 assume that (1) denotes the identity function `I`.
640 If we use (1) in a context in which *this sentence* refers to the
641 sentence in which the demonstrative occurs, then we must find a
642 meaning `m` such that `I m = I`. But since in this context `m` is the
643 same as the meaning `I`, so we have `m = I m`. In other words, `m` is
644 a fixed point for the denotation of the sentence (when used in the
645 appropriate context).
647 That means that in a context in which *this sentence* refers to the
648 sentence in which it occurs, the sentence denotes a fixed point for
649 the identity function. Here's a fixed point for the identity
653 (\h. (\u. h (u u)) (\u. h (u u))) I ~~>
654 (\u. I (u u)) (\u. I (u u))) ~~>
655 (\u. (u u)) (\u. (u u))) ≡
659 Oh. Well! That feels right. The meaning of *This sentence is true*
660 in a context in which *this sentence* refers to the sentence in which
661 it occurs is `Ω`, our prototypical infinite loop...
663 What about the liar paradox?
665 (2) This sentence is false.
667 Used in a context in which *this sentence* refers to the utterance of
668 (2) in which it occurs, (2) will denote a fixed point for `\f. neg f`,
669 or `\f l r. f r l`, which is the `C` combinator. So in such a
670 context, (2) might denote
673 (\h. (\u. h (u u)) (\u. h (u u))) C
674 (\u. C (u u)) (\u. C (u u)))
675 C ((\u. C (u u)) (\u. C (u u)))
676 C (C ((\u. C (u u)) (\u. C (u u))))
677 C (C (C ((\u. C (u u)) (\u. C (u u)))))
680 And infinite sequence of `C`s, each one negating the remainder of the
681 sequence. Yep, that feels like a reasonable representation of the
684 See Barwise and Etchemendy's 1987 OUP book, [The Liar: an essay on
685 truth and circularity](http://tinyurl.com/2db62bk) for an approach
686 that is similar, but expressed in terms of non-well-founded sets
687 rather than recursive functions.
691 You should be cautious about feeling too comfortable with
692 these results. Thinking again of the truth-teller paradox, yes,
693 `Ω` is *a* fixed point for `I`, and perhaps it has
694 some privileged status among all the fixed points for `I`, being the
695 one delivered by `Y` and all (though it is not obvious why `Y` should have
696 any special status, versus other fixed point combinators).
698 But one could ask: look, literally every formula is a fixed point for
703 for any choice of `X` whatsoever.
705 So the `Y` combinator is only guaranteed to give us one fixed point out
706 of infinitely many --- and not always the intuitively most useful
707 one. (For instance, the squaring function `\x. mul x x` has `0` as a fixed point,
708 since `0 * 0 = 0`, and `1` as a fixed point, since `1 * 1 = 1`, but `Y
709 (\x. mul x x)` doesn't give us `0` or `1`.) So with respect to the
710 truth-teller paradox, why in the reasoning we've
711 just gone through should we be reaching for just this fixed point at
714 One obstacle to thinking this through is the fact that a sentence
715 normally has only two truth values. We might consider instead a noun
718 (3) the entity that this noun phrase refers to
720 The reference of (3) depends on the reference of the embedded noun
721 phrase *this noun phrase*. It's easy to see that any object is a
722 fixed point for this referential function: if this pen cap is the
723 referent of *this noun phrase*, then it is the referent of (3), and so
726 The chameleon nature of (3), by the way (a description that is equally
727 good at describing any object), makes it particularly well suited as a
728 gloss on pronouns such as *it*. In the system of
729 [Jacobson 1999](http://www.springerlink.com/content/j706674r4w217jj5/),
730 pronouns denote (you guessed it!) identity functions...
732 Ultimately, in the context of this course, these paradoxes are more
733 useful as a way of gaining leverage on the concepts of fixed points
734 and recursion, rather than the other way around.