3 Sometimes when you type in a web search, Google will suggest
4 alternatives. For instance, if you type in "Lingusitics", it will ask
5 you "Did you mean Linguistics?". But the engineers at Google have
6 added some playfulness to the system. For instance, if you search for
7 "anagram", Google asks you "Did you mean: nag a ram?" And if you
8 [search for "recursion"](http://www.google.com/search?q=recursion), Google asks: "Did you mean: recursion?"
10 ## What is the "rec" part of "letrec" doing? ##
12 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:
14 > the empty list has length 0
16 > any non-empty list has length 1 + (the length of its tail)
18 In OCaml, you'd define that like this:
20 let rec length = fun xs ->
22 else 1 + length (List.tl xs)
23 in ... (* here you go on to use the function "length" *)
25 In Scheme you'd define it like this:
27 (letrec [(length (lambda (xs)
29 (+ 1 (length (cdr xs))) )))]
30 ... ; here you go on to use the function "length"
33 Some comments on this:
35 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)`).
37 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.) `List.tl` is the function that gets the tail of an OCaml list.
39 3. We alternate between `[ ]`s and `( )`s in the Scheme code just to make it more readable. These have no syntactic difference.
42 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?
44 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:
46 let rec length = fun xs ->
48 else 1 + length (List.tl xs)
50 (* this evaluates to 2 *)
54 (letrec [(length (lambda (xs)
56 (+ 1 (length (cdr xs))) )))]
57 (length (list 20 30)))
60 If you instead use an ordinary `let` (or `let*`), here's what would happen, in OCaml:
62 let length = fun xs ->
64 else 1 + length (List.tl xs)
66 (* fails with error "Unbound value length" *)
70 (let* [(length (lambda (xs)
72 (+ 1 (length (cdr xs))) )))]
73 (length (list 20 30)))
74 ; fails with error "reference to undefined identifier: length"
76 Why? Because we said that constructions of this form:
82 really were just another way of saying:
86 and so the occurrences of `length` in A *aren't bound by the `\length` that wraps B*. Those occurrences are free.
88 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`:
90 let length = fun xs -> 99
91 in let length = fun xs ->
93 else 1 + length (List.tl xs)
95 (* evaluates to 1 + 99 *)
97 Here the use of `length` in `1 + length (List.tl xs)` can clearly be seen to be bound by the outermost `let`.
99 And indeed, if you tried to define `length` in the Lambda Calculus, how would you do it?
101 \xs. (empty? xs) 0 (succ (length (tail xs)))
103 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.
105 What we really want to do is something like this:
107 \xs. (empty? xs) 0 (succ ((...) (tail xs)))
109 where this very same formula occupies the `...` position:
111 \xs. (empty? xs) 0 (succ (
112 \xs. (empty? xs) 0 (succ ((...) (tail xs)))
115 but as you can see, we'd still have to plug the formula back into itself again, and again, and again... No dice.
117 So how could we do it? And how do OCaml and Scheme manage to do it, with their `let rec` and `letrec`?
119 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.
121 2. If you tried this in Scheme:
123 (define length (lambda (xs)
125 (+ 1 (length (cdr xs))) )))
127 (length (list 20 30))
129 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.
131 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`:
133 \xs. xs (\x sofar. succ sofar) 0
135 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.
137 We can use similar techniques to define many recursive operations on
138 lists and numbers. The reason we can do this is that our
139 fold-based encoding of lists, and Church's encodings of
140 numbers, have a internal structure that *mirrors* the common recursive
141 operations we'd use lists and numbers for. In a sense, the recursive
142 structure of the `length` operation is built into the data
143 structure we are using to represent the list. The non-recursive
144 definition of length, above, exploits this embedding of the recursion into
147 This illustrates what will be one of the recurring themes of the course: using data structures to
148 encode the state of some recursive operation. See our discussions later this semester of the
149 [[zipper]] technique, and [[defunctionalization]].
151 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.
153 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.
155 ## Some functions require full-fledged recursive definitions ##
157 However, some computable functions are just not definable in this
158 way. We can't, for example, define a function that tells us, for
159 whatever function `f` we supply it, what is the smallest natural number `x`
160 where `f x` is `true` (even if `f` itself is a function we do already know how to define).
162 Neither do the resources we've so far developed suffice to define the
163 [[!wikipedia Ackermann function]]. In OCaml:
165 let rec A = fun (m,n) ->
167 else if n = 0 then A(m-1,1)
168 else A(m-1, A(m,n-1));;
174 A(4,y) = 2^(2^(2^...2)) (* where there are y+3 2s *) - 3
177 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.
179 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.
181 The example used to illustrate this in Chapter 9 of *The Little Schemer* is a function `looking` where:
183 (looking '(6 2 4 caviar 5 7 3))
185 returns `#t`, because if we follow the path from the head of the list argument, `6`, to the sixth element of the list, `7` (the authors of that book count positions starting from 1, though generally Scheme follows the convention of counting positions starting from 0), and then proceed to the seventh element of the list, `3`, and then proceed to the third element of the list, `4`, and the proceed to the fourth element of the list, we find the `'caviar` we are looking for. On other hand, if we say:
187 (looking '(6 2 grits caviar 5 7 3))
189 our path will take us from `6` to `7` to `3` to `grits`, which is not a number but not the `'caviar` we were looking for either. So this returns `#f`. It's not clear how to define such functions without recourse to something like `letrec` or `define`, or the techniques developed below (and also in that chapter of *The Little Schemer*).
191 *The Little Schemer* also mentions the Ackermann function, as well as the interesting [[!wikipedia Collatz conjecture]]. They also point out that functions like their `looking` never return any value --- neither `#t` nor `#f` --- for some arguments, as in the example:
193 (looking '(7 1 2 caviar 5 6 3))
195 Here our path takes us from `7` to `3` to `2` to `1` back to `7`, and the cycle repeats. So in this case, the `looking` function never returns any value.
197 We've already tacitly been dealing with functions that we assumed to be defined only for expressions representing booleans, or only for expressions representing numbers. But in all such cases we could specify in advance what the intended domain of the function was. With examples like the above, it's not clear how to specify the domain in advance, in such a way that our function will still give a definite result for every argument in the domain. Instead, the capacity for fully general recursion brings with it also the downside that some functions will be only **partially defined**, even over restricted domains we're able to define in advance. We will see more extreme examples of this below.
199 (Being only definable with the power of fully general recursion doesn't by itself render you only partially defined: the Ackermann function is total. The downside is rather that there's no way to let fully general recursion in, while limiting its use to just the cases where a definite value will be returned for every argument.)
202 ## Using fixed-point combinators to define recursive functions ##
206 In mathematics, a **fixed point** of a function `f` is any value `ξ`
207 such that `f ξ` is equivalent to `ξ`. For example,
208 consider the squaring function `square` that maps natural numbers to their squares.
209 `square 2 = 4`, so `2` is not a fixed point. But `square 1 = 1`, so `1` is a
210 fixed point of the squaring function. (Can you think of another?)
212 There are many beautiful theorems guaranteeing the existence of a
213 fixed point for various classes of interesting functions. For
214 instance, imagine that you are looking at a map of Manhattan, and you
215 are standing somewhere in Manhattan. Then the [[!wikipedia Brouwer
216 fixed-point theorem]] guarantees that there is a spot on the map that is
217 directly above the corresponding spot in Manhattan. It's the spot on the map
218 where the blue you-are-here dot should go.
220 Whether a function has a fixed point depends on the domain of arguments
221 it is defined for. For instance, consider the successor function `succ`
222 that maps each natural number to its successor. If we limit our
223 attention to the natural numbers, then this function has no fixed
224 point. (See the discussion below concerning a way of understanding
225 the successor function on which it *does* have a fixed point.)
227 In the Lambda Calculus, we say a fixed point of a term `f` is any *term* `ξ` such that:
231 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.
233 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.
235 You should be able to immediately provide a fixed point of the
236 identity combinator `I`. In fact, you should be able to provide a
237 whole bunch of distinct fixed points.
239 With a little thought, you should be able to provide a fixed point of
240 the false combinator, `KI`. Here's how to find it: recall that `KI`
241 throws away its first argument, and always returns `I`. Therefore, if
242 we give it `I` as an argument, it will throw away the argument, and
243 return `I`. So `KII` ~~> `I`, which is all it takes for `I` to qualify as a
246 What about `K`? Does it have a fixed point? You might not think so,
247 after trying on paper for a while.
249 However, it's a theorem of the Lambda Calculus that *every* lambda term has
250 a fixed point. Even bare variables like `x`! In fact, they will have infinitely many, non-equivalent
251 fixed points. And we don't just know that they exist: for any given
252 formula, we can explicit define many of them.
254 (As we mentioned, even the formula that you're using the define
255 the successor function will have a fixed point. Isn't that weird? There's some `ξ` such that it is equivalent to `succ ξ`?
256 Think about how it might be true. We'll return to this point below.)
259 ### How fixed points help define recursive functions ###
261 Recall our initial, abortive attempt above to define the `length` function in the Lambda Calculus. We said:
263 > What we really want to do is something like this:
265 > \xs. (empty? xs) 0 (succ ((...) (tail xs)))
267 > where this very same formula occupies the `...` position...
269 Imagine replacing the `...` with some expression `LENGTH` that computes the
270 length function. Then we have
272 \xs. (empty? xs) 0 (succ (LENGTH (tail xs)))
274 (More generally, we might have some lambda term `Φ[...SELF...]` where we
275 want the contained `SELF` to refer to that very lambda term `Φ[...SELF...]`.)
277 At this point, we have a definition of the length function, though
278 it's not complete, since we don't know what value to use for the
279 symbol `LENGTH`. Technically, it has the status of an unbound
283 Imagine now binding the mysterious variable, and calling the resulting
286 h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
288 (More generally, convert `Φ[...SELF...]` to `\body. Φ[...body...]`, where
289 the variable `body` wants to be bound to the very lambda term that is that abstract's body.)
291 Now we have no unbound variables, and we have complete non-recursive
292 definitions of each of the other symbols (`empty?`, `0`, `succ`, and `tail`).
294 So `h` takes a `length` argument, and returns a function that accurately
295 computes the length of a list --- as long as the argument we supply is
296 already the length function we are trying to define. (Dehydrated
297 water: to reconstitute, just add water!)
299 Here is where the discussion of fixed points becomes relevant. Saying
300 that `h` is looking for an argument (call it `LENGTH`) that has the same
301 behavior as the result of applying `h` to `LENGTH` is just another way of
302 saying that we are looking for a fixed point for `h`:
306 Replacing `h` with its definition, we have:
308 (\xs. (empty? xs) 0 (succ (LENGTH (tail xs)))) <~~> LENGTH
310 If we can find a value for `LENGTH` that satisfies this constraint, we'll
311 have a function we can use to compute the length of an arbitrary list.
312 All we have to do is find a fixed point for `h`.
314 Let's reinforce this. The left-hand side has the form:
316 (\body. Φ[...body...]) LENGTH
318 which beta-reduces to:
322 where that whole formula is convertible with the term `LENGTH` itself. In other words, the term `Φ[...LENGTH...]` contains (a term that convertible with) itself --- despite being only finitely long. (If it had to contain a term *syntactically identical to* itself, this could not be achieved.)
324 The key to achieving all this is finding a fixed point for `h`. The strategy we will present will turn out to be a general way of
325 finding a fixed point for any lambda term.
328 <a id=deriving-y></a>
329 ## Deriving Y, a fixed point combinator ##
331 How shall we begin? Well, we need to find an argument to supply to
332 `h`. The argument has to be a function that computes the length of a
333 list. The function `h` is *almost* a function that computes the
334 length of a list. Let's try applying `h` to itself. It won't quite
335 work, but examining the way in which it fails will lead to a solution.
337 h h <~~> \xs. (empty? xs) 0 (succ (h (tail xs)))
339 The problem is that in the subexpression `h (tail xs)`, we've
340 applied `h` to a list, but `h` expects as its first argument the
343 So let's adjust `h`, calling the adjusted function `H`. (We'll use `u` as the variable
344 that expects to be bound to the as-yet-*unknown* argument, rather than `length`. This will make it easier
345 to discuss generalizations of this strategy.)
347 h ≡ \length \xs. (empty? xs) 0 (succ (length (tail xs)))
348 H ≡ \u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))
350 (We'll discuss the general case, when you're starting from `\body. Φ[...body...]` rather than this specific `h`, below.)
352 Shifting to `H` is the key creative step. Instead of applying `u` to a list, as happened
353 when we self-applied `h`, `H` applies its argument `u` first to *itself*: `u u`.
354 After `u` gets an argument, the *result* is ready to apply to a list, so we've solved the problem noted above with `h (tail xs)`.
355 We're not done yet, of course; we don't yet know what argument `u` to give
356 to `H` that will behave in the desired way.
358 So let's reason about `H`. What exactly is `H` expecting as its first
359 argument? Based on the excerpt `(u u) (tail xs)`, it appears that
360 `H`'s argument, `u`, should be a function that is ready to take itself
361 as an argument, and that returns a function that takes a list as an
362 argument. `H` itself fits the bill:
364 H H <~~> (\u \xs. (empty? xs) 0 (succ ((u u) (tail xs)))) H
366 <~~> \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
368 <~~> \xs. (empty? xs) 0 (succ ((
369 \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
371 <~~> \xs. (empty? xs) 0 (succ (
372 (empty? (tail xs)) 0 (succ ((H H) (tail (tail xs))))
374 <~~> \xs. (empty? xs) 0 (succ (
375 (empty? (tail xs)) 0 (succ (
376 \xs. (empty? xs) 0 (succ ((H H) (tail xs)))
379 <~~> \xs. (empty? xs) 0 (succ (
380 (empty? (tail xs)) 0 (succ (
381 (empty? (tail (tail xs))) 0 (succ ((H H) (tail (tail (tail xs)))))
388 ### How does the recursion work? ###
390 We've defined `H` in such a way that `H H` turns out to be the length function.
391 That is, `H H` is the `LENGTH` we were looking for.
392 In order to evaluate `H H`, we substitute `H` into the body of the
393 lambda term `H`. Inside that lambda term, once the substitution has
394 occurred, we are once again faced with evaluating `H H`. And so on.
396 We've got the (potentially) infinite regress we desired, defined in terms of a
397 finite lambda term with no undefined symbols.
399 Since `H H` turns out to be the length function, we can think of `H`
400 by itself as *half* of the length function (which is why we called it
401 `H`, of course). (Thought exercise: Can you think up a recursion strategy that involves
402 "dividing" the recursive function into equal thirds `T`, such that the
403 length function <~~> `T T T`?)
405 We've starting with a particular recursive definition, and arrived at
406 a fixed point for that definition.
407 What's the general recipe?
409 1. Start with a formula `h` that takes the recursive function you're seeking as an argument: `h ≡ \length. ...length...` (This is what we also called `\body. Φ[...body...]`.)
410 2. Next, define `H ≡ \u. h (u u)`
411 3. Then compute `H H ≡ ((\u. h (u u)) (\u. h (u u)))`
412 4. That's the fixed point of `h`, the recursive function you're seeking.
414 Expressed in terms of a single formula, here is this method for taking an arbitrary `h`-style term and returning
415 that term's fixed point, which will be the recursive function that term expects as an argument:
417 Y ≡ \h. (\u. h (u u)) (\u. h (u u))
419 Let's test that `Y h` will indeed be `h`'s fixed point:
421 Y h ≡ (\h. (\u. h (u u)) (\u. h (u u))) h
422 ~~> (\u. h (u u)) (\u. h (u u))
423 ~~> h ((\u. h (u u)) (\u. h (u u)))
425 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`.
429 ## A fixed point for K? ##
431 Let's do one more example to illustrate. We'll do `K` (boolean true), since we
432 wondered above whether it had a fixed point.
434 Before we begin, we can reason a bit about what the fixed point must
435 be like. We're looking for a fixed point for `K`, i.e., `\x y. x`. The term `K`
436 ignores its second argument. That means that no matter what we give
437 `K` as its first argument, the result will ignore the next argument
438 (that is, `K ξ` ignores its first argument, no matter what `ξ` is). So
439 if `K ξ <~~> ξ`, `ξ` had also better ignore its first argument. But we
440 also have `K ξ ≡ (\x y. x) ξ ~~> \y. ξ`. This means that if `ξ` ignores
441 its first argument, then `K ξ <~~> \y. ξ` will ignore its first two arguments.
442 So once again, if `K ξ <~~> ξ`, `ξ` also had better ignore (at least) its
443 first two arguments. Repeating this reasoning, we realize that `ξ` here
444 must be a function that ignores as many arguments as you give it.
446 Our expectation, then, is that our recipe for finding fixed points
447 will build us a term that somehow manages to ignore arbitrarily many arguments.
451 ≡ \u. (\x y. x) (u u)
453 H H ~~> (\u y. u u) (\u y. u u)
454 ~~> \y. (\u y. u u) (\u y. u u)
456 Let's check that it is in fact a fixed point for `K`:
458 K (H H) ~~> (\x y. x) ((\u y. u u) (\u y. u u))
459 ~~> \y. (\u y. u u) (\u y. u u)
461 Yep, `H H` and `K (H H)` both reduce to the same term.
463 To see what this fixed point does, let's reduce it a bit more:
465 H H ~~> (\u y. u u) (\u y. u u)
466 ~~> \y. (\u y. u u) (\u y. u u)
467 ~~> \y. \y. (\u y. u u) (\u y. u u)
468 ~~> \y. \y. \y. (\u y. u u) (\u y. u u)
470 Sure enough, this fixed point ignores an endless, arbitrarily-long series of
471 arguments. It's a write-only memory, a black hole.
473 Now that we have one fixed point, we can find others, for instance,
475 (\uy.[uu]u) (\uy.uuu)
476 ~~> \y. [(\uy.uuu) (\uy.uuu)] (\uy.uuu)
477 ~~> \y. [\y. (\uy.uuu) (\uy.uuu) (\uy.uuu)] (\uy.uuu)
478 ~~> \yyy. (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu) (\uy.uuu)
480 Continuing in this way, you can now find an infinite number of fixed
481 points, all of which have the crucial property of ignoring an infinite
484 ## A fixed point for succ? ##
486 As we've seen, the recipe just given for finding a fixed point worked
487 great for our `h`, which we wrote as a definition for the length
488 function. But the recipe doesn't make any assumptions about the
489 internal structure of the term it works with. That means it can
490 find a fixed point for literally any lambda term whatsoever.
492 In particular, what could the fixed point for (our encoding of) the
493 successor function possibly be like?
495 Well, you might think, only some of the formulas that we might give to `succ` as arguments would really represent numbers. If we said something like:
499 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.
501 Yes! That's exactly right. And which formula this is will depend on the particular way you've encoded the successor function.
503 One (by now obvious) upshot is that the recipes that enable us to name
504 fixed points for any given formula `h` aren't *guaranteed* to give us
505 *terminating, normalizing* fixed points. They might give us formulas `ξ` such that
506 neither `ξ` nor `h ξ` have normal forms. (Indeed, what they give us
507 for the `square` function isn't any of the Church numerals, but is
508 rather an expression with no normal form.) However, if we take care we
509 can ensure that we *do* get terminating fixed points. And this gives
510 us a principled, fully general strategy for doing recursion. It lets
511 us define even functions like the Ackermann function, which were until
512 now out of our reach. It would also let us define list
513 functions on [[the encodings we discussed last week|week3_lists#other-lists]], where it
514 wasn't always clear how to force the computation to "keep going."
516 ## Varieties of fixed-point combinators ##
518 Many fixed-point combinators have been discovered. (And as we've seen, some
519 fixed-point combinators give us models for building infinitely many
520 more, non-equivalent fixed-point combinators.)
524 Θ′ ≡ (\u h. h (\n. u u h n)) (\u h. h (\n. u u h n))
525 Y′ ≡ \h. (\u. h (\n. u u n)) (\u. h (\n. u u n))
527 Applying either of these to a term `h` gives a fixed point `ξ` for `h`, meaning that `h ξ` <~~> `ξ`. The combinator `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.
529 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′`?
531 Indeed you can, getting the simpler:
533 Θ ≡ (\u h. h (u u h)) (\u h. h (u u h))
534 Y ≡ \h. (\u. h (u u)) (\u. h (u u))
536 We stated the more complex formulas for the following reason: in a language whose evaluation order is *call-by-value*, the evaluation of `Θ (\body. BODY)` and `Y (\body. BODY)` won't terminate. But evaluation of the eta-unreduced primed versions may.
538 Of course, if you define your `\body. BODY` stupidly, your formula won't terminate, no matter what fixed point combinator you use. For example, let `Ψ` be any fixed point combinator in:
540 Ψ (\body. \n. body n)
542 When you try to evaluate the application of that to some argument `M`, it's going to try to give you back:
546 where `BODY` is equivalent to the very formula `\n. BODY n` that contains it. So the evaluation will proceed:
554 You've written an infinite loop! (This is like the function `eternity` in Chapter 9 of *The Little Schemer*.)
556 However, when we evaluate the application of our:
558 Ψ (\body. (\xs. (empty? xs) 0 (succ (body (tail xs))) ))
560 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:
562 \xs. (empty? xs) 0 (succ (body (tail xs)))
564 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.
567 ## Fixed-point Combinators Are a Bit Intoxicating ##
569 [[tatto|/images/y-combinator-fixed.png]]
571 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.
573 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.
575 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:
577 \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))
579 then this is a fixed-point combinator:
581 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
584 ## Sink: watching Y in action ##
586 For those of you who like to watch ultra slow-mo movies of bullets
587 piercing apples, here's a stepwise computation of the application of a
588 recursive function. We'll use a function `sink`, which takes one
589 argument. If the argument is boolean true (i.e., `\y n. y`), it
590 returns itself (a copy of `sink`); if the argument is boolean false
591 (`\y n. n`), it returns `I`. That is, we want the following behavior:
594 sink true false <~~> I
595 sink true true false <~~> I
596 sink true true true false <~~> I
598 To get this behavior, we want `sink` to be the fixed point
599 of `\sink. \b. b sink I`. That is, `sink ≡ Y (\sb.bsI)`:
603 3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) false
604 4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) false
605 5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
606 6. (\b. b [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] I) false
607 7. false [(\u. (\sb.bsI) (u u))(\u. (\sb.bsI) (u u))] I
608 --------------------------------------------
611 So far so good. The crucial thing to note is that as long as we
612 always reduce the outermost redex first, we never have to get around
613 to computing the underlined redex: because `false` ignores its first
614 argument, we can throw it away unreduced.
616 Now we try the next most complex example:
619 2. Y (\sb.bsI) true false
620 3. (\h. (\u. h [u u]) (\u. h (u u))) (\sb.bsI) true false
621 4. (\u. (\sb.bsI) [u u]) (\u. (\sb.bsI) (u u)) true false
622 5. (\sb.bsI) [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] true false
623 6. (\b. b [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I) true false
624 7. true [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] I false
625 8. [(\u. (\sb.bsI) (u u)) (\u. (\sb.bsI) (u u))] false
627 We've now arrived at line (4) of the first computation, so the result
630 You should be able to see that `sink` will consume as many `true`s as
631 we throw at it, then turn into the identity function when it
632 encounters the first `false`.
634 The key to the recursion is that, thanks to `Y`, the definition of
635 `sink` contains within it the ability to fully regenerate itself as
636 many times as is necessary. The key to *ending* the recursion is that
637 the behavior of `sink` is sensitive to the nature of the input: if the
638 input is the magic function `false`, the self-regeneration machinery
639 will be discarded, and the recursion will stop.
641 That's about as simple as recursion gets.
644 ## Base cases, and their lack ##
646 As any functional programmer quickly learns, writing a recursive
647 function divides into two tasks: figuring out how to handle the
648 recursive case, and *remembering to insert a base case*. The
649 interesting and enjoyable part is figuring out the recursive pattern,
650 but the base case cannot be ignored, since leaving out the base case
651 creates a program that runs forever. For instance, consider computing
652 a factorial: `n!` is `n * (n-1) * (n-2) * ... * 1`. The recursive
653 case says that the factorial of a number `n` is `n` times the
654 factorial of `n-1`. But if we leave out the base case, we get
656 3! = 3 * 2! = 3 * 2 * 1! = 3 * 2 * 1 * 0! = 3 * 2 * 1 * 0 * -1! ...
658 That's why it's crucial to declare that `0!` = `1`, in which case the
659 recursive rule does not apply. In our terms,
661 fact ≡ Y (\fact n. (zero? n) 1 (fact (pred n)))
663 If `n` is `0`, `fact` reduces to `1`, without computing the recursive case.