PL: Lecture #11  Tuesday, February 23rd
(text)

Recursion, Recursion, Recursion

PLAI §9

There is one major feature that is still missing from our language: we have no way to perform recursion (therefore no kind of loops). So far, we could only use recursion when we had names. In FLANG, the only way we can have names is through with which not good enough for recursion.

To discuss the issue of recursion, we switch to a “broken” version of (untyped) Racket — one where a define has a different scoping rules: the scope of the defined name does not cover the defined expression. Specifically, in this language, this doesn’t work:

#lang pl broken
(define (fact n)
  (if (zero? n) 1 (* n (fact (- n 1)))))
(fact 5)

In our language, this translation would also not work (assuming we have if etc):

{with {fact {fun {n}
              {if {= n 0} 1 {* n {call fact {- n 1}}}}}}
  {call fact 5}}

And similarly, in plain Racket this won’t work if let is the only tool you use to create bindings:

(let ([fact (lambda (n)
              (if (zero? n) 1 (* n (fact (- n 1)))))])
  (fact 5))

In the broken-scope language, the define form is more similar to a mathematical definition. For example, when we write:

(define (F x) x)
(define (G y) (F y))
(G F)

it is actually shorthand for

(define F (lambda (x) x))
(define G (lambda (y) (F y)))
(G F)

we can then replace defined names with their definitions:

(define F (lambda (x) x))
(define G (lambda (y) (F y)))
((lambda (y) (F y)) (lambda (x) x))

and this can go on, until we get to the actual code that we wrote:

((lambda (y) ((lambda (x) x) y)) (lambda (x) x))

This means that the above fact definition is similar to writing:

fact := (lambda (n)
          (if (zero? n) 1 (* n (fact (- n 1)))))
(fact 5)

which is not a well-formed definition — it is meaningless (this is a formal use of the word “meaningless”). What we’d really want, is to take the equation (using = instead of :=)

fact = (lambda (n)
        (if (zero? n) 1 (* n (fact (- n 1)))))

and find a solution which will be a value for fact that makes this true.

If you look at the Racket evaluation rules handout on the web page, you will see that this problem is related to the way that we introduced the Racket define: there is a hand-wavy explanation that talks about knowing things.

The big question is: can we define recursive functions without Racket’s magical define form?

Note: This question is a little different than the question of implementing recursion in our language — in the Racket case we have no control over the implementation of the language. As it will eventually turn out, implementing recursion in our own language will be quite easy when we use mutation in a specific way. So the question that we’re now facing can be phrased as either “can we get recursion in Racket without Racket’s magical definition forms?” or “can we get recursion in our interpreter without mutation?”.

Recursion without the Magic

PLAI §22.4 (we go much deeper)

Note: This explanation is similar to the one you can find in “The Why of Y”, by Richard Gabriel.

To implement recursion without the define magic, we first make an observation: this problem does not come up in a dynamically-scoped language. Consider the let-version of the problem:

#lang pl dynamic
(let ([fact (lambda (n)
              (if (zero? n) 1 (* n (fact (- n 1)))))])
  (fact 5))

This works fine — because by the time we get to evaluate the body of the function, fact is already bound to itself in the current dynamic scope. (This is another reason why dynamic scope is perceived as a convenient approach in new languages.)

Regardless, the problem that we have with lexical scope is still there, but the way things work in a dynamic scope suggest a solution that we can use now. Just like in the dynamic scope case, when fact is called, it does have a value — the only problem is that this value is inaccessible in the lexical scope of its body.

Instead of trying to get the value in via lexical scope, we can imitate what happens in the dynamically scoped language by passing the fact value to itself so it can call itself (going back to the original code in the broken-scope language):

(define (fact self n)  ;***
  (if (zero? n) 1 (* n (self (- n 1)))))
(fact fact 5)          ;***

except that now the recursive call should still send itself along:

(define (fact self n)
  (if (zero? n) 1 (* n (self self (- n 1))))) ;***
(fact fact 5)

The problem is that this required rewriting calls to fact — both outside and recursive calls inside. To make this an acceptable solution, calls from both places should not change. Eventually, we should be able to get a working fact definition that uses just

(lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))

The first step in resolving this problem is to curry the fact definition.

(define (fact self)                  ;***
  (lambda (n)                        ;***
    (if (zero? n)
      1
      (* n ((self self) (- n 1)))))) ;***
((fact fact) 5)                      ;***

Now fact is no longer our factorial function — it’s a function that constructs it. So call it make-fact, and bind fact to the actual factorial function.

(define (make-fact self)            ;***
  (lambda (n)
    (if (zero? n) 1 (* n ((self self) (- n 1))))))
(define fact (make-fact make-fact))  ;***
(fact 5)                            ;***

We can try to do the same thing in the body of the factorial function: instead of calling (self self), just bind fact to it:

(define (make-fact self)
  (lambda (n)
    (let ([fact (self self)])      ;***
      (if (zero? n)
        1
        (* n (fact (- n 1)))))))    ;***
(define fact (make-fact make-fact))
(fact 5)

This works fine, but if we consider our original goal, we need to get that local fact binding outside of the (lambda (n) ...) — so we’re left with a definition that uses the factorial expression as is. So, swap the two lines:

(define (make-fact self)
  (let ([fact (self self)])  ;***
    (lambda (n)              ;***
      (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact (make-fact make-fact))
(fact 5)

But the problem is that this gets us into an infinite loop because we’re trying to evaluate (self self) too ea(ge)rly. In fact, if we ignore the body of the let and other details, we basically do this:

(define (make-fact self) (self self)) (make-fact make-fact)
--reduce-sugar-->
(define make-fact (lambda (self) (self self))) (make-fact make-fact)
--replace-definition-->
((lambda (self) (self self)) (lambda (self) (self self)))
--rename-identifiers-->
((lambda (x) (x x)) (lambda (x) (x x)))

And this expression has an interesting property: it reduces to itself, so evaluating it gets stuck in an infinite loop.

So how do we solve this? Well, we know that (self self) should be the same value that is the factorial function itself — so it must be a one-argument function. If it’s such a function, we can use a value that is equivalent, except that it will not get evaluated until it is needed, when the function is called. The trick here is the observation that (lambda (n) (add1 n)) is really the same as add1 (provided that add1 is a one-argument function), except that the add1 part doesn’t get evaluated until the function is called. Applying this trick to our code produces a version that does not get stuck in the same infinite loop:

(define (make-fact self)
  (let ([fact (lambda (n) ((self self) n))]) ;***
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact (make-fact make-fact))
(fact 5)

Continuing from here — we know that

(let ([x v]) e)  is the same as  ((lambda (x) e) v)

(remember how we derived fun from a with), so we can turn that let into the equivalent function application form:

(define (make-fact self)
  ((lambda (fact)                    ;***
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))
  (lambda (n) ((self self) n))))    ;***
(define fact (make-fact make-fact))
(fact 5)

And note now that the (lambda (fact) …) expression is everything that we need for a recursive definition of fact — it has the proper factorial body with a plain recursive call. It’s almost like the usual value that we’d want to define fact as, except that we still have to abstract on the recursive value itself. So lets move this code into a separate definition for fact-step:

(define fact-step                  ;***
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-fact self)
  (fact-step                        ;***
  (lambda (n) ((self self) n))))
(define fact (make-fact make-fact))
(fact 5)

We can now proceed by moving the (make-fact make-fact) self application into its own function which is what creates the real factorial:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-fact self)
  (fact-step
  (lambda (n) ((self self) n))))
(define (make-real-fact) (make-fact make-fact)) ;***
(define fact (make-real-fact))                  ;***
(fact 5)

Rewrite the make-fact definition using an explicit lambda:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define make-fact                    ;***
  (lambda (self)                      ;***
    (fact-step
    (lambda (n) ((self self) n)))))
(define (make-real-fact) (make-fact make-fact))
(define fact (make-real-fact))
(fact 5)

and fold the functionality of make-fact and make-real-fact into a single make-fact function by just using the value of make-fact explicitly instead of through a definition:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-real-fact)
  (let ([make (lambda (self)                      ;***
                (fact-step                        ;***
                (lambda (n) ((self self) n))))]) ;***
    (make make)))
(define fact (make-real-fact))
(fact 5)

We can now observe that make-real-fact has nothing that is specific to factorial — we can make it take a “core function” as an argument:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-real-fact core)                    ;***
  (let ([make (lambda (self)
                (core                            ;***
                (lambda (n) ((self self) n))))])
    (make make)))
(define fact (make-real-fact fact-step))          ;***
(fact 5)

and call it make-recursive:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-recursive core)            ;***
  (let ([make (lambda (self)
                (core
                (lambda (n) ((self self) n))))])
    (make make)))
(define fact (make-recursive fact-step))  ;***
(fact 5)

We’re almost done now — there’s no real need for a separate fact-step definition, just use the value for the definition of fact:

(define (make-recursive core)
  (let ([make (lambda (self)
                (core
                (lambda (n) ((self self) n))))])
    (make make)))
(define fact
  (make-recursive
  (lambda (fact)                                          ;***
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))) ;***
(fact 5)

turn the let into a function form:

(define (make-recursive core)
  ((lambda (make) (make make))              ;***
  (lambda (self)                          ;***
    (core (lambda (n) ((self self) n)))))) ;***
(define fact
  (make-recursive
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))))
(fact 5)

do some renamings to make things simpler — make and self turn to x, and core to f:

(define (make-recursive f)                  ;***
  ((lambda (x) (x x))                        ;***
  (lambda (x) (f (lambda (n) ((x x) n)))))) ;***
(define fact
  (make-recursive
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))))
(fact 5)

or we can manually expand that first (lambda (x) (x x)) application to make the symmetry more obvious (not really surprising because it started with a let whose purpose was to do a self-application):

(define (make-recursive f)
  ((lambda (x) (f (lambda (n) ((x x) n))))  ;***
  (lambda (x) (f (lambda (n) ((x x) n)))))) ;***
(define fact
  (make-recursive
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))))
(fact 5)

And we finally got what we were looking for: a general way to define any recursive function without any magical define tricks. This also work for other recursive functions:

#lang pl broken
(define (make-recursive f)
  ((lambda (x) (f (lambda (n) ((x x) n))))
  (lambda (x) (f (lambda (n) ((x x) n))))))
(define fact
  (make-recursive
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))))
(fact 5)
(define fib
  (make-recursive
  (lambda (fib)
    (lambda (n) (if (<= n 1) n (+ (fib (- n 1)) (fib (- n 2))))))))
(fib 8)
(define length
  (make-recursive
  (lambda (length)
    (lambda (l) (if (null? l) 0 (+ (length (rest l)) 1))))))
(length '(x y z))

A convenient tool that people often use on paper is to perform a kind of a syntactic abstraction: “assume that whenever I write (twice foo) I really meant to write (foo foo)”. This can often be done as plain abstractions (that is, using functions), but in some cases — for example, if we want to abstract over definitions — we just want such a rewrite rule. (More on this towards the end of the course.) The broken-scope language does provide such a tool — rewrite extends the language with a rewrite rule. Using this, and our make-recursive, we can make up a recursive definition form:

(rewrite (define/rec (f x) E)
      => (define f (make-recursive (lambda (f) (lambda (x) E)))))

In other words, we’ve created our own “magical definition” form. The above code can now be written in almost the same way it is written in plain Racket:

#lang pl broken
(define (make-recursive f)
  ((lambda (x) (f (lambda (n) ((x x) n))))
  (lambda (x) (f (lambda (n) ((x x) n))))))
(rewrite (define/rec (f x) E)
      => (define f (make-recursive (lambda (f) (lambda (x) E)))))
;; examples
(define/rec (fact n) (if (zero? n) 1 (* n (fact (- n 1)))))
(fact 5)
(define/rec (fib n) (if (<= n 1) n (+ (fib (- n 1)) (fib (- n 2)))))
(fib 8)
(define/rec (length l) (if (null? l) 0 (+ (length (rest l)) 1)))
(length '(x y z))

Finally, note that make-recursive is limited to 1-argument functions only because of the protection from eager evaluation. In any case, it can be used in any way you want, for example,

(make-recursive (lambda (f) (lambda (x) f)))

is a function that returns itself rather than calling itself. Using the rewrite rule, this would be:

(define/rec (f x) f)

which is the same as:

(define (f x) f)

in plain Racket.

The Core of make-recursive

As in Racket, being able to express recursive functions is a fundamental property of the language. It means that we can have loops in our language, and that’s the essence of making a language powerful enough to be TM-equivalent — able to express undecidable problems, where we don’t know whether there is an answer or not.

The core of what makes this possible is the expression that we have seen in our derivation:

((lambda (x) (x x)) (lambda (x) (x x)))

which reduces to itself, and therefore has no value: trying to evaluate it gets stuck in an infinite loop. (This expression is often called “Omega”.)

This is the key for creating a loop — we use it to make recursion possible. Looking at our final make-recursive definition and ignoring for a moment the “protection” that we need against being stuck prematurely in an infinite loop:

(define (make-recursive f)
  ((lambda (x) (x x)) (lambda (x) (f (x x)))))

we can see that this is almost the same as the Omega expression — the only difference is that application of f. Indeed, this expression (the result of (make-recursive F) for some F) reduces in a similar way to Omega:

((lambda (x) (x x)) (lambda (x) (F (x x))))
((lambda (x) (F (x x))) (lambda (x) (F (x x))))
(F ((lambda (x) (F (x x))) (lambda (x) (F (x x)))))
(F (F ((lambda (x) (F (x x))) (lambda (x) (F (x x))))))
(F (F (F ((lambda (x) (F (x x))) (lambda (x) (F (x x)))))))
...

which means that the actual value of this expression is:

(F (F (F ...forever...)))

This definition would be sufficient if we had a lazy language, but to get things working in a strict one we need to bring back the protection. This makes things a little different — if we use (protect f) to be a shorthand for the protection trick,

(rewrite (protect f) => (lambda (x) (f x)))

then we have:

(define (make-recursive f)
  ((lambda (x) (x x)) (lambda (x) (f (protect (x x))))))

which makes the (make-recursive F) evaluation reduce to

(F (protect (F (protect (F (protect (...forever...)))))))

and this is still the same result (as long as F is a single-argument function).

(Note that protect cannot be implemented as a plain function!)

Denotational Explanation of Recursion

Note: This explanation is similar to the one you can find in “The Little Schemer” called “(Y Y) Works!”, by Dan Friedman and Matthias Felleisen.

The explanation that we have now for how to derive the make-recursive definition is fine — after all, we did manage to get it working. But this explanation was done from a kind of an operational point of view: we knew a certain trick that can make things work and we pushed things around until we got it working like we wanted. Instead of doing this, we can re-approach the problem from a more declarative point of view.

So, start again from the same broken code that we had (using the broken-scope language):

(define fact
  (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))

This is as broken as it was when we started: the occurrence of fact in the body of the function is free, which means that this code is meaningless. To avoid the compilation error that we get when we run this code, we can substitute anything for that fact — it’s even better to use a replacement that will lead to a runtime error:

(define fact
  (lambda (n) (if (zero? n) 1 (* n (777 (- n 1)))))) ;***

This function will not work in a similar way to the original one — but there is one case where it does work: when the input value is 0 (since then we do not reach the bogus application). We note this by calling this function fact0:

(define fact0                ;***
  (lambda (n) (if (zero? n) 1 (* n (777 (- n 1))))))

Now that we have this function defined, we can use it to write fact1 which is the factorial function for arguments of 0 or 1:

(define fact0
  (lambda (n) (if (zero? n) 1 (* n (777 (- n 1))))))
(define fact1
  (lambda (n) (if (zero? n) 1 (* n (fact0 (- n 1))))))

And remember that this is actually just shorthand for:

(define fact1
  (lambda (n)
    (if (zero? n)
      1
      (* n ((lambda (n)
              (if (zero? n)
                1
                (* n (777 (- n 1)))))
            (- n 1))))))

We can continue in this way and write fact2 that will work for n<=2:

(define fact2
  (lambda (n) (if (zero? n) 1 (* n (fact1 (- n 1))))))

or, in full form:

(define fact2
  (lambda (n)
    (if (zero? n)
      1
      (* n ((lambda (n)
              (if (zero? n)
                1
                (* n ((lambda (n)
                        (if (zero? n)
                          1
                          (* n (777 (- n 1)))))
                      (- n 1)))))
            (- n 1))))))

If we continue this way, we will get the true factorial function, but the problem is that to handle any possible integer argument, it will have to be an infinite definition! Here is what it is supposed to look like:

(define fact0 (lambda (n) (if (zero? n) 1 (* n (777 (- n 1))))))
(define fact1 (lambda (n) (if (zero? n) 1 (* n (fact0 (- n 1))))))
(define fact2 (lambda (n) (if (zero? n) 1 (* n (fact1 (- n 1))))))
(define fact3 (lambda (n) (if (zero? n) 1 (* n (fact2 (- n 1))))))
...

The true factorial function is fact-infinity, with an infinite size. So, we’re back at the original problem…

To help make things more concise, we can observe the repeated pattern in the above, and extract a function that abstracts this pattern. This function is the same as the fact-step that we have seen previously:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact0 (fact-step 777))
(define fact1 (fact-step fact0))
(define fact2 (fact-step fact1))
(define fact3 (fact-step fact2))
...

which is actually:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact0 (fact-step 777))
(define fact1 (fact-step (fact-step 777)))
(define fact2 (fact-step (fact-step (fact-step 777))))
...
(define fact
  (fact-step (fact-step (fact-step (... (fact-step 777) ...)))))

Do this a little differently — rewrite fact0 as:

(define fact0
  ((lambda (mk) (mk 777))
  fact-step))

Similarly, fact1 is written as:

(define fact1
  ((lambda (mk) (mk (mk 777)))
  fact-step))

and so on, until the real factorial, which is still infinite at this stage:

(define fact
  ((lambda (mk) (mk (mk (... (mk 777) ...))))
  fact-step))

Now, look at that (lambda (mk) ...) — it is an infinite expression, but for every actual application of the resulting factorial function we only need a finite number of mk applications. We can guess how many, and as soon as we hit an application of 777 we know that our guess is too small. So instead of 777, we can try to use the maker function to create and use the next.

To make things more explicit, here is the expression that is our fact0, without the definition form:

((lambda (mk) (mk 777))
fact-step)

This function has a very low guess — it works for 0, but with 1 it will run into the 777 application. At this point, we want to somehow invoke mk again to get the next level — and since 777 does get applied, we can just replace it with mk:

((lambda (mk) (mk mk))
fact-step)

The resulting function works just the same for an input of 0 because it does not attempt a recursive call — but if we give it 1, then instead of running into the error of applying 777:

(* n (777 (- n 1)))

we get to apply fact-step there:

(* n (fact-step (- n 1)))

and this is still wrong, because fact-step expects a function as an input. To see what happens more clearly, write fact-step explicitly:

((lambda (mk) (mk mk))
(lambda (fact)
  (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))

The problem is in what we’re going to pass into fact-step — its fact argument will not be the factorial function, but the mk function constructor. Renaming the fact argument as mk will make this more obvious (but not change the meaning):

((lambda (mk) (mk mk))
(lambda (mk)
  (lambda (n) (if (zero? n) 1 (* n (mk (- n 1)))))))

It should now be obvious that this application of mk will not work, instead, we need to apply it on some function and then apply the result on (- n 1). To get what we had before, we can use 777 as a bogus function:

((lambda (mk) (mk mk))
(lambda (mk)
  (lambda (n) (if (zero? n) 1 (* n ((mk 777) (- n 1)))))))

This will allow one recursive call — so the definition works for both inputs of 0 and 1 — but not more. But that 777 is used as a maker function now, so instead, we can just use mk itself again:

((lambda (mk) (mk mk))
(lambda (mk)
  (lambda (n) (if (zero? n) 1 (* n ((mk mk) (- n 1)))))))

And this is a working version of the real factorial function, so make it into a (non-magical) definition:

(define fact
  ((lambda (mk) (mk mk))
  (lambda (mk)
    (lambda (n) (if (zero? n) 1 (* n ((mk mk) (- n 1))))))))

But we’re not done — we “broke” into the factorial code to insert that (mk mk) application — that’s why we dragged in the actual value of fact-step. We now need to fix this. The expression on that last line

(lambda (n) (if (zero? n) 1 (* n ((mk mk) (- n 1)))))

is close enough — it is (fact-step (mk mk)). So we can now try to rewrite our fact as:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact
  ((lambda (mk) (mk mk))
  (lambda (mk) (fact-step (mk mk)))))

… and would fail in a familiar way! If it’s not familiar enough, just rename all those mks as xs:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact
  ((lambda (x) (x x))
  (lambda (x) (fact-step (x x)))))

We’ve run into the eagerness of our language again, as we did before. The solution is the same — the (x x) is the factorial function, so protect it as we did before, and we have a working version:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define fact
  ((lambda (x) (x x))
  (lambda (x) (fact-step (lambda (n) ((x x) n))))))

The rest should not be surprising now… Abstract the recursive making bit in a new make-recursive function:

(define fact-step
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1)))))))
(define (make-recursive f)
  ((lambda (x) (x x))
  (lambda (x) (f (lambda (n) ((x x) n))))))
(define fact (make-recursive fact-step))

and now we can do the first reduction inside make-recursive and write the fact-step expression explicitly:

#lang pl broken
(define (make-recursive f)
  ((lambda (x) (f (lambda (n) ((x x) n))))
  (lambda (x) (f (lambda (n) ((x x) n))))))
(define fact
  (make-recursive
  (lambda (fact)
    (lambda (n) (if (zero? n) 1 (* n (fact (- n 1))))))))

and this is the same code we had before.