The U combinator

:: computer, maths, lisp

The U combinator allows you to define recursive functions and I think it is simpler to understand than the Y combinator.

It’s not obvious how things like letrec get defined in Scheme, without using secret assignment. In fact I think they are defined using secret assignment:

(letrec ([f (λ (...) ... (f ...) ...)])
  ...)

turns into

(let ([f ...])
  (set! f (λ (...) ... (f ...) ...))
  ...)

But it’s interesting to see how you can define recursive functions without relying on assignment, including mutually-recursive collections of functions. One way is using the U combinator.

I suspect that there is lots of information about this out there, but it’s seriously hard to search for anything which looks like ’*-combinator’ now (even now I am starting a set of companies called ‘integration by parts’, ‘the quotient rule’ &c).

You can famously do this with the Y combinator, but I didn’t want to do that because Y is something I find I can understand for a few hours at a time and then I have to work it all out again. But it turns out that you can use something much simpler: the U combinator. It seems to be even harder to search for this than Y, but here is a quote about it:

In the theory of programming languages, the U combinator, \(U\), is the mathematical function that applies its argument to its argument; that is \(U(f) = f(f)\), or equivalently, \(U = \lambda f \cdot f(f)\).

Self-application permits the simulation of recursion in the λ-calculus, which means that the U combinator enables universal computation. (The U combinator is actually more primitive than the more well-known fixed-point Y combinator.)

The expression \(U(U)\) is the smallest non-terminating program.

(Text mildly edited from here, which unfortunately is not a site all about the U combinator other than this quote.)

Prerequisites

All of the following code samples are in Racket. The macros are certainly Racket-specific and some of the other code probably is as well. To make the macros work you will need syntax-parse via:

(require (for-syntax syntax/parse))

However note that my use of syntax-parse is naïve in the extreme: I’m really just an unfrozen CL caveman pretending to understand Racket’s macro system.

Also note I have not ruthlessly turned everything into λ: Rather than ((λ (...) ...) ...) there is (let ([... ...] ...) ...) in this code; there is use of multiple values including let-values; there is (define (f ...) ...) rather than (define f (λ (...) ...)) and so on.

Two versions of U

The first version of U is the obvious one:

(define (U f)
  (f f))

But this will run into some problems with an applicative-order language, which Racket is by default. To avoid that we can make the assumption that (f f) is going to be a function, and wrap that form in another function to delay its evaluation until it’s needed: this is the standard trick that you have to do for Y in an applicative-order language as well. I’m only going to use the applicative-order U when I have to, so I’ll give it a different name:

(define (U/ao f)
  (λ args (apply (f f) args)))

Note also that I’m allowing more than one argument rather than doing the pure-λ-calculus thing.

Using U to construct a recursive functions

To do this we do a similar trick that you do with Y: write a function which, if given a function as argument which deals with the recursive cases, will return a recursive function. And obviously I’ll use the Fibonacci function as the canonical recursive function.

So, consider this thing:

(define fibber
  (λ (f)
    (λ (n)
      (if (<= n 2)
          1
          (+ ((U f) (- n 1))
             ((U f) (- n 2)))))))

This is a function which, given another function, U of which computes smaller Fibonacci numbers, will return a function which will compute the Fibonacci number for n.

In other words, U of this function is the Fibonacci function!

And we can test this:

> (define fibonacci (U fibber))
> (fibonacci 10)
55

So that’s very nice.

Wrapping U in a macro

So, to hide all this the first thing to do is to remove the explicit calls to U in the recursion. We can lift them out of the inner function completely:

(define fibber/broken
  (λ (f)
    (let ([fib (U f)])
      (λ (n)
        (if (<= n 2)
            1
            (+ (fib (- n 1))
               (fib (- n 2))))))))

Don’t try to compute U of this: it will recurse endlessly because (U fibber/broken) -> (fibber/broken fibber/broken) and this involves computing (U fibber/broken), and we’re doomed.

Instead we can use U/ao:

(define fibber
  (λ (f)
    (let ([fib (U/ao f)])
      (λ (n)
        (if (<= n 2)
            1
            (+ (fib (- n 1))
               (fib (- n 2))))))))

And this is all fine ((U fibber) 10) is 55 (and terminates!).

Purists can then turn let into λ in the usual way:

(define fibber
  (λ (f)
    ((λ (fib)
       (λ (n)
         (if (<= n 2)
             1
             (+ (fib (- n 1))
                (fib (- n 2))))))
     (U/ao f))))

And this is really all you need to be able to write the macro:

(define-syntax (with-recursive-binding stx)
  (syntax-parse stx
    [(_ (name:id value:expr) form ...+)
     #'(let ([name (U (λ (f)
                        (let ([name (U/ao f)])
                          value)))])
         form ...)]))

Or, for the pure of heart:

(define-syntax (with-recursive-binding stx)
  (syntax-parse stx
    [(_ (name:id value:expr) form ...+)
     #'((λ (name)
          form ...)
        (U (λ (f)
             ((λ (name)
                value)
              (U/ao f)))))]))

And this works fine:

(with-recursive-binding (fib (λ (n)
                               (if (<= n 2)
                                   1
                                   (+ (fib (- n 1))
                                      (fib (- n 2))))))
  (fib 10))

A caveat on bindings

One fairly obvious thing here is that there are two bindings constructed by this macro: the outer one, and an inner one of the same name. And these are not bound to the same function in the sense of eq?:

(with-recursive-binding (ts (λ (it)
                              (eq? ts it)))
  (ts ts))

is #f. This matters only in a language where bindings can be mutated: a language with assignment in other words. Both the outer and inner bindings, unless they have been mutated, are to functions which are identical as functions: they compute the same values for all values of their arguments. In fact, it’s hard to see what purpose eq? would serve in a language without assignment.

This caveat will apply below as well.

Two versions of U for many functions

The obvious generalization of U, U*, to many functions is that \(U^*(f_1, \ldots, f_n)\) is the tuple \((f_1(f_1, \ldots, f_n), f_2(f_1, \ldots, f_n), \ldots)\). And a nice way of expressing that in Racket is to use multiple values:

(define (U* . fs)
  (apply values (map (λ (f)
                       (apply f fs))
                     fs)))

And we need the applicative-order one as well:

(define (U*/ao . fs)
  (apply values (map (λ (f)
                       (λ args (apply (apply f fs) args)))
                     fs)))

Note that U* is a true generalization of U: (U f) and (U* f) are the same.

Using U* to construct mutually-recursive functions

I’ll work with a trivial pair of functions:

  • an object is a numeric tree if it is a cons and its car and cdr are numeric objects;
  • an objct is a numeric object if it is a number, or if it is a numeric tree.

So we can define ‘maker’ functions (with an ’-er’ convention: a function which makes an x is an xer, or, if x has hyphens in it, an x-er) which will make suitable functions:

(define numeric-tree-er
  (λ (nter noer)
    (λ (o)
      (let-values ([(nt? no?) (U* nter noer)])
        (and (cons? o)
             (no? (car o))
             (no? (cdr o)))))))

(define numeric-object-er
  (λ (nter noer)
    (λ (o)
      (let-values ([(nt? no?) (U* nter noer)])
        (cond
          [(number? o) #t]
          [(cons? o) (nt? o)]
          [else #f])))))

Note that for both of these I’ve raised the call to U* a little, simply to make the call to the appropriate value of U* less opaque.

And this works:

(define-values (numeric-tree? numeric-object?)
  (U* numeric-tree-er numeric-object-er))

And now:

> (numeric-tree? 1)
#f
> (numeric-object? 1)
#t
> (numeric-tree? '(1 . 2))
#t
> (numeric-tree? '(1 2 . (3 4)))
#f

Wrapping U* in a macro

The same problem as previously happens when we raise the inner call to U* with the same result: we need to use U*/ao. In addition the macro becomes significantly more hairy and I’m moderately surprised that I got it right so easily. It’s not conceptually hard: it’s just not obvious to me that the pattern-matching works.

(define-syntax (with-recursive-bindings stx)
  (syntax-parse stx
    [(_ ((name:id value:expr) ...) form ...+)
     #:fail-when (check-duplicate-identifier (syntax->list #'(name ...)))
     "duplicate variable name"
     (with-syntax ([(argname ...) (generate-temporaries #'(name ...))])
       #'(let-values
             ([(name ...) (U* (λ (argname ...)
                                (let-values ([(name ...)
                                              (U*/ao argname ...)])
                                  value)) ...)])
           form ...))]))

And now, in a shower of sparks, we can write:

(with-recursive-bindings ((numeric-tree?
                           (λ (o)
                             (and (cons? o)
                                  (numeric-object? (car o))
                                  (numeric-object? (cdr o)))))
                          (numeric-object?
                           (λ (o)
                             (cond [(number? o) #t]
                                   [(cons? o) (numeric-tree? o)]
                                   [else #f]))))
  (numeric-tree? '(1 2 3 (4 (5 . 6) . 7) . 8)))

and get #t.

As I said, I am sure there are well-known better ways to do this, but I thought this was interesting enough not to lose. This originated as an answer to this Stack Overflow question.