[racket-dev] using the Y combinator to...

Look up the difference between the call by name and call by value Y combinator.

Jay

On Wed, Aug 4, 2010 at 10:20 AM, The Configurator
<configurator at gmail.com> wrote:
> I'll hijack the thread since I've been meaning to ask about the Y
> combinator.
> From SICP I've seen that the Y combinator is the function
> (define (y f)
>   ((lambda (x) (f (x x)))
>    (lambda (x) (f (x x)))))
> This makes mathematical sense, since
> (y f)
> = ((lambda (x) (f (x x))) (lambda (x) (f (x x))))
> = (f ((lambda (x) (f (x x))) (lambda (x) (f (x x)))))
> = (f (y f))
> But when actually applying it:
> // g improves a function to get a ceiling function. The fixed point would be
> similar to (lambda (x) (inexact->exact (ceiling x)))
> (define (g f)
>   (lambda (x)
>     (if (<= x 0)
>         0
>         (+ 1 (f (- x 1))))))
>> ((g (g (g (g (g #f))))) 3.3)
> 4
> Now let's step through
> ((y g) 3.3)
> = (((lambda (x) (g (x x))) (lambda (x) (g (x x)))) 3.3)
> = ((g ((lambda (x) (g (x x))) (lambda (x) (g (x x))))) 3.3)
> now, to get the (g ...) result we need to apply g to it's operand. For that,
> we need to evaluate it:
> = ((g (g ((lambda (x) (g (x x))) (lambda (x) (g (x x)))))) 3.3)
> = ((g (g (g ((lambda (x) (g (x x))) (lambda (x) (g (x x)))))))) 3.3)
> = ((g (g (g (g ((lambda (x) (g (x x))) (lambda (x) (g (x x))))))))) 3.3)
> And I just ran out of memory.
> What am I missing here?
> It seems obvious to me that if we add a delay/force or somesuch thing it
> solves the problem. But that's not the y-combinator I was shown.
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-- 
Jay McCarthy <jay at cs.byu.edu>
Assistant Professor / Brigham Young University
http://teammccarthy.org/jay

"The glory of God is Intelligence" - D&C 93