[racket] sicp exercise 2.20

Short after sending my message, I realized that I have shown too much.
Apology. Wont happen again.
Jos.


> -----Original Message-----
> From: Martin DeMello [mailto:martindemello at gmail.com] 
> Sent: 20 July 2010 17:21
> To: Jos Koot
> Cc: PLT-Scheme Mailing List
> Subject: Re: [racket] sicp exercise 2.20
> 
> Ah, nice :) Thanks!
> 
> martin
> 
> On Tue, Jul 20, 2010 at 8:39 PM, Jos Koot 
> <jos.koot at telefonica.net> wrote:
> > Yet another one using -apply-:
> >
> > (define (same-parity x . xs)
> >  (cond
> >   ((null? xs) (list x))
> >   ((equal? (even? x) (even? (car xs))) (cons x (apply same-parity 
> > xs)))
> >   (else (apply same-parity x (cdr xs)))))
> >
> > Jos
> >
> >> -----Original Message-----
> >> From: users-bounces at racket-lang.org
> >> [mailto:users-bounces at racket-lang.org] On Behalf Of Martin DeMello
> >> Sent: 20 July 2010 15:37
> >> To: PLT-Scheme Mailing List
> >> Subject: [racket] sicp exercise 2.20
> >>
> >> By way of a puzzle, I've been trying to solve SICP exercise 2.20
> >> ----------------------------
> >> Using the (define (f x . args)) notation, write a procedure 
> >> "same-parity" that takes one or more integers and returns 
> a list of 
> >> all the arguments that have the same even-odd parity as the first 
> >> argument. For example, (same-parity 1 2 3 4 5 6 7)
> >> (1 3 5 7)
> >> (same-parity 2 3 4 5 6 7)
> >> (2 4 6)
> >> ----------------------------
> >> using "straight recursion", that is, without using any `let` or 
> >> `define` constructs. Still not managed to find the trick that will 
> >> tack on the first argument as the head of the list.
> >> Anyone have a hint?
> >>
> >> martin
> >> _________________________________________________
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> >
> >
> >