I am not too sure if this is less than quadratic ??
(define points (list->vector '(a b c d e f)))
(define l (vector-length points))
(define (gen org start end)
(for/list ([n (in-range start end )])
(cons (vector-ref points org ) (vector-ref points n ))))
(for/fold ([accum empty]) ([i (in-range l)])
(append accum (gen i (+ i 1) l)))
On Mon, Dec 5, 2011 at 9:10 PM, Sam Tobin-Hochstadt <sa...@ccs.neu.edu> wrote:
> On Mon, Dec 5, 2011 at 10:00 AM, Geoffrey S. Knauth <ge...@knauth.org> wrote:
>> I'm wondering if there is something in the now very rich set of Racket
>> libraries that already does this. Let's say I have 5 points {A,B,C,D,E}. I
>> want to interconnect all of them:
>>
>> {AB,AC,AD,AE,AF,BC,BD,BE,BF,CD,CE,CF,DE,DF,EF}
>>
>> That's 15 edges rather than the 5x5=25 that a dumb interconnect would do.
>> To start, I just need to track and sort the edge weights, and AB is the same
>> as BA.
>
> Here's the start of an answer (sadly quadratic, but for N=121 I don't
> think that will matter much):
>
> #lang racket
> (require unstable/sequence)
> (define (subsets-of-size-2 l)
> (for*/lists (r) ([i l] [j l]
> #:unless (eq? i j)
> #:unless (member (cons j i) r))
> (cons i j)))
>
> (for ([(p q) (in-pairs (subsets-of-size-2 '(A B C D E F)))])
> (printf "~a <-> ~a\n" p q))
>
> --
> sam th
> sa...@ccs.neu.edu
>
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