On 12/05/2011 09:20 AM, Maurizio Giordano GMAIL wrote:
On Mon, 2011-12-05 at 09:00 -0700, Ryan Culpepper wrote:
On 12/05/2011 08:40 AM, Sam Tobin-Hochstadt 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))


That looks quartic in the length of l, because of the member check.

Here's a quadratic version:

(require srfi/1)
(define (subsets-of-size-2 l)
    (for*/list ([ne-sublist (pair-fold-right cons null l)]
                [b (in-list (cdr ne-sublist))])
      (cons (car ne-sublist) b)))

Note: (pair-fold-right cons null l) produces a list of the non-empty
sublists of l.

Hi Ryan,
some time ago I was looking for an efficient algorithm in Racket to
extract k-sized combinations of elements from a list l... Could you
provide a generalisation of your code, I mean "subset-of-size-k"
where k<  (lenght l) ?

Here's the basic idea:

;; subsets-of-size : nat list -> (listof list)
(define (subsets-of-size n l)
  (cond [(zero? n) (list null)]
        [else
         (for*/list ([ne-list (pair-fold-right cons null l)]
                     [subsetN-1
                      (in-list (subsets-of-size (sub1 n)
                                                (cdr ne-list)))])
           (cons (car ne-list) subsetN-1))]))

This has bad complexity, I believe, although I haven't calculated it. If you add in memoization, though, you should get good (optimal?) complexity and optimal tail-sharing.

Ryan
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