[PRL] Stooge sort - Wikipedia, the free encyclopedia

[J. Ian Johnson]

On Wed, Jun 22, 2011 at 11:31 AM, J. Ian Johnson <ianj at ccs.neu.edu> wrote:
>> I've been spending a little time on this in the mornings, and I can tell you that a formal proof is no simple matter.
>> In ACL2, the straightforward implementation is formally impossible to prove terminating since you must know that the output is a >list of the same size as the input before admitting the function. Thus, you must add "no-ops" like (if (= (length output) (length >input)) [... rest of program ...] 'impossible)

>Why do you need this?  You should be able to compute N = 2/3 the size
>of the input list once, and any time you recur, do it by taking the
>first or last N elements of some intermediate list.  That should make
>the size of the input to each recursion N, and proving totality should
>be straightforward.

(defun stsort (x)
  (cond ((endp x) nil)
        (t (let ((l (if (< (last x) (first x))
                           (swap-ends x)
                         x)))
                (if (< (length l) 3)
                    l
                  (let* ((l (append (stsort (lower2/3 l))
                                    (upper1/3 l)))
                         (l (append (lower1/3 l)
                                    (stsort (upper2/3 l)))) ;; oh bugger
                         (l (append (stsort (lower2/3 l))
                                    (upper1/3 l))))
                    l))))))
The intermediate lists depend on the output of the function. We need to know that the length of the first (and second) append is equal to the length of x.

>> Once you do that, proving that taking the different thirds of the list for input longer than 2 also requires multiple lemmas >>about floor, ceiling, take and nthcdr.

>Now this, I agree with.  Reasoning about fractions of list lengths in
>ACL2 is where I would expect the difficulty to arise.

The lemmas you need aren't too bad, but trying to prove
(implies (and (posp n) (integerp x))
         (equal (+ (floor x n) (ceiling x (- 1 (/ 1 n))))
                x))
[or something like it, that lemma is on my home computer]
then with arithmetic-5, you get a rewriting loop. Automated math is hard.

-Ian
> ----- Original Message -----
> From: "Paul A. Steckler" <steck at stecksoft.com>
> To: "Mitchell Wand" <wand at ccs.neu.edu>
> Cc: "Dan Friedman" <dfried00 at gmail.com>, "PRL" <prl at lists.ccs.neu.edu>
> Sent: Friday, June 17, 2011 12:33:41 AM GMT -05:00 US/Canada Eastern
> Subject: Re: [PRL] Stooge sort - Wikipedia, the free encyclopedia
>
> On Fri, Jun 17, 2011 at 11:46 AM, Mitchell Wand <wand at ccs.neu.edu> wrote:
>> Anybody want to formalize this in Coq/ACL2/whatever ?
>
> In Sydney, we're organizing a CoqFight, where contestants will be
> given problems
> to formalize in a competitive setting. This problem might be useful in
> that context (if
> in no other).
>
> -- Paul
>
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