[PRL] Stooge sort - Wikipedia, the free encyclopedia

Matthias Felleisen matthias at ccs.neu.edu
Wed Jun 22 11:59:56 EDT 2011


On Jun 22, 2011, at 11:54 AM, J. Ian Johnson wrote:

> 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)


Could we please use (or (empty? (cdr l)) (empty? (cddr l))) please? Pretty?!?! 




>                    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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> 
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