Re[2]: sorted?

------ Original Message ------
From [email protected]
To "Stefan Schmiedl" <[email protected]>
Cc [email protected]
Date 09.12.2024 15:10:18
Subject Re: sorted?

>On Mon, Dec 09, 2024 at 01:54:47PM +0000, Stefan Schmiedl wrote:
>>  ------ Original Message ------
>>  > From [email protected]
>>  To [email protected]
>>  Date 09.12.2024 12:42:22
>>  Subject Re: sorted?
>>
>>  > On Mon, Dec 09, 2024 at 11:37:33AM +0000, Ricardo G. Herdt wrote:
>>  > >  Hi Jeremy,
>>  > >
>>  > >  Am 09.12.2024 11:21 schrieb Jeremy Korwin-Zmijowski:
>>  > >  > The reference says :
>>  > >  >
>>  > >  >    Scheme Procedure: *sorted?* items less
>>  > >  >    C Function: *scm_sorted_p* (items, less)
>>  > >  >
>>  > >  >        Return |#t| if items is a list or vector such that, for each
>>  > >  >        element x and the next element y of items, |(less y x)| returns
>>  > >  >        |#f|. Otherwise return |#f|.
>>  > >  >
>>  > >  > I think the description should be :
>>  > >  >
>>  > >  >    Return |#t| if items is a list or vector such that, for each element
>>  > >  >    x and the next element y of items, |(less y x)| returns |#t|.
>>  > >  >    Otherwise return |#f|.
>>  > >
>>  > >  Actually no, since less is applied to y and x in that order. This way
>>  > >  (sorted? '(1 1) <) correctly returns #t as your experiments show.
>>  >
>>  > I don't get it. (< 1 1) is /always/ #f, regardless of the order of the
>>  > ones?
>>  >
>>  > I'm as confused as Jeremy is.
>>  >
>>
>>  I understand the reference text as "Return #t if the list is _not
>>  unsorted_".
>>  Since (< 1 1) returns #f, '(1 1) is _not unsorted_ and all is well.
>
>This seems the intention. But since it accepts an arbitrary "less"
>function, it ends being iffy. How do you go from some "less" to a
>"less-or-equal" without running into undecidability dark alleys?
>
Well, the iffyness is managed by the programmer, who chooses the
comparison function.

Do you have an example of such an iffy set?

s.