Hi Cev and Paulo!
Is this what you want?
(define condeo
(lambda (ls x)
(fresh (a d)
(== `(,a . ,d) ls)
(conde
((== a x))
((condeo d x))))))
> (run* (q) (condeo '(1 2 3) q))
(1 2 3)
Cheers,
--Will
On Wed, Mar 8, 2017 at 8:44 PM, Paulo César Cuneo
<[email protected]> wrote:
> Hi! Im not a minikanren authority.
> Anyway I cant make sense of what you want to do.
> As i undestand conde expresses alternative "executions", not a list of
> results
>
> See what this outputs
>
> (run* (x)
> (conde
> ((== x "1"))
> ((== x "2"))
> ((== x "3")))
> (conde
> ((== x "4"))
> ((== x "5"))
> ((== x "6"))))
>
> Any way i guess ill try something like this:
> (I niether a schemer :P, and probably ahead comes buggy code)
>
> (define condeo (lambda in out)
> (conde
> ((== in '()) ;; the case in is empty cant unify/relate
> out to nothing
> #f) ;; must fail or out will be free
> ((fresh (head tail)
> (conso head tail in) ;; the case in is not
> empty splits in 2
> (conde ((== out head)) ;; theres a case
> where out is the first element
> ((condeo tail out))))))) ;; theres case
> where out has some value from the rest of the list.
>
>
> Cheers.
>
>
> On Wednesday, March 8, 2017 at 1:49:31 PM UTC-3, Cev Ing wrote:
>>
>> I am new to miniKanren and still struggling with the basics. Maybe someone
>> can help me.
>>
>> I try to write a function taking a variable number of arguments and
>> behaving like conde.
>>
>> This is my cond example:
>>
>> (run* (x)
>> (conde
>> ((== x "1"))
>> ((== x "2"))
>> ((== x "3")))) ;; => ("1" "2" "3")
>>
>> I can write a macro, which takes a variable number of arguments and
>> converts it into a conde:
>>
>> (define-syntax conde-list
>> (syntax-rules ()
>> ((_ (a0 a1 ...) x)
>> (conde
>> ((== x a0))
>> ((== x a1))
>> ...))))
>>
>> (run* (x)
>> (conde-list ("1" "2" "3") x)) ;; => ("1" "2" "3")
>>
>> Now I tried to write a similar function condeo in miniKanren using
>> recursion so that the following expression is true.
>>
>> (run* (x)
>> (condeo '("1" "2" "3") x)) ;; => ("1" "2" "3")
>>
>> This is what I did, but it does not work:
>>
>> (define condeo
>> (lambda (in out)
>> (fresh (hi ti)
>> (== `(,hi . ,ti) in)
>> (fresh (ho to)
>> (== `(,ho . ,to) out)
>> (conde
>> ((== hi ho))
>> ((condeo ti to)))))))
>>
>> Like the appendo example in the tutorials I split the input list in head
>> and tail and the same for the output list. And then I use conde for head and
>> tail, putting the recursion at the end. But when I try it, I get funny
>> results:
>>
>> (run 4 (x)
>> (condeo '("1" "2" "3") x)) ;; => (("1" . _.0) (_.0 "2" . _.1) (_.0 _.1
>> "3" . _.2))
>>
>> It iterates somehow but why are there so many free variables?
>>
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