[racket] handy trick for "amb-collect" desired

Hello,

of course I have to use  amb-collect.

Maybe I didnt describe my problem properly.

My intention is to pass the arguments to choose from in a list.

The teaching material I got from university gives the following exanple for the usage
of amb-collect:

***************************************************************************************

;;get all pairs of values von a and b,
;;which have sum 7

(amb-collect)
( let ( ( a (amb 1 2 3 4 5 6 7 ) )
           (b (amb 2 4 6 8 ) ) )
( amb-assert (= (+ a b) 7 ) )
( cons a b ) ) )


-------->( ( 1 . 6) (3 . 4) (5 . 2 ) )

***************************************************************************************

I'm looking for a way to use two lists instead to provide these values: (1 2 3 4 5 6 7) and (2 4 6 8)


Thanks,
Stefan




Am 14.09.2010 17:50, schrieb Thomas Chust:
> 2010/9/14 Stefan Busch<stefan_busch_ at arcor.de>:
>    
>> [...]
>> Your solution looked convincing to me,
>> but when I try e.g.
>>
>> (amb list (list 1 2 3 4))
>>
>> it's not evaluated to the desired
>>
>> 1 2 3 4,
>>
>> but to
>>
>> 1.
>> [...]
>>      
> Hello,
>
> I assume you meant to say (amb-list ...) instead of (amb list ...). Of course
> this expression evaluates to the first item in the list. This is the whole
> point of amb: To return the result *one* of its arguments evaluates to, but to
> provide the possibility of backtracking and choosing another argument.
>
> If you want a list of all possible return values of an expression, you *must*
> wrap it in an amb-collect. The implementation of amb provided by
> (planet murphy/amb:1:1/amb) will even signal an error if the dynamic context
> of an ambiguous expression is not enclosed by something equivalent to amb-find
> or amb-collect.
>
> Ciao,
> Thomas
>
>
>