Hello Will,
thanks for your suggestion.
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.
I think the problem is that the recursion doesn't work,
because if in
(amb (car list)
(amb-list (cdr list)))))
the first line doesnt' lead to a fail,
then the second line isn't evaluated.
greetings,
Stefan
Am 12.09.2010 20:16, schrieb Will M. Farr:
Stefan,
amb is a macro that chooses among the results of evaluating the expressions it
is provided, so
(amb *someList*)
evaluates *someList*, resulting in a list, and chooses that. You could define
a function
(define (amb-list list)
(if (null? list)
(amb)
(amb (car list)
(amb-list (cdr list)))))
It's also possible that the amb library provides such a function, which may be
implemented more efficiently; I don't know which amb library you're using, so I
can't help you there.
Note that the simple solution of using
(apply amb *someList*)
will fail, since amb is a macro, not an ordinary function. Good luck!
Will
On Sep 12, 2010, at 9:16 AM, Stefan Busch wrote:
Hello racket community,
though it may be impolite to enter the community with a wish,
I'd like to ask for help with the following problem:
I want a and b to "run through" all elements of two given lists,
and then to catch the combinations that fullfill a desired condition with
an "amb-assert" statement.
If I try this in the following way
(amb-collect
(let ((a (amb *someList*)
(b (amb *someDifferentList*))...............
it turns out that not the list elemets are considered possible values for a or
b,
but the lists themselves are treated as the only possible value.
In all examples I could find for the uage of "amb-collect" the values to run
through were
recited, but in my case these are quite many, so would be grateful for a
handier solution.
thanks,
Stefan
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