I've used clojure for a while but just starting out with core.logic.
Given a domain like 1 2 3 4 I want to get a vector of two vectors back like ([[1 2] [3 4]]).
Note: This is just a simplified version of what I'm really trying to do. :) See: https://github.com/adamhoward/lineup
I found this definition for not-membero on the web:
(defne not-membero [x l]
([_ []])
([_ [?y . ?r]]
(!= x ?y)
(not-membero x ?r)))
And I'm trying to use it like this:
(run 1 [q]
(fresh [w x
y z]
(== q [[w x]
[y z]])
(infd w x y z (domain 1 2 3 4))
(everyg distinctfd [[w x] [y z]])
(everyg #(not-membero % [y z]) [w x])))
Running this in Emacs gives me an Evaluation aborted. message.
When I try switching out membero for not-membero I get back ([[1 2] [1 2]]) which makes sense to me. Every element in the first vector [w x] is a member of the second vector [y z].
But, when I call run 2 I get back ([[1 2] [1 2]] [[1 2] [1 3]]). I don't understand how [[1 2] [1 3]] can be correct for the rules above. Am I not understanding everyg correctly? Any guidance (including rtfmanual, rtfbook, rtfdissertation) will be appreciated.
Thanks.
Edit: May have solved this.
Still not sure about the strange results from membero but instead of the not-membero goal I found that I could do this:
(everyg #(distinctfd (conj [y z] %)) [w x])
Each element of [w x] conj'd to [y z] contains all distinct values. This may be less efficient than not-membero so I'm still open to any help.
Your example seems to be doing the same as this:
But seeing the soccer problem...