Re: [rules-users] Can I solve the 'Zebra Puzzle' in drools?

Hi, This is a puzzling one... but then again, I believe that is the idea. I should have a very close look at it, but a loop always suggests that there are rules that re-activate themselves or eachother. For example: rule "抽kools牌的香烟的人与养马的人是邻居" when $h1:House(cigarette == Cigarette.kools) $h2:House(pet == Pet.horse) $h3:House(eval(position - $h2.position == 1) || eval(position - $h2.position == -1)) eval($h1.position - $h2.position != 1 && $h1.position - $h2.position != -1) then System.out.println("抽kools牌的香烟的人与养马（horse）的人是邻居"); modify($h1){setCigarette($h3.cigarette)}; modify($h3){setCigarette(Cigarette.kools)}; end In this rule, $h3 and $h1 might be the same house OR $h1 and $h2 might be the same house. I think you need to add extra constraints to make sure that all 3 houses are different. But still, I haven't had a thorough look at all the rules neither did I refresh my knowledge of the zebrapuzzle, so I am not sure whether this is a complete answer (if any). Please let us know. Regards, Frank -- View this message in context: http://drools.46999.n3.nabble.com/rules-users-Can-I-solve-the-Zebra-Puzzl... Sent from the Drools: User forum mailing list archive at Nabble.com.