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

[FrankVhh]

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


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