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

[Wolfgang Laun]

This puzzle can be solved using forward chaining, i.e., by using Drools or
any similar RBS, but it isn't possible by implementing the givens as rules,
expecting that an increase in the filled-in values will result in a (the)
solution. Certainly, it is possible to write a rule such as

rule EnglishRed
when
    $h: House( nationality == null && colour == Colour.RED ||
               nationality == Nationality.ENGLISHMAN && colour == null )
then
    modify( $h ){
        setNationality( Nationality.ENGLISHMAN ),
        setColour( Colour.RED )
    }
end

which adds "red" or "Englishman" to a House as soon as the other property is
present. But this approach comes to a standstill as soon as more
sophisticated mental processes are required in the manual solution
technique.

One approach I've tried successfully is to generate all permutations of
animals, colours, drinks, nationalities and smokes (120 each), insert them
as facts, and to write one big rule according to the givens. A similar
approach uses facts consisting of an attribute indication (animal,...
smoke), a value and a reference to a House, again starting with all possible
associations and using one rule to select the attribute combinations
satisfying the givens.

I'm not quite sure how the rules presented by Miles should work, but I don't
think that swapping values between House facts is going to work, even when
more defensive strategies against looping are employed.

-W







On 15 June 2011 14:47, FrankVhh <frank.vanhoenshoven at agserv.eu> wrote:

> 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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> Sent from the Drools: User forum mailing list archive at Nabble.com.
>
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