3:25 a.m.
hello everybody,
I have some trouble in solving the 'Zebra
Puzzle<http://en.wikipedia.org/wiki/Zebra_Puzzle>'
using drools5.
I defined a class named 'House' to represent the data, and the corresponding
enums as well : Color.java, Nation.java, Pet.java, Drink.java,
Cigarette.java, here are those sources:
House.java:
public class House {
// 颜色
public Color color;
// 房子从左数的位置 1-5
public int position;
public Nation nation;
public Pet pet;
public Drink drink;
public Cigarette cigarette;
......
Color.java:
public enum Color {
red, yellow, blue, green, ivory
}
Nation.java:
public enum Nation {
englishman, spaniard, norwegian, ukrainian, japanese
}
Pet.java:
public enum Pet {
dog, fox, snails, horse, zebra
}
Drink.java:
public enum Drink {
orangeJuice, tea, coffee, milk, water
}
Cigarette.java:
public enum Cigarette {
kools, chesterfields, winston, luckyStrike, parliaments
}
And this is my drl file:
package com.sample
import com.sample.fh.*
rule "英国人住在红色的房子里"
when
$h1:House(nation == Nation.englishman, color != Color.red)
$h2:House(color == Color.red)
then
System.out.println("英国人（englishman）住在红色（red）的房子里");
modify($h2){setColor($h1.color)};
modify($h1){setColor(Color.red)};
end
rule "西班牙人养了一条狗"
when
$h1:House(nation == Nation.spaniard)
$h2:House(pet == Pet.dog, nation != Nation.spaniard)
then
System.out.println("西班牙人（spaniard）养了一条狗（dog）");
modify($h2){setPet($h1.pet)};
modify($h1){setPet(Pet.dog)};
end
rule "挪威人住在左边的第一个房子里"
when
$h1:House(position == 1, nation != Nation.norwegian)
$h2:House(nation == Nation.norwegian)
then
System.out.println("挪威人（norwegian）住在左边的第一个房子里");
modify($h2){setNation($h1.nation)};
modify($h1){setNation(Nation.norwegian)};
end
rule "黄房子里的人喜欢抽kools牌的香烟"
when
$h1:House(color == Color.yellow, cigarette != Cigarette.kools)
$h2:House(cigarette == Cigarette.kools)
then
System.out.println("黄房子（yellow）里的人喜欢抽kools牌的香烟");
modify($h2){setCigarette($h1.cigarette)};
modify($h1){setCigarette(Cigarette.kools)};
end
rule "抽chesterfields牌香烟的人与养狐狸的人是邻居"
when
$h1:House(cigarette == Cigarette.chesterfields, pet != Pet.fox)
$h2:House(pet == Pet.fox)
$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chesterfields牌香烟的人与养狐狸（fox）的人是邻居");
int tmp = $h3.position;
modify($h3){setPosition($h1.position)};
modify($h1){setPosition(tmp)};
end
rule "抽chesterfields牌香烟的人与养狐狸的人是同一人"
when
$h1:House(cigarette == Cigarette.chesterfields, pet == Pet.fox)
$h2:House(eval(position - $h1.position == 1) || eval(position -
$h1.position == -1))
then
System.out.println("抽chesterfields牌香烟的人与养狐狸的人是同一人");
modify($h1){setPet($h2.pet)};
modify($h2){setPet(Pet.fox)};
end
rule "挪威人住在蓝色的房子旁边"
when
$h1:House(nation == Nation.norwegian)
$h2:House(color == Color.blue)
$h3:House(eval(position - $h2.position == 1) || eval(position -
$h2.position == -1))
eval($h1.position - $h2.position != 1 && $h1.position - $h2.position
!= -1)
then
System.out.println("挪威人（norwegian）住在蓝色（blue）的房子旁边");
modify($h1){setNation($h3.nation)};
modify($h3){setNation(Nation.norwegian)};
end
rule "抽winston牌香烟的人养了一只蜗牛"
when
$h1:House(cigarette == Cigarette.winston)
$h2:House(pet == Pet.snails, cigarette != Cigarette.winston)
then
System.out.println("抽winston牌香烟的人养了一只蜗牛（Snails）");
modify($h2){setPet($h1.pet)};
modify($h1){setPet(Pet.snails)};
end
rule "抽LuckyStrike牌香烟的人喜欢喝桔子汁"
when
$h1:House(cigarette == Cigarette.luckyStrike)
$h2:House(drink == Drink.orangeJuice, cigarette !=
Cigarette.luckyStrike)
then
System.out.println("抽Lucky Strike牌香烟的人喜欢喝桔子汁（orange juice）");
modify($h2){setDrink($h1.drink)};
modify($h1){setDrink(Drink.orangeJuice)};
end
rule "乌克兰人喜欢喝茶"
when
$h1:House(nation == Nation.ukrainian)
$h2:House(drink == Drink.tea, nation != Nation.ukrainian)
then
System.out.println("乌克兰人（ukrainian）喜欢喝茶（tea）");
modify($h2){setDrink($h1.drink)};
modify($h1){setDrink(Drink.tea)};
end
rule "日本人抽parliaments牌的烟"
when
$h1:House(nation == Nation.japanese)
$h2:House(cigarette == Cigarette.parliaments, nation !=
Nation.japanese)
then
System.out.println("日本人（japanese）抽parliaments牌的烟");
modify($h2){setCigarette($h1.cigarette)};
modify($h1){setCigarette(Cigarette.parliaments)}
end
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
rule "喜欢喝咖啡的人住在绿房子里"
when
$h1:House(drink == Drink.coffee)
$h2:House(color == Color.green, drink != Drink.coffee)
then
System.out.println("喜欢喝咖啡（coffee）的人住在绿（green）房子里");
modify($h2){setColor($h1.color)};
modify($h1){setColor(Color.green)};
end
rule "绿房子在象牙白房子的右边"
when
$h1:House(color == Color.green, position != 1)
$h2:House(color == Color.ivory, position != 5)
$h3:House(eval(position == $h2.position + 1), color != Color.green)
then
System.out.println("绿（green）房子在象牙白（ivory）房子的右边（图中的右边）");
modify($h1){setColor($h3.color)};
modify($h3){setColor(Color.green)};
end
rule "白房子已经在最右边了"
when
$h1:House(color == Color.ivory, position == 5)
$h2:House(color == Color.green, position != 1)
$h3:House(eval(position == $h2.position -1))
then
System.out.println("白房子已经在最右边了");
$h3.setPosition($h1.position);
$h1.setPosition($h2.position - 1);
update($h3);
update($h1);
end
rule "绿房子已经在最左边了"
when
$h1:House(color == Color.green, position == 1)
$h2:House(color == Color.ivory, position != 5)
$h3:House(eval(position == $h2.position + 1))
then
System.out.println("绿房子已经在最左边了");
modify($h3){setPosition($h1.position)};
modify($h1){setPosition($h2.position + 1)};
end
rule "绿房子在最左边且白房子在最右边"
when
$h1:House(color == Color.green, position == 1)
$h2:House(color == Color.ivory, position == 5)
$h3:House()
$h4:House(eval(position - $h3.position == 1))
then
System.out.println("绿房子在最左边且白房子在最右边");
modify($h2){setPosition($h3.position)};
modify($h3){setPosition(5)};
modify($h1){setPosition($h4.position)};
modify($h4){setPosition(1)};
end
rule "中间那个房子里的人喜欢喝牛奶"
when
$h1:House(position == 3, drink != Drink.milk)
$h2:House(drink == Drink.milk)
then
System.out.println("中间那个房子里的人喜欢喝牛奶");
modify($h2){setDrink($h1.drink))};
modify($h1){setDrink(Drink.milk)};
end
rule "output"
when
$h1:House(position == 1)
$h2:House(position == 2)
$h3:House(position == 3)
$h4:House(position == 4)
$h5:House(position == 5)
House(nation == Nation.englishman, color == Color.red)
House(nation == Nation.spaniard, pet == Pet.dog)
House(position == 1, nation == Nation.norwegian)
House(color == Color.yellow, cigarette == Cigarette.kools)
$h6:House(cigarette == Cigarette.chesterfields)
$h7:House(pet == Pet.fox)
eval($h6.position - $h7.position == 1 || $h6.position - $h7.position
== -1)
$h8:House(nation == Nation.norwegian)
$h9:House(color == Color.blue)
eval($h8.position - $h9.position == 1 || $h8.position - $h9.position
== -1)
House(cigarette == Cigarette.winston, pet == Pet.snails)
House(cigarette == Cigarette.luckyStrike, drink ==
Drink.orangeJuice)
House(nation == Nation.ukrainian, drink == Drink.tea)
House(cigarette == Cigarette.parliaments, nation == Nation.japanese)
$h10:House(cigarette == Cigarette.kools)
$h11:House(pet == Pet.horse)
eval($h10.position - $h11.position == 1 || $h10.position -
$h11.position == -1)
House(drink == Drink.coffee, color == Color.green)
$h12:House(color == Color.green)
$h13:House(color == Color.ivory)
eval($h12.position - $h13.position == 1)
House(position == 3, drink == Drink.milk)
then
System.out.println("从左到右:");
System.out.println("第一间房子是"+$h1.color+"色的;住着"+$h1.nation+"人;养"+$h1.pet+"宠物;抽"+$h1.cigarette+"牌香烟;喝"+$h1.drink+"饮料;");
System.out.println("第二间房子是"+$h2.color+"色的;住着"+$h2.nation+"人;养"+$h2.pet+"宠物;抽"+$h2.cigarette+"牌香烟;喝"+$h2.drink+"饮料;");
System.out.println("第三间房子是"+$h3.color+"色的;住着"+$h3.nation+"人;养"+$h3.pet+"宠物;抽"+$h3.cigarette+"牌香烟;喝"+$h3.drink+"饮料;");
System.out.println("第四间房子是"+$h4.color+"色的;住着"+$h4.nation+"人;养"+$h4.pet+"宠物;抽"+$h4.cigarette+"牌香烟;喝"+$h4.drink+"饮料;");
System.out.println("第五间房子是"+$h5.color+"色的;住着"+$h5.nation+"人;养"+$h5.pet+"宠物;抽"+$h5.cigarette+"牌香烟;喝"+$h5.drink+"饮料;");
end
There's some unicode character in the file.But it shouldn't bother the
compilation>...<
The rule named 'output' describes when the computation should stop. The
other rules are absorbed from the original 'Zebra puzzle' problem and each
making a little change to the working memory in order to 'approach' the
solution...(I don't know if I'm doing this wrong..)
No matter how I modify this program, it always result in an infinite loop...
For example, the console prints these messages repeatedly:
白房子已经在最右边了
抽chesterfields牌香烟的人与养狐狸的人是同一人
抽kools牌的香烟的人与养马（horse）的人是邻居
黄房子（yellow）里的人喜欢抽kools牌的香烟
抽kools牌的香烟的人与养马（horse）的人是邻居
黄房子（yellow）里的人喜欢抽kools牌的香烟
抽kools牌的香烟的人与养马（horse）的人是邻居
黄房子（yellow）里的人喜欢抽kools牌的香烟
抽kools牌的香烟的人与养马（horse）的人是邻居
黄房子（yellow）里的人喜欢抽kools牌的香烟
抽kools牌的香烟的人与养马（horse）的人是邻居
黄房子（yellow）里的人喜欢抽kools牌的香烟
......................
I worked quite much time on this problem till now, and I referenced another
solution which is programmed in prolog, but still I didn't figured out the
drools solution to this problem.
Could somebody give a hint? Any help is appreciated,
Thanks!
--
Regards.
Miles. Wen
7:47 a.m.
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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9:48 a.m.
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(a)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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10 a.m.
Hi,
Just a last quick reply before getting stuck in a traffic jam.
As said, I haven't had a real good look at the rules, but imo they might be
able to work when you feed the engine an initial solution. I.e. insert five
house objects into working memory with a random assignment of the
"attributes". The engine could then re-arrange the objects until all
constraints are satisfied.
But again, I am not really certain about these statements, but it is an
interesting topic to think about...
Regards,
Frank
Wolfgang Laun-2 wrote:
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 &lt;frank.vanhoenshoven(a)agserv.eu&gt;
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
>
>
> --
> 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.
>
> _______________________________________________
> rules-users mailing list
> rules-users(a)lists.jboss.org
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>
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4913
days inactive
4913
days old
3 comments
3 participants
participants (3)
-
FrankVhh -
Miles Wen -
Wolfgang Laun