Why cant i get the answer to the zebra puzzle in prolog?

问题

this is my code currently, I am trying to solve the zebra puzzle.

exists(A,(A,_,_,_,_)).
exists(A,(_,A,_,_,_)).
exists(A,(_,_,A,_,_)).
exists(A,(_,_,_,A,_)).
exists(A,(_,_,_,_,A)).

rightOf(A,B,(B,A,_,_,_)).
rightOf(A,B,(_,B,A,_,_)).
rightOf(A,B,(_,_,B,A,_)).
rightOf(A,B,(_,_,_,B,A)).

middleHouse(A,(_,_,A,_,_)).

firstHouse(A,(A,_,_,_,_)).

nextTo(A,B,(B,A,_,_,_)).
nextTo(A,B,(_,B,A,_,_)).
nextTo(A,B,(_,_,B,A,_)).
nextTo(A,B,(_,_,_,B,A)).
nextTo(A,B,(A,B,_,_,_)).
nextTo(A,B,(_,A,B,_,_)).
nextTo(A,B,(_,_,A,B,_)).
nextTo(A,B,(_,_,_,A,B)).

:- Houses = (house(N1,P1,S1,D1,C1),house(N2,P2,S2,D2,C2),house(N3,P3,S3,D3,C3),house(N4,P4,S4,D4,C4),house(N5,P5,S5,D5,C5)),
exists(house(english,_,_,_,red),Houses),
exists(house(spainish,dog,_,_,_),Houses),
exists(house(_,_,_,coffee,green),Houses),
exists(house(ukrainian,_,_,tea,_),Houses),
rightOf(house(_,_,_,_,green),house(_,_,_,_,ivory),Houses),
exists(house(_,dog,oldgold,_,_),Houses),
exists(house(_,_,kools,_,yellow),Houses),
middleHouse(house(_,_,_,milk,_),Houses),
firstHouse(house(norwegian,_,_,_,_),Houses),
nextTo(house(_,_,chesterfields,_,_),house(_,fox,_,_,_),Houses),
nextTo(house(_,_,kools,_,_),house(_,horse,_,_,_),Houses),
exists(house(_,_,luckystike,orangejuice,_),Houses),
exists(house(japanise,_,parliments,_,_),Houses),
nextTo(house(norwegian,_,_,_,_),house(_,_,_,_,blue),Houses),
exists(house(WaterDrinker,_,_,water,_),Houses),
exists(house(ZebraOwner,zebra,_,_,_),Houses).

I have typed this up and saved it as zebra.pl, this I open it and enter [zebra] into SWI-prolog, it returns a warning message about the singleton use of N1,P1,C1 etc.. and returns true, then i ask it to print water drinker using print(WaterDrinker) and it returns _G317 and true,

why is it doing this and not returning the answer which could be norwegian, it does the same if i ask it return any variable like C3 or ZebraOwner

回答1:

The main problem is that you cannot write a goal like :- Houses = ... in the middle of your program. Rather, you should write something like

solution(WaterDrinker, ZebraOwner) :-
    Houses = ...

and then after the program has been loaded type

 solution(W, Z).

at the -? prompt to compute the solution.

You also don't specify that all the values should be distinct. If you do that you will use the variables a second time and the warning will go away.

If you really need a variable only once you can prepend it with an underscore to make the warning go away. Or you can just use an underscore, like you already did many times.

来源：https://stackoverflow.com/questions/7338225/why-cant-i-get-the-answer-to-the-zebra-puzzle-in-prolog

标签

prolog

logic

logic-programming

zebra-puzzle