Zebra puzzle

Zebra puzzle is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The Zebra puzzle, a.k.a. Einstein's Riddle, is a logic puzzle which is to be solved programmatically. It has several variants, one of them this:

1. There are five houses.
2. The English man lives in the red house.
3. The Swede has a dog.
4. The Dane drinks tea.
5. The green house is immediately to the left of the white house.
6. They drink coffee in the green house.
7. The man who smokes Pall Mall has birds.
8. In the yellow house they smoke Dunhill.
9. In the middle house they drink milk.
10. The Norwegian lives in the first house.
11. The man who smokes Blend lives in the house next to the house with cats.
12. In a house next to the house where they have a horse, they smoke Dunhill.
13. The man who smokes Blue Master drinks beer.
14. The German smokes Prince.
15. The Norwegian lives next to the blue house.
16. They drink water in a house next to the house where they smoke Blend.

The question is, who owns the zebra?

cf. Dinesman's multiple-dwelling problem

Prolog

In Prolog we can specify the domain by selecting from it, making mutually exclusive choices for efficiency:

<lang Prolog>select([A|As],S):- select(A,S,S1),select(As,S1). select([],_).

next_to(A,B,C):- left_of(A,B,C),left_of(B,A,C). left_of(A,B,C):- append(_,[A,B|_],C).

zebra(Owns, HS):-  % color,nation,pet,drink,smokes

     HS = [h(_,norwegian,_,_,_),h(blue,_,_,_,_),h(_,_,_,milk,_),_,_], 
     select( [h(red,eng,_,_,_),h(_,swede,dog,_,_),h(_,dane,_,tea,_),
              h(_,german,_,_,prince)], HS),
     select( [h(_,_,birds,_,pallmall),h(yellow,_,_,_,dunhill),
              h(_,_,_,beer,bluemaster)], HS), 
     left_of( h(green,_,_,coffee,_),h(white,_,_,_,_), HS),
     next_to( h(_,_,_,_,dunhill),h(_,_,horse,_,_), HS),
     next_to( h(_,_,_,_,blend),  h(_,_,cats, _,_), HS),
     next_to( h(_,_,_,_,blend)  ,h(_,_,_,water,_), HS),
     member(  h(_,Owns,zebra,_,_), HS).

- findall(Who,zebra(Who,_),L).</lang>

Output: <lang Prolog> L = [german].</lang>

Works with SWI-Prolog.