Interpreters riddle in prolog

Question

Here's a riddle's info

Six interpreters: Fran French, Geraldine German, Dudley Dutch, Spike Spanish, Polly Polish and Romanian Ron. Everyone speaks two languages, among them indicating their surnames, namely French (French), Deutsch (German), Dutch (Dutch), Spanish (Spanish), Polish (Polish) and Romanian ( Romanian). The following applies:

No interpreter speak the language corresponding to his surname.
No two interpreters speak the same two languages
Each of the six languages is spoken by exactly two interpreters.
Spike speaks Dutch and German.
There is an interpreter who speaks Dutch and Polish.
Fran and Dudley speak four languages (two each) that do not correspond to their surnames.
Interpreters who have as surnames the languages spoken by Dudley, speak French.
There is no interpreter who can speak German and Polish.

What languages are spoken by every interpreter?

This is what I've done until now. Rules 1,4 and 8 are returning right results.

interpreter(fran,french).
interpreter(geraldine,german).
interpreter(dudley,dutch).
interpreter(spike,spanish).
interpreter(polly,polish).
interpreter(ron,romanian).

language(french).
language(german).
language(dutch).
language(spanish).
language(polish).
language(romanian).



%rule 1                                                                     OK
rule1([X,Y,Z,W]):- interpreter(X,Y), language(Z), language(W),   
                not(Z=W;Y=Z;Y=W),
                Z\=W, Z@<W.


%rule 2    
    %rule2([X,Y,Z,W]):- interpreter(X,Y), language(Z), language(W).
                    %interpreter(I2,S2), language(Z2), language(W2).
                    %not(X1=X2),
                    %((Z1=Z2) -> not(W1=W2)).



%rule 3
rule3([X,Y,Z,W]):- interpreter(X,Y), language(Z1), language(W), 
                interpreter(X,Y), language(Z1), language(W), 
                interpreter(X,Y), language(Z2), language(W), 
                not(Z1=Z2).

%rule 4                                                                 OK
rule4([X,Y,Z,W]):-  (X=spike -> Z=dutch,W=german;Z=_,W=_).

%rule 5
rule5([X,Y,dutch,polish]).

%rule 6
rule6a([fran,french,Z,W]):- interpreter(X,Y), language(Z1), language(W), 
                not(Z=dutch).
rule6b([dudley,dutch,Z,W]):- interpreter(X,Y), language(Z1), language(W), 
                not(Z=french).
    /*      
%rule 7
rule7([dudley,dutch,Z,W]):- rule7a(X,Z,L1,french),
                        rule7b(X,Z,french,L2),
                        rule7c(X,W,L1,french),
                        rule7d(X,W,french,L2).
                        */
%rule 8                                                                 OK
rule8([X,Y,Z,W]):- interpreter(X,Y), language(Z), language(W), 
                ((Z=german)-> not(W=polish);Z=_,W=_).   


solution(X):- rule1(X), rule2(X), rule3(X), rule4(X), rule5(X), rule6(X), rule7(X), rule8(X), .

I need help mostly with rules 2 and 7. Any help appreciated.

A general observation: most rules forget that Z and W can be interchanged. E.g. Rule 5 can be both `speaks(X,Y,dutch,polish)`and `speaks(X,Y,polish,dutch)`. You might consider imposing an order over languages, something like `speaks(X,Y,Z,W):- interpreter(X,Y), language(Z), language(W), Z — Lyth, Apr 19 '17 at 16:18
Note: Prolog is like SQL. You **must** join a relation arguments among **other relations** arguments. Otherwise, the whole is - most probably - **totally useless** — CapelliC, Apr 19 '17 at 18:34

score 0 · Answer 1 · answered Dec 03 '19 at 17:53

I tried to solve this puzzle here

I find this way of putting constraints in problems like these easier.

:- use_rendering(table,
         [header(h('Name','Surname','L1','L2'))]).

interpreters(Is) :-
    length(Is,6),
    member(h(fran,french,_,_), Is),
    member(h(geraldine,german,_,_), Is),
    member(h(dudley,dutch,_,_), Is),
    member(h(spike,spanish,_,_), Is),
    member(h(polly,polish,_,_), Is),
    member(h(ron,romanian,_,_), Is),
    member(h(_,_,french,_), Is),member(h(_,_,_,french), Is),
    member(h(_,_,german,_), Is),member(h(_,_,_,german), Is),
    member(h(_,_,dutch,_), Is),member(h(_,_,_,dutch), Is),
    member(h(_,_,spanish,_), Is),member(h(_,_,_,spanish), Is),
    member(h(_,_,polish,_), Is),member(h(_,_,_,polish), Is),
    member(h(_,_,romanian,_), Is),member(h(_,_,_,romanian), Is),
    \+member(h(_,X,X,_), Is),
    \+member(h(_,Y,_,Y), Is),
    (member(h(spike,spanish,dutch,german), Is);member(h(spike,spanish,german,dutch), Is)),
    (member(h(_,_,dutch,polish), Is);member(h(_,_,polish,dutch), Is)),
    member(h(fran,french,E,F), Is),member(h(dudley,dutch,G,H), Is),E \= G, F \= G, E \= H, F \= H,
    member(h(dudley,dutch,I,J), Is), (member(h(_,I,_,french), Is);member(h(_,I,french,_), Is)), (member(h(_,J,_,french), Is);member(h(_,J,french,_), Is)),
    \+member(h(_,_,german,polish), Is),
    \+member(h(_,_,polish,german), Is),
    \+member(h(_,_,K,K), Is).

One of the outputs is

Interpreters riddle in prolog

1 Answers1

Linked