Prolog logic puzzle & constraint programming

Question

I have a homework assignment to define 10 facts in prolog to solve this puzzle.

Five translators are working in an international organization:
Spaniard, Englishman, Frenchman, German and Russian.
Each of them speaks its native language and also two
languages from the native languages of other translators. Find
the languages speaked by each translator if it is known that
1. The Englishman speaks German.
2. The Spaniard speaks French.
3. The German does not speak Spanish.
4. The Frenchman does not speak German.
5. Spaniard and Englishman can talk with each other in German.
6. Englishman and Frenchman can talk with each other in two
languages.
7. Four translators speak Russian.
8. Exactly two translators speak French.
9. Only one translator who speaks Spanish speaks also Russian.
10. Russian and Englishman have no common languages except
their native languages.

I already have defined 8 constraints and reduces my solution space to 15. But I have no idea how to define the last two facts.

% lahendusstruktuur
structure(
  [translator(inglane,_,_,inglise),
   translator(sakslane,_,_,saksa),
   translator(hispaanlane,_,_,hispaania),
   translator(prantslane,_,_,prantsuse),
   translator(venelane,_,_,vene)]).

% abipredikaadid
nationality(translator(B,_,_,_),B).
language1(  translator(_,B,_,_),B).
language2(  translator(_,_,B,_),B).
native_language(
            translator(_,_,_,B),B).

% keelte faktid (andmebaas)
possible_language(vene).
possible_language(inglise).
possible_language(hispaania).
possible_language(prantsuse).
possible_language(saksa).

% tabeli täitmine
solve(S):-
    structure(S),
    fill_structure(S),
    fact1(S),
    fact2(S),
    fact3(S),
    fact4(S),
    fact5(S),
    fact6(S),
    fact7(S),
    fact8(S),
    %fact9(S),
    %fact10(S),
    true.

fill_structure([A,B,C,D,E]):-
    fill_line(A,inglane),
    fill_line(B,sakslane),
    fill_line(C,hispaanlane),
    fill_line(D,prantslane),
    fill_line(E,venelane).

fill_line(X,Nationality):-
    nationality(X,Nationality),
    possible_language(Keel1),
    language1(X,Keel1),
    possible_language(Keel2),
    language2(X,Keel2),
    native_language(X,Native),
    Keel1\=Native,
    Keel2\=Native,
    Keel1 @> Keel2.

% keelte generaator
speaks(Nationality,Language,S):-
    member(X,S),
    nationality(X,Nationality),
    language(X,Language).

language(X,L):-native_language(X,L).
language(X,L):-language1(X,L).
language(X,L):-language2(X,L).

% faktid 1-10
%1. The Englishman speaks German.
fact1(X):-
   speaks(inglane,saksa,X).
%2. The Spaniard speaks French.
fact2(X):-
   speaks(hispaanlane,saksa,X).
%3. The German does not speak Spanish.
fact3(X):-
   \+speaks(sakslane,hispaania,X).

% The Frenchman does not speak German.
fact4(X):-
     \+speaks(prantslane,saksa,X).

% Spaniard and Englishman can talk with each other in German.
fact5(X):-
  speaks(hispaanlane,saksa,X),
  speaks(inglane,saksa,X).

%6. Englishman and Frenchman can talk with each other in two languages.
fact6(X):-
    findall(Keel,
      (
      speaks(inglane,Keel,X),
      speaks(prantslane,Keel,X)
      ),
      Keelte_hulk),
      Keelte_hulk=[_,_].

%% Four translators speak Russian.
fact7(X):-
  findall(Inimene,
  (
  speaks(Inimene,vene,X),
  speaks(Inimene,vene,X),
  speaks(Inimene,vene,X),
  speaks(Inimene,vene,X)),Inimeste_hulk),
  Inimeste_hulk=[_,_,_,_].
 /*
  (speaks(inglane,vene,X), 
  speaks(sakslane, vene,X), 
  speaks(hispaanlane, vene,X), 
  \+ speaks(prantslane, vene,X));
  (speaks(inglane,vene,X), 
  speaks(sakslane, vene,X), 
  \+speaks(hispaanlane, vene,X), 
  speaks(prantslane, vene,X));
  (speaks(inglane,vene,X), 
  \+speaks(sakslane, vene,X), 
  speaks(hispaanlane, vene,X), 
  speaks(prantslane, vene,X));
  (speaks(inglane,vene,X), 
  \+speaks(sakslane, vene,X), 
  speaks(hispaanlane, vene,X), 
  speaks(prantslane, vene,X)). */

  %Exactly two translators speak French.
fact8(X):-
    findall(Inimene,
  (
  speaks(Inimene,prantsuse,X),
  speaks(Inimene,prantsuse,X)),Inimeste_hulk),
  Inimeste_hulk=[_,_].

For example, I tried such solution for the fact 10.

   fact10(X):-
     speaks(inglane,inglise,X), speaks(venelane,inglise,X),
     speaks(venelane,inglise,X), speaks(venelane,vene,X),
     \=(speaks(inglane,hispaania,X),speaks(venelane, hispaania,X),
     \=(speaks(inglane,prantsuse,X),speaks(venelane, prantsuse,X),
     \=(speaks(inglane,saksa,X),speaks(venelane,saksa,X).

This reduces the solution space, but after that it still contains such elements which shouldn't be in the reduced solution set.

I'm using this function to get my solution set size

findall(_, solve(X), Solutions), length(Solutions,N).

And this for solution set elements

solve(X).

I have no ideas left how to describe these two facts. If someone could help, I would appreciate it :).

Sorry for my English, it's not my first language.

score 0 · Answer 1 · answered Oct 08 '14 at 17:40

A different approach from yours is to use constraint programming constructs with finite domains (i.e. Prolog's clpfd library). Here's two MiniZinc models of this problem which might give you some inspiration. Since it's a homework, I won't even try to solve it in "pure" Prolog, i.e. without clpfd.

Here's a model using an array of sets: http://hakank.org/minizinc/five_translators.mzn

And the same principal approach but using a 0/1 matrix instead: http://hakank.org/minizinc/five_translators.mzn

(I'm not sure at all that this will help you, but it's a fun problem...)

Topological Sort · Answer 2 · 2014-11-20T06:57:58.783

I think this relation will fix your problem with rule 9: http://www.swi-prolog.org/pldoc/man?predicate=bagof%2f3

Something like

bagof(Translator,(some condition, possibly with commas or semicolons in it, 
                  verifying that Translator speaks those two languages), 
      Translators),

The bagof clause here

identifies a variable (first argument) to represent the kind of thing we're interested in
sets a condition involving that variable, to solve in any way Prolog can
collects all the variables for which that condition applies, into a list (the 3rd argument)

Then you verify that list is of length 1.

I don't want to do your homework for you, so I won't fill in the condition, but if you were looking for all Translators that have giraffes and drink vodka, it would be

(has(Translator,giraffe),drinks(Translator,vodka))

bagof can also be used for condition 10.

Prolog logic puzzle & constraint programming

2 Answers2