Prolog 逻辑难题和约束编程

Question

我有一项家庭作业，要在序言中定义10个事实来解决这个难题。

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.

我已经定义了8个约束，并将我的解决方案空间减少到15个。但我不知道如何定义最后两个事实。

% 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=[_,_].

例如，我尝试了这样的解决方案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).

这减少了解决方案空间，但在那之后，它仍然包含不应该在减少的解决方案集中的元素。

我正在使用此功能获取我的解决方案集大小

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

这适用于解决方案集元素

solve(X).

我不知道该如何描述这两个事实。如果有人能帮忙，我将不胜感激：）。

对不起，我的英语不是我的第一语言。

Answer 1

与您不同的方法是使用具有有限域的约束编程构造（即Prolog的clpfd库）。以下是关于这个问题的两个MiniZinc模型，可能会给你一些启发。由于这是一个家庭作业，我甚至不会尝试在"纯"Prolog中解决它，即没有clpfd。

下面是一个使用集合数组的模型：http://hakank.org/minizinc/five_translators.mzn

使用相同的主要方法，但使用0/1矩阵：http://hakank.org/minizinc/five_translators.mzn

（我不确定这是否会对你有所帮助，但这是一个有趣的问题…）

Answer 2

我认为这个关系将解决您使用规则9的问题：http://www.swi-prolog.org/pldoc/man?predicate=bagof%2f3

类似的东西

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

此处的bagof条款

• 确定一个变量（第一个自变量）来表示我们感兴趣的事物
• 设置一个涉及该变量的条件，以Prolog可以的任何方式求解
• 将该条件适用的所有变量收集到列表中（第三个参数）

然后验证该列表的长度为1。

我不想为你做作业，所以我不会填写条件，但如果你要找所有有长颈鹿和喝伏特加的翻译，那就是

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

bagof也可以用于条件10。