Prolog 逻辑谜题和约束编程

Posted 2023-05-08

【中文标题】Prolog 逻辑谜题和约束编程

【英文标题】：Prolog logic puzzle & constraint programming

【发布时间】：2014-12-03 07:41:24

【问题描述】：

我有一个家庭作业，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，我为事实 10 尝试了这样的解决方案。

   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，但之后它仍然包含不应该在缩减解决方案集中的元素。

我正在使用这个函数来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).

我不知道如何描述这两个事实。如果有人可以提供帮助，我将不胜感激:)。

对不起，我的英语不是我的母语。

【参考方案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）。这是这个问题的两个 MiniZinc 模型，可能会给你一些启发。由于这是一个家庭作业，我什至不会尝试在“纯”Prolog 中解决它，即没有 clpfd。

这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

（我完全不确定这是否会对您有所帮助，但这是一个有趣的问题......）

【参考方案2】：

我认为这种关系将解决您对规则 9 的问题：http://www.swi-prolog.org/pldoc/man?predicate=bagof%2f3

类似

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

这里的bagof子句

标识一个变量（第一个参数）来表示我们感兴趣的事物的种类 设置涉及该变量的条件，以 Prolog 可以解决的任何方式 将适用该条件的所有变量收集到一个列表中（第三个参数）

然后您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，但如果你要寻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。