prolog 为不同的参数打印出相同的答案

Question

我正在编写一个解决逻辑难题的 Prolog 程序。当试图打印出答案时，它给了我所有答案的相同答案。逻辑谜题的答案是：

• genevieve 买了一个 4 号的 manzarita
• lucia 买了 7 号 Graffetz
• 肖娜买了 5 码的威利福德
• vanessa 买了 6 号 Abbot Hill

我的代码如下：

customer(genevieve).
customer(lucia).
customer(shawna).
customer(vanessa).

shoesize(4).
shoesize(5).
shoesize(6).
shoesize(7).

manufactorer(abbothill).
manufactorer(manzarita).
manufactorer(graffetz).
manufactorer(williford).

solve :-
   shoesize(GenevieveShoesize),
   shoesize(LuciaShoesize),
   shoesize(ShawnaShoesize),
   shoesize(VanessaShoesize),
   all_different([GenevieveShoesize, LuciaShoesize, ShawnaShoesize,     VanessaShoesize]),
   manufactorer(AbbotHillManufactorer),
   manufactorer(ManzaritaManufactorer),
   manufactorer(GraffetzManufactorer),
   manufactorer(WillifordManufactorer),

   all_different([AbbotHillManufactorer, ManzaritaManufactorer, 
    GraffetzManufactorer, WillifordManufactorer]),
   List = [ [genevieve,GenevieveShoesize,AbbotHillManufactorer],
      [lucia,LuciaShoesize,ManzaritaManufactorer],
      [shawna,ShawnaShoesize,GraffetzManufactorer],
      [vanessa,VanessaShoesize,WillifordManufactorer]],

   \+(member([_,5,manzarita],List)),
   \+(member([_,6,manzarita],List)),
   \+(member([lucia,5,_],List)),
   \+(member([lucia,6,_],List)),
   \+(member([genevieve,_,abbothill],List)),
   \+(member([shawna,_,graffetz],List)),
   (member([vanessa,_,abbothill],List)),

   tell(genevieve,GenevieveShoesize,GeneieveManufactorer),
   tell(lucia,LuciaShoesize,LuciaManufactorer),
   tell(shawna,ShawnaShoesize,ShawnaManufactorer),
   tell(vanessa,VanessaShoesize,VanessaManufactorer).

all_different([H | T]) :- member(H,T), !, fail.
all_different([_ | T]) :- all_different(T).
all_different([]).


tell(X,Y,Z) :-
   customer(X),
   shoesize(Y),
   manufactorer(Z),
   write(X), write(' got their shoes from'),
   write(Y), write(' and is a size '), write(Z), nl.

当我去 SWI 并要求它：

-告诉(X,Y,Z) 它输出： X 等于吉纳维芙， Y 等于 4， Z 等于方丈山。

如果我指定变量之一，例如 告诉（凡妮莎，Y，Z） 它会打印出来：vanessa买了一个4号的方丈山；保持 Y 和 Z 的值不变。

Answer 1

逻辑难题的答案是：-genevieve 买了 4 码的 manzarita -lucia 买了 7 码 Graffetz -shawna 买了 5 码 williford -vanessa 买了 6 码 Abbot Hill"

如果这些是谜题的唯一答案，那么您的逻辑是错误的，您的代码会生成比这更多的答案。

以下是根据您的程序的一些解决方案：

X = genevieve,
Y = 4,
Z = abbothill ;
genevieve got their shoes from4 and is a size manzarita
X = genevieve,
Y = 4,
Z = manzarita ;
genevieve got their shoes from4 and is a size graffetz
X = genevieve,
Y = 4,
Z = graffetz ;
genevieve got their shoes from4 and is a size williford
X = genevieve,
Y = 4,
Z = williford ;
genevieve got their shoes from5 and is a size abbothill
X = genevieve,
Y = 5,
Z = abbothill ;
genevieve got their shoes from5 and is a size manzarita
X = genevieve,
Y = 5,
Z = manzarita ;
genevieve got their shoes from5 and is a size graffetz
X = genevieve,
Y = 5,
Z = graffetz ;
genevieve got their shoes from5 and is a size williford
X = genevieve,
Y = 5,
Z = williford ;

如果您想找到所有独特的解决方案，您可以运行：setof((X,Y,Z), tell(X,Y,Z), Y).

根据您的程序生成 64 个有效的不同解决方案。

如果我指定变量之一，例如tell(vanessa,Y,Z)，它将打印出：vanessa买了一个4号的方丈山；保持 Y 和 Z 的值不变。

它不会使Y 和Z 的值相同，它只是意味着Y = 4, Z = abbothill 是X = vanessa 的第一个 解决方案，您可以生成下一个解决方案; 你会看到Y 和Z 并不总是相同的，例如：

?- tell(vanessa,Y,Z).
vanessa got their shoes from4 and is a size abbothill
Y = 4,
Z = abbothill ;
vanessa got their shoes from4 and is a size manzarita
Y = 4,
Z = manzarita ;
vanessa got their shoes from4 and is a size graffetz
Y = 4,
Z = graffetz ;
vanessa got their shoes from4 and is a size williford
Y = 4,
Z = williford ;
vanessa got their shoes from5 and is a size abbothill
Y = 5,
Z = abbothill ;

编辑：如果你的目标是解决这个逻辑难题：

有 4 位客户：Genevieve、Lucia、Shawna、Vanessa

有 4 种鞋码：4、5、6、7

有 4 家鞋厂：Abbott Hill、Manzarita、Graffetz、 威利福德

1. 在 Manzarita 鞋和 Lucia 的那双鞋中，一个是 7 码，另一个是 4 码。

2. Genevieve 的这双鞋比 Abbott Hill 的鞋小 2 码。

3. Vanessa 的那双比 Genevieve 的大 2 码。

4. Graffetz 的鞋比 Shawna 的鞋稍大。

对于每位顾客，请显示他们的姓名、鞋码和鞋码 制造商的格式与此类似：

乔买了 13 号耐克。

Tom 买了 12 码的 Reebok。

你可以这样使用：

customer(genevieve).
customer(lucia).
customer(shawna).
customer(vanessa).

shoesize(4).
shoesize(5).
shoesize(6).
shoesize(7).

manufactorer(abbothill).
manufactorer(manzarita).
manufactorer(graffetz).
manufactorer(williford).

four_or_seven(L):-
    member((lucia, 7, _), L),
    member((_, 4, manzarita), L).
four_or_seven(L):-
    member((lucia, 4, _), L),
    member((_, 7, manzarita), L).

less_than_abbott_hill(L):-
    member((genevieve, S0, _), L),
    member((_, S1, abbothill), L),
    S0 is S1-2.

vanessa_larger_than_genevieve(L):-
    member((vanessa, S0, _), L),
    member((genevieve, S1, _), L),
    S0 is S1+2.

graffetz_larger_than_shawna(L):-
    member((shawna, S0, _), L),
    member((_, S1, graffetz), L),
    S1 > S0.

solve:-
    shoesize(S0), manufactorer(M0),
    shoesize(S1), manufactorer(M1),
    shoesize(S2), manufactorer(M2),
    shoesize(S3), manufactorer(M3),
    all_different([M0,M1,M2,M3]),
    L = [(genevieve, S0, M0), (lucia, S1, M1), (shawna, S2, M2), (vanessa, S3, M3)],
    four_or_seven(L),
    less_than_abbott_hill(L),
    vanessa_larger_than_genevieve(L),
    graffetz_larger_than_shawna(L),
    print(L).

all_different([H | T]) :- member(H,T), !, fail.
all_different([_ | T]) :- all_different(T).
all_different([]).

print([(C, S, M)|T]):-
    format("~p bhought a size ~p ~p~n", [C,S,M]),
    print(T).

print([]).

请注意，这个谜题有多种解决方案，示例运行：

?- [puzzle].
true.

?- solve.
genevieve bhought a size 4 manzarita
lucia bhought a size 7 graffetz
shawna bhought a size 4 williford
vanessa bhought a size 6 abbothill
true ;
genevieve bhought a size 4 manzarita
lucia bhought a size 7 graffetz
shawna bhought a size 5 williford
vanessa bhought a size 6 abbothill
true