prolog打印出不同参数的相同答案

我正在写一个Prolog程序来解决一个逻辑难题。当试图打印出答案时，它会给我所有答案都相同的答案。逻辑之谜的答案是：

• genevieve买了一件4号的manzarita
• 露西亚买了一件7号的格拉芙茨
• shawna买了一件5号的williford
• 瓦妮莎买了一个6号的方丈山

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

如果我指定一个变量，例如 告诉（瓦内萨，Y，Z） 它将打印出来：瓦妮莎买了一个4号的方丈山；使Y和Z的值保持不变

逻辑难题的答案是：-genevieve买了一件4号的manzarita-lucia买了一件7号的Graffetz-shawna买了一件5号的williford-vanessa买了一件6号的Abbot Hill”

如果这些是谜题的唯一答案，那么您的逻辑是错误的，您的代码生成的答案远不止这些

根据您的计划，以下是一些解决方案：

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 ;

如果您想找到所有可以运行的唯一解决方案：

setof（（X，Y，Z），tell（X，Y，Z），Y）。

根据您的程序生成有效的不同解决方案

如果我指定一个变量，例如tell（vanessa，Y，Z），它将打印出来：vanessa买了一个尺寸为4的abbot hill；Y和Z的值保持不变

它没有使值

Y

和

Z

相同，它只是意味着

Y=4，Z=abbothill

是

X=vanesa

的第一个解决方案，您可以使用

生成下一个解决方案；

，您将看到

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：如果您的目标是解决这个逻辑难题：

共有4位客户：Genevieve、Lucia、Shawna、Vanessa

共有4种鞋码：4、5、6、7

有4家制鞋厂：雅培山、曼扎里塔、格拉菲茨、， 威利福德

在Manzarita和Lucia的鞋子中，一双是7号的，另一双是4号的

Genevieve的鞋款比Abbott Hill鞋款小2码

瓦内萨的那对比吉纳维芙的那对大两码

Graffetz鞋比Shawna的鞋稍大一些

为每位客户显示他们的姓名、鞋码和鞋号 制造商的格式与此类似：

乔买了一件13号的耐克

汤姆买了一辆12码的锐步

您可以使用以下内容：

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，此难题有多种解决方案，例如：

?- [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

---

谢谢！这是我第一次用prolog编程，我的逻辑哪里出错了？我发现这很令人困惑。今晚我有时间的时候会看一看，但你应该继续尝试，我想经过一些练习，你会得到正确的答案。