Question

我尝试使用类似于此链接的代码：

Solving a textual logic puzzle in Prolog - Find birthday and month

我试图解决的问题是（电话对话）。 http://www.cis.upenn.edu/~matuszek/cis554-2012/Assignments/prolog-01-logic-puzzle.html

我的代码：

dated(Date):-
member(Date,[1928,1929,1932,1935]).
exchanged(Exchange):-
member(Exchange,[al,be,pe,sl]).

solve(X):-
X=[[gertie,Exchange1,Date1],
   [herbert,Exchange2,Date2],
   [miriam,Exchange3,Date3],
   [wallace,Exchange4,Date4]],

exchanged(Exchange1), exchanged(Exchange2), exchanged(Exchange3), exchanged(Exchange4),
Exchange1 \== Exchange2, Exchange1 \== Exchange3, Exchange1 \== Exchange4,
Exchange2 \== Exchange1, Exchange2 \== Exchange3, Exchange2 \== Exchange4,
Exchange3 \== Exchange1, Exchange3 \== Exchange2, Exchange3 \== Exchange4,
Exchange4 \== Exchange1, Exchange4 \== Exchange2, Exchange4 \== Exchange3,

dated(Date1), dated(Date2), dated(Date3), dated(Date4),
Date1 \== Date2, Date1 \== Date3, Date1 \== Date4,
Date2 \== Date1, Date2 \== Date3, Date2 \== Date4,
Date3 \== Date1, Date3 \== Date2, Date3 \== Date4,
Date4 \== Date1, Date4 \== Date2, Date4 \== Date3,

%Herbet's first exchange wasn't for BE
Exchange2 \== be,

%The Person whose first exchange was SL wasn't Getie or Herbert
Exchange1 \== sl,
Exchange2 \== sl,

%The person whose first exchange was BE didn't get the phone in 1935
member([_,be, \+1935], X),

%The person who got the first phone in 1932 didn't have an exchange for AL or BE
member([_, \+al, 1932], X),
member([_, \+be, 1932],X),

%The person who got the first phone in 1928 had an exchange for PE
member([_,pe,1929], X),

%Wallace first exchange was AL
Exchange4 == al.

我的问题是：

?- solve(X).
 false.

Answer 1

所以问题是，您的solve谓词找不到任何解决方案。这意味着，找到解决方案的先决条件之一对于解决方案树中的所有可能路径都会失败。

你真的试着搜索它是哪一个吗？当然不是，否则你会注意到这一点：

member([_,be,\+1935],X)

总是失败。为什么？什么是\+/1？如果无法证明目标，则“\+ : 目标为真。”换句话说，您无法使用\+进行匹配。相反，你可以写：

\+ member([_,be,1935),X).

所有更正：

?- solve(X).
X = [[gertie, be, 1928], [herbert, pe, 1929], [miriam, sl, 1932], [wallace, al, 1935]] ;
false.

假设程序的其余部分是正确的。

使用stackoverflow作为调试代码的替代方法真的很糟糕。

Answer 2

而不是

exchanged(Exchange1), exchanged(Exchange2), exchanged(Exchange3), exchanged(Exchange4),
Exchange1 \== Exchange2, Exchange1 \== Exchange3, Exchange1 \== Exchange4,
Exchange2 \== Exchange1, Exchange2 \== Exchange3, Exchange2 \== Exchange4,
Exchange3 \== Exchange1, Exchange3 \== Exchange2, Exchange3 \== Exchange4,
Exchange4 \== Exchange1, Exchange4 \== Exchange2, Exchange4 \== Exchange3,

dated(Date1), dated(Date2), dated(Date3), dated(Date4),
Date1 \== Date2, Date1 \== Date3, Date1 \== Date4,
Date2 \== Date1, Date2 \== Date3, Date2 \== Date4,
Date3 \== Date1, Date3 \== Date2, Date3 \== Date4,
Date4 \== Date1, Date4 \== Date2, Date4 \== Date3,

你可以写

permutation([al,be,pe,sl], [Exchange1, Exchange2, Exchange3, Exchange4]),
permutation([1928,1929,1932,1935], [Date1, Date2, Date3, Date4]),

在prolog中解决导致错误？

2 个答案: