在序言中解决导致错误？答案

【问题标题】：Solve in prolog resulting in false?在序言中解决导致错误？
【发布时间】：2013-04-06 09:26:02
【问题描述】：

我尝试过使用类似于此链接的代码：

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

我要解决的问题是这个（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.

我的问题是这样的：

?- solve(X).
 false.

【问题讨论】：

标签： prolog

【解决方案1】：

所以您的问题是，您的 solve 谓词找不到任何解决方案。这意味着，对于解决方案树中的所有可能路径，找到解决方案的先决条件之一都失败了。

您是否真的尝试搜索它是哪一个？当然不是，否则你会注意到这个：

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

总是失败。为什么？ \+/1 是什么？ “如果 Goal 无法证明，\+ :Goal 为真”。换句话说，您不能使用\+ 进行匹配。相反，你可以写：

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

所以有了所有的更正：

?- 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。

【讨论】：

请原谅我不知道如何使用 SWI prolog 进行精确调试。我对声明式编程完全陌生！感谢您的回复 - 是的，我也尝试过搜索
@AlexCutajar 我试图展示我发现你的程序有什么问题的方式，这样你就可以自己做。实际上，无论使用哪种语言，您都必须学会在程序中发现问题。

【解决方案2】：

代替

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]),

【讨论】：