6

我正在开发一款名为“狼山羊白菜”的益智游戏。编程语言是 Prolog。

change(e,w).
change(w,e).
move([X,X,Goat,Cabbage],wolf,[Y,Y,Goat,Cabbage]) :- change(X,Y).
move([X,Wolf,X,Cabbage],goat,[Y,Wolf,Y,Cabbage]) :- change(X,Y).
move([X,Wolf,Goat,X],cabbage,[Y,Wolf,Goat,Y]) :- change(X,Y).
move([X,Wolf,Goat,Cabbage],nothing,[Y,Wolf,Goat,Cabbage]) :- change(X,Y).

oneeq(X,X,WW).
oneeq(X,WWW,X).

safe([Man,Wolf,Goat,Cabbage]) :-
        oneeq(Man,Goat,Wolf),
        oneeq(Man,Goat,Cabbage).

wgc([e,e,e,e],[]).

wgc(Config,[FirstMove|OtherMoves]) :-
        move(Config,FirstMove,NextConfig),
        safe(NextConfig),
        wgc(NextConfig,OtherMoves).

为了让它工作，我打电话给 length(X,7),wgc([w,w,w,w],X).它，它显示了结果。问题是它多次显示第一个结果，然后显示第二个结果：

X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
false.

如何只显示一次两个结果？我试图添加！wgc的符号：

wgc(Config,[FirstMove|OtherMoves]) :-
        move(Config,FirstMove,NextConfig),
        safe(NextConfig),
        wgc(NextConfig,OtherMoves), !.

...但它只显示第一个结果一次。任何想法如何解决它？

4

2 回答 2

7

你在这里得到多余的答案/解决方案。要消除 a (终止) 的冗余答案，只需使用 注意最后一个参数是 justGoal包装该目标，您也可以改为编写。setof(t, Goal, _)._[t]

?- 长度(X,7), setof(t, wgc([w,w,w,w],X), _)。
X = [山羊，无，卷心菜，山羊，狼，无，山羊]；
X = [山羊，无，狼，山羊，卷心菜，无，山羊]。

只要答案是基本答案，这就是有效的。

正如@SergeyDymchenko 建议的那样，明确地在列表中收集解决方案需要为该列表提供一个新的变量名称，并将所有解决方案表示在一个列表中，这可能比将其留给实现更昂贵。然而，在这种具体情况下，并没有本质上的区别。

于 2013-01-28T07:55:40.480 回答
4

您可以使用setof谓词来获取唯一解决方案的列表：

setof(X, (length(X,7),wgc([w,w,w,w],X)), Sols).

X = X
Sols = [[goat, nothing, cabbage, goat, wolf, nothing, goat], [goat, nothing, wolf, goat, cabbage, nothing, goat]]
Yes (0.00s cpu)
于 2013-01-27T21:08:02.317 回答