序言-狼山羊白菜

Question

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).

为了让它发挥作用，我调用 length(X,7),wgc([w,w,w,w],X).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？我试着补充! wgc的符号:

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

...但它一次只显示第一个结果。
任何想法如何解决它？

Answer 1

你在这里得到了多余的答案/解决方案。消除(终止)Goal 的冗余答案简单地用 setof(t, Goal, _). 包装这个目标请注意，最后一个参数只是 _ , 你也可以写 [t]反而。

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

只要答案是基本答案，这就会起作用。

按照@SergeyDymchenko 的建议，在列表中明确收集解决方案，需要为该列表提供一个新变量名称，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，没有内在的区别。