Prolog - 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).

为了使它工作,我打电话 length(X,7),wgc([w,w,w,w],X).,它显示结果.问题是它显示了很多次第一个结果,然后是第二个结果:

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.

如何只显示两次结果？我试着补充一下!符号到wgc:

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

......但它只显示了一次的第一个结果.任何想法如何解决它？

Answer 1

你来到这里多余的答案/解决方案.为了消除(终止)的冗余答案,Goal只需将setof(t, Goal, _). 最后一个参数包含在内_,然后您也可以编写[t].

?- length(X,7), setof(t,wgc([w,w,w,w],X),_).
X = [goat, nothing, cabbage, goat, wolf, nothing, goat] ;
X = [goat, nothing, wolf, goat, cabbage, nothing, goat].

只要答案是基础答案,这就有效.

在@SergeyDymchenko建议中明确地收集列表中的解决方案,需要为该列表提供一个新的变量名称,并在单个列表中表示所有解决方案,这可能比将其留给实现更昂贵.然而,在这个具体案例中,没有固有的区别.

Answer 2

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)