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).

为了使其有效，我致电length(X,7),wgc([w,w,w,w],X).并显示结果。问题是它首先显示了第一个结果，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), !.

...但它只显示了第一次结果。任何想法如何解决它？

Answer 1

你来到这里是多余的答案/解决方案。要消除（终止）Goal的冗余答案，只需用setof(t, Goal, _).包裹该目标请注意，最后一个参数只是_，您也可以改为编写[t]。 / p>

?- 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，并没有固有的差异。

Answer 2

您可以使用setof谓词来获取唯一解决方案列表：

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)

Prolog - 狼山羊白菜

2 个答案: