Prolog 从顶点到所有其他可到达节点的所有简单路径

我对Prolog完全是新手，并且一直在研究图表。我在网上发现一个问题，要求我指定一个节点，然后列出从该节点可以访问的所有简单路径。没有目标节点，只需尝试所有可能性并返回所有路径

我将图形表示为路径（X，Y），表示从X到Y的有向边

我建立了这个简单的知识库，它是循环的：

path(a, b).
path(b, c).
path(c, d).
path(d, a).
path(d, e).
path(d, f).
path(f, g).

如果我查询所有_路径（a，p），那么p应该是（假设；被垃圾处理，直到所有选项都用尽）

我写了这样的东西作为开场白：

all_paths(Source, P) :- all_paths(Source, P, []).
all_paths(_, [], _).
all_paths(Source, [Source | P], Visited) :-
    path(Source, Node),
    \+ memberchk(Node, Visited),
    all_paths(Node, P, [Node | Visited]).

好的，稍微改变一下，现在我回来了：

X = [] ? ;
X = [a] ? ;
X = [a,b] ? ;
X = [a,b,c] ? ;
X = [a,b,c,d] ? ; <- Here it does not pick up e
X = [a,b,c,d] ? ;
X = [a,b,c,d] ? ;
X = [a,b,c,d,f] ? ;

X=[]；
X=[a]；
X=[a，b]；
X=[a，b，c]；
X=[a，b，c，d] 不需要重新发明轮子

首先，我们将谓词
path/2
重命名为
edge/2
：

edge(a, b).
edge(b, c).
edge(c, d).
edge(d, a).
edge(d, e).
edge(d, f).
edge(f, g).

?- path(edge,Path,From,To).
  Path = [To],          From = To
; Path = [a,b],         From = a, To = b
; Path = [a,b,c],       From = a, To = c
; Path = [a,b,c,d],     From = a, To = d
; Path = [a,b,c,d,e],   From = a, To = e
; Path = [a,b,c,d,f],   From = a, To = f
; Path = [a,b,c,d,f,g], From = a, To = g
; Path = [b,c],         From = b, To = c
; Path = [b,c,d],       From = b, To = d
; Path = [b,c,d,a],     From = b, To = a
; Path = [b,c,d,e],     From = b, To = e
; Path = [b,c,d,f],     From = b, To = f
; Path = [b,c,d,f,g],   From = b, To = g
; Path = [c,d],         From = c, To = d
; Path = [c,d,a],       From = c, To = a
; Path = [c,d,a,b],     From = c, To = b
; Path = [c,d,e],       From = c, To = e
; Path = [c,d,f],       From = c, To = f
; Path = [c,d,f,g],     From = c, To = g
; Path = [d,a],         From = d, To = a
; Path = [d,a,b],       From = d, To = b
; Path = [d,a,b,c],     From = d, To = c
; Path = [d,e],         From = d, To = e
; Path = [d,f],         From = d, To = f
; Path = [d,f,g],       From = d, To = g
; Path = [f,g],         From = f, To = g
; false.

然后，我们结合使用
edge/2
：

edge(a, b).
edge(b, c).
edge(c, d).
edge(d, a).
edge(d, e).
edge(d, f).
edge(f, g).

?- path(edge,Path,From,To).
  Path = [To],          From = To
; Path = [a,b],         From = a, To = b
; Path = [a,b,c],       From = a, To = c
; Path = [a,b,c,d],     From = a, To = d
; Path = [a,b,c,d,e],   From = a, To = e
; Path = [a,b,c,d,f],   From = a, To = f
; Path = [a,b,c,d,f,g], From = a, To = g
; Path = [b,c],         From = b, To = c
; Path = [b,c,d],       From = b, To = d
; Path = [b,c,d,a],     From = b, To = a
; Path = [b,c,d,e],     From = b, To = e
; Path = [b,c,d,f],     From = b, To = f
; Path = [b,c,d,f,g],   From = b, To = g
; Path = [c,d],         From = c, To = d
; Path = [c,d,a],       From = c, To = a
; Path = [c,d,a,b],     From = c, To = b
; Path = [c,d,e],       From = c, To = e
; Path = [c,d,f],       From = c, To = f
; Path = [c,d,f,g],     From = c, To = g
; Path = [d,a],         From = d, To = a
; Path = [d,a,b],       From = d, To = b
; Path = [d,a,b,c],     From = d, To = c
; Path = [d,e],         From = d, To = e
; Path = [d,f],         From = d, To = f
; Path = [d,f,g],       From = d, To = g
; Path = [f,g],         From = f, To = g
; false.


编辑
如果我们只对从
a
开始的路径感兴趣，我们只需编写：

?- path(edge,Path,a,To).
  Path = [a],                To = a
; Path = [a, b],             To = b
; Path = [a, b, c],          To = c
; Path = [a, b, c, d],       To = d
; Path = [a, b, c, d, e],    To = e
; Path = [a, b, c, d, f],    To = f
; Path = [a, b, c, d, f, g], To = g
; false.

“交换”节点和源

all_paths(_, [], _).
all_paths(Source, [Node | P], Visited) :-
    path(Source, Node),
    \+ memberchk(Node, Visited),
    all_paths(Node, P, [Source | Visited]).

屈服

?- all_paths(a, P).
P = [] ;
P = [b] ;
P = [b, c] ;
P = [b, c, d] ;
P = [b, c, d, e] ;
P = [b, c, d, f] ;
P = [b, c, d, f, g] ;
false.

?- all_paths(a, P).
P = [a] ;
P = [a, b] ;
P = [a, b, c] ;
P = [a, b, c, d] ;
P = [a, b, c, d, e] ;
P = [a, b, c, d, f] ;
P = [a, b, c, d, f, g] ;
false.

它缺少start节点，我只需在“driver”谓词中添加该节点：

all_paths(Source, [Source|P]) :- all_paths(Source, P, []).

屈服

?- all_paths(a, P).
P = [] ;
P = [b] ;
P = [b, c] ;
P = [b, c, d] ;
P = [b, c, d, e] ;
P = [b, c, d, f] ;
P = [b, c, d, f, g] ;
false.

?- all_paths(a, P).
P = [a] ;
P = [a, b] ;
P = [a, b, c] ;
P = [a, b, c, d] ;
P = [a, b, c, d, e] ;
P = [a, b, c, d, f] ;
P = [a, b, c, d, f, g] ;
false.

风格提示：如果我们遵循一些关于IO参数的规则，代码的可读性会更好。输出参数应该在输入参数之后。嗯，这并不总是适用的…
对不起，我可能没有说清楚。除了成员函数之外，我不想使用任何库，而且我的路径必须始终从a开始，即，我希望所有简单路径都从a开始到其他一些节点。例如，我不需要从节点b开始的路径。@Pterodactylus。有一个很常见的成语：用它！我很感谢您的输入，但是我如何自己实现这个路径函数呢？我对黑匣子本身很感兴趣，看起来我正在朝着正确的方向前进，但我写的逻辑中缺少了一些东西。@Pterodactylus。重新开始。请看“代码>路径/ 4 /代码>（））的摇滚实作。感谢链接，我将调查它。考虑此修复：<代码> ALLYPATH（X，[X-XS]）：-ALLYROPATH（X，XS，[X]）。所有路径（u，[]，u）。所有路径（源，[节点|已访问]）：路径（源，节点），映射列表（dif（节点），已访问），所有路径（节点，节点，[节点|已访问]）。

非常有用。我需要训练我的眼睛以防犯这样的错误。你可能帮我省了几个小时的检查时间。谢谢，投票表决，我接受答案。