如何处理在序言图的遍历

Question

我已经写在序言的路径：

edge(x, y). 
edge(y, t). 
edge(t, z). 
edge(y, z). 
edge(x, z). 
edge(z, x). 

path(Start, End, Path) :- 
    path3(Start, End, [Start], Path). 

path3(End, End, RPath, Path) :- 
    reverse(RPath, Path). 
path3(A,B,Path,[B|Path]) :- 
    edge(A,B), 
    !. 
path3(A, B, Done, Path) :- 
    edge(A, Next), 
    \+ memberchk(Next, Done), 
    path3(Next, B, [Next|Done], Path). 

其照顾循环图，以及，我得到一个不规则的输出，当我尝试遍历从同一个节点相同的节点。

如：path(x,x,P). 预期的输出应该是：

P = [x, z, t, y, x] 
P = [x, z, y, x] 
P = [x, z, x] 

但是，我得到的输出：

p = [x]    ------------> wrong case 
P = [x, z, t, y, x] 
P = [x, z, y, x] 
P = [x, z, x] 

我怎样才能摆脱这种有害的情况。 感谢

Answer 1

path(Start, End, Path) :- 
    edge(Start,First), 
    path3(Start, End, [Start,First], Path). 

应该工作

Answer 2

我们meta-predicatepath/4一起使用与edge/2：

 
?- path(edge,Path,x,Last), edge(Last,x). 
    Last = z, Path = [x,y,t,z] 
; Last = z, Path = [x,y,z] 
; Last = z, Path = [x,z] 
; false. 

好吧！以上三个答案正是OP在问题中所期望的。

只是为了好玩我们来看看全部根据edge/2的可能路径！

?- path(edge,Path,From,To). 
    From = To  , Path = [To] 
; From = x, To = y, Path = [x,y] 
; From = x, To = t, Path = [x,y,t] 
; From = x, To = z, Path = [x,y,t,z] 
; From = x, To = z, Path = [x,y,z] 
; From = y, To = t, Path = [y,t] 
; From = y, To = z, Path = [y,t,z] 
; From = y, To = x, Path = [y,t,z,x] 
; From = t, To = z, Path = [t,z] 
; From = t, To = x, Path = [t,z,x] 
; From = t, To = y, Path = [t,z,x,y] 
; From = y, To = z, Path = [y,z] 
; From = y, To = x, Path = [y,z,x] 
; From = x, To = z, Path = [x,z] 
; From = z, To = x, Path = [z,x] 
; From = z, To = y, Path = [z,x,y] 
; From = z, To = t, Path = [z,x,y,t] 
; false.