3

我在 Prolog 中写过：

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).

它也处理循环图，当我尝试从同一个节点遍历同一个节点时，我得到一个不规则的输出。

例如：path(x,x,P). 预期输出应该是：

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不需要的情况。谢谢

4

2 回答 2

3

我们将 path/4与edge/2：

?-路径（边缘，路径，x，最后），边缘（最后，x）。
  最后 = z，路径 = [x,y,t,z]
; 最后 = z，路径 = [x,y,z]
; 最后 = z，路径 = [x,z]
; 错误的。

好吧！以上三个答案正是 OP 在问题中所希望的。

只是为了好玩，让我们看看所有可能的路径基于edge/2！

?- 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.
于 2015-08-20T09:55:01.103 回答
0 
path(Start, End, Path) :-
    edge(Start,First),
    path3(Start, End, [Start,First], Path).

应该管用

于 2014-12-03T10:17:28.310 回答