如何处理Prolog图遍历中的路径 [英] How to handle a path in Prolog graph traversal

问题描述

我用Prolog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).

它也照顾循环图，当我尝试从同一节点遍历同一节点时，我得到了不规则的输出.

Its taking care of cyclic graphs as well, I am getting an irregular output when I try to traverse same node from same node.

例如:path(x,x,P). 预期输出应为:

eg: path(x,x,P). expected output should be:

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

但是，我得到的输出是

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

如何摆脱这种不必要的情况. 谢谢

How can I get rid of this unwanted case. Thanks

推荐答案

我们使用元谓词 path/4 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.

好的！以上三个答案正是OP希望在问题中得到的.

Alright! Above three answers are exactly what the OP wished for in the question.

只是为了好玩，让我们看看基于edge/2的所有可能路径！

Just for fun let's look at all possible paths based on 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.

这篇关于如何处理Prolog图遍历中的路径的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！