Question

我有以下PROLOG查询及其数据库。

r(X,Y), s(Y,Z), not(r(Y,X)), not(s(Y,Y).

r(a,b). r(a,c). r(b,a). r(a,d).
s(b,c). s(b,d). s(c,c). s(d,e).

在这个例子中PROLOG如何回溯？我认为它会是这样的：

1-  Unifies X with 'a' and Y with 'b'

2-  Unifies Y with 'b' and Z with 'c'

3-  This goal means that there mustn't be any other clause in the database where 
   this happens: r(a,b) r(b,a). 
    My doubt lies here. Does prolog advance to the next goals or does it verify 
   the other r(X,Y) database clauses to check for a match and possibly invalidate 
   the solution?

    I guess that what Prolog does is the following:
   - Verifies the rest of the r(X,Y) clauses to check for a r(Y,X) match and if 
   there is one, then it backtracks to the 2nd step (s(Y,Z)). 
    This will obviously generate unnecessary tree branches which do not need to be 
   tested since the 1st goal is always the same.

4. Verifies if S(X,Y), X == Y. Backtracks to step 1 with new values (a & c) and so on.

我说错了吗？如果有人可以根据这个问题绘制一棵树，我真的很赞成，所以我可以完全理解它。

Answer 1

我建议您使用跟踪器来查看证明树（有些人认为这是一种不好的做法，但如果您遇到困难，它确实有助于理解执行模型）。所以你去了（用not/1取代\+/1）：

?- trace(r/1), trace(s/1).
true.

[debug]  ?- r(X, Y), s(Y, Z), \+ r(Y, X), \+ s(Y, Y).
 T Call: (7) r(_G341, _G342)
 T Exit: (7) r(a, b)
 T Call: (7) s(b, _G345)
 T Exit: (7) s(b, c)
 T Call: (7) r(b, a)
 T Exit: (7) r(b, a)
 T Redo: (7) s(b, _G345)
 T Exit: (7) s(b, d)
 T Call: (7) r(b, a)
 T Exit: (7) r(b, a)
 T Redo: (7) r(_G341, _G342)
 T Exit: (7) r(a, c)
 T Call: (7) s(c, _G345)
 T Exit: (7) s(c, c)
 T Call: (7) r(c, a)
 T Fail: (7) r(c, a)
 T Call: (7) s(c, c)
 T Exit: (7) s(c, c)
 T Redo: (7) r(_G341, _G342)
 T Exit: (7) r(b, a)
 T Call: (7) s(a, _G345)
 T Fail: (7) s(a, _G345)
 T Redo: (7) r(_G341, _G342)
 T Exit: (7) r(a, d)
 T Call: (7) s(d, _G345)
 T Exit: (7) s(d, e)
 T Call: (7) r(d, a)
 T Fail: (7) r(d, a)
 T Call: (7) s(d, d)
 T Fail: (7) s(d, d)
X = a,
Y = d,
Z = e.

有你的证明树。 Redo是Prolog回溯的地方。当呼叫成功时，\+失败，Prolog在Redo之后执行Exit。当目标失败时，\+会在Fail之后成功。

如何在PROLOG中回溯

1 个答案: