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我有这个序言问题，我无法解决。我想要实现的是断言 FACT A，当我有输入时收回 Fact B：接受并断言 Fact B 并在我输入输入时收回 Fact A。

IE：

:- dynamic s/2.
:- dynamic s/3.

s(P0, s(V, NP)) --> v(P1, V), np(P2, NP), {P0 is P1*P2*0.35}.

s(P0, s(V, NP, PP)) --> v(P1, V), np(P2, NP), pp(P3, PP), {P0 is P1*P2*P3*0.65}.
s(P0, s(V, NP)) --> v(P1, V), np(P2, NP), {V == take -> P0 is P1*P2*0.35; P0 is 0}.
s(P0, s(V, NP, PP)) --> v(P1, V), np(P2, NP), pp(P3, PP), {V == put -> P0 is P1*P2*P3*0.65; P0 is 0}.

np(P0, np(D, N)) --> det(P1, D), n(P2, N), {P0 is P1*P2*0.36}.
np(P0, np(D, A, N)) --> det(P1, D), a(P2, A), n(P3, N), {P0 is P1*P2*P3*0.46}.
np(P0, np(D, N, PP)) --> det(P1, D), n(P2, N), pp(P3, PP), {P0 is P1*P2*P3*0.13}.
np(P0, np(D, A, N, PP)) --> det(P1, D), a(P2, A), n(P3, N), pp(P4, PP), {P0 is P1*P2*P3*P4*0.05}.

pp(P0, pp(P, NP)) --> p(P1, P), np(P2, NP), {P0 is P1*P2*1.0}.

v(0.65, v(put)) --> {retract(s(V, NP))}, [put].
v(0.35, v(take)) --> {retract(s(V, NP, PP))}, [take].

n(0.23, n(block)) --> [block].
n(0.25, n(circle)) --> [circle].
n(0.15, n(cone)) --> [cone].
n(0.12, n(cube)) --> [cube].
n(0.25, n(square)) --> [square].

a(0.56, a(blue)) --> [blue].
a(0.27, a(green)) --> [green].
a(0.17, a(red)) --> [red].

det(1.0, det(the)) --> [the].

p(1.0, p(on)) --> [on].

我无法让它工作：任何帮助将不胜感激。

编辑：发布的所有代码

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3 回答 3

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我怀疑问题出在您代码的其他地方。这对我来说适用于 SWI：

:- dynamic s/1.

foo --> "hello",   { retractall(s(_)), asserta(s(hi))  }.
foo --> "goodbye", { retractall(s(_)), asserta(s(bye)) }.

例如：

?- s(X).
false.

?- phrase(foo, "hello").
true .

?- phrase(foo, "hello").
true ;
false.

?- s(X).
X = hi.

?- phrase(foo, "goodbye").
true.

?- s(X).
X = bye.

我很好奇你为什么要这样做。在所有条件相同的情况下，我的倾向是用你断言的信息来增加你正在生成的 AST。再说一次，我对动态商店有偏见。

于 2013-02-24T03:56:00.397 回答
1

这就是我最终做的事情，我使用约束来选择某个规则而不是另一个规则：这是代码：

s(P0, s(V, NP)) --> v(P1, V), np(P2, NP), {P0 is P1*P2*0.35, V == v(take)}.
s(P0, s(V, NP, PP)) --> v(P1, V), np(P2, NP), pp(P3, PP), {P0 is P1*P2*P3*0.65, V == v(put)}.

np(P0, np(D, N)) --> det(P1, D), n(P2, N), {P0 is P1*P2*0.36}.
np(P0, np(D, A, N)) --> det(P1, D), a(P2, A), n(P3, N), {P0 is P1*P2*P3*0.46}.
np(P0, np(D, N, PP)) --> det(P1, D), n(P2, N), pp(P3, PP), {P0 is P1*P2*P3*0.13}.
np(P0, np(D, A, N, PP)) --> det(P1, D), a(P2, A), n(P3, N), pp(P4, PP), {P0 is P1*P2*P3*P4*0.05}.

pp(P0, pp(P, NP)) --> p(P1, P), np(P2, NP), {P0 is P1*P2*1.0, NP \= np(_, _ , _, _)}.

v(0.65, v(put)) --> [put].
v(0.35, v(take)) --> [take].

n(0.23, n(block)) --> [block].
n(0.25, n(circle)) --> [circle].
n(0.15, n(cone)) --> [cone].
n(0.12, n(cube)) --> [cube].
n(0.25, n(square)) --> [square].

a(0.56, a(blue)) --> [blue].
a(0.27, a(green)) --> [green].
a(0.17, a(red)) --> [red].

det(1.0, det(the)) --> [the].

p(1.0, p(on)) --> [on].
于 2013-03-11T05:45:42.353 回答
1

我不完全理解你的代码，但我有这样的感觉：

v(0.65, v(put)) --> {retract(s(V, NP))}, [put].

v(0.35, v(take)) --> {retract(s(V, NP, PP))}, [take].

应该：

v(0.65, v(put)) --> [put], {retract(s(V, NP))}.

v(0.35, v(take)) --> [take], {retract(s(V, NP, PP))}.

但是，为什么V, NP,PP没有实例化？如果你想删除所有你应该使用的事件retractall/1；如果只出现一次，我建议使用swipl 的全局变量。无论如何，在 DCG 中使用副作用就像是在与魔鬼做交易。我在我的编译器中完成了它，这是一个调试地狱 XD

于 2013-02-24T12:37:12.983 回答