Prolog根據輸入決定一定的規則選擇

Question

我有這個序言問題，我不能解決。我試圖達到的目的是斷言FACT A，當我輸入時收回事實B：取得並斷言事實B，並在輸入時撤銷事實A.

即：

:- 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]. 

我無法得到它的工作：任何幫助，將不勝感激。

編輯：所有的代碼張貼

Answer 1

這裏就是我終於實現了，我用的約束，選擇了另一一定的規則： 這裏是代碼：

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]. 

Answer 2

我懷疑問題出在你的代碼的其他地方。這個工作對我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，爲什麼你這樣做雖然。我的傾向，所有的事情都是平等的，將會增加你用你所宣稱的信息產生的AST。再一次，我偏向於動態商店。

Answer 3

我不完全理解你的代碼，但我有一種感覺，這一點：

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

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

應該是：

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

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

但是，爲什麼在V，NP，PP非實例化？如果你想刪除所有的事件，你應該使用retractall/1;如果只有一次發生，我建議使用swipl's global variables。無論如何，在DCG中使用副作用就像是與魔鬼打交道;我做到了我的編譯器，這是一個調試運行地獄XD