我在Prolog中定义了一个STRIPS Planner来解决逻辑问题。经过一些其他更简单问题的试用后，我开始研究它是否可以解决更复杂的问题。我给了他一个STRIPS定义的peg纸牌，英文版，考虑到我们不能进行对角移动，最后一个球最终会在棋盘的中心进行尝试，程序会进入循环。问题在于：https://en.wikipedia.org/wiki/Peg_solitaire
这是我的解决方案：
%%%%%%%%%%%%%%%%%%%%%% PLAN %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
accao(nome : move(Xi,Yi,Xf,Yf),
condicoes : [empty(Xf,Yf),ball(Xi,Yi), ball(Xm,Ym)],
efeitos : [ball(Xf,Yf), -ball(Xm,Ym),-ball(Xi,Yi), empty(Xi,Yi), empty(Xm,Ym), -empty(Xf,Yf)],
restricoes : [abs(Xf-Xi)+abs(Yf-Yi)=:=2, abs(Xf-Xi)*abs(Yf-Yi)=:=0, Xi=<Xm, Xm=<Xf, Yi=<Ym, Ym=<Yf]).
inicial([empty(5,5), ball(1,4), ball(1,5), ball(1,6),
ball(2,4), ball(2,5), ball(2,6),
ball(3,4), ball(3,5), ball(3,6),
ball(4,1), ball(4,2), ball(4,3),ball(4,4), ball(4,5), ball(4,6),ball(4,7), ball(4,8), ball(4,9),
ball(5,1), ball(5,2), ball(5,3),ball(5,4), ball(5,6),ball(5,7), ball(5,8), ball(5,9),
ball(6,1), ball(6,2), ball(6,3),ball(6,4), ball(6,5), ball(6,6),ball(6,7), ball(6,8), ball(6,9),
ball(7,4), ball(7,5), ball(7,6),
ball(8,4), ball(8,5), ball(8,6),
ball(9,4), ball(9,5), ball(9,6)]).
objectivos([ball(5,5), empty(1,4), empty(1,5), empty(1,6),
empty(2,4), empty(2,5), empty(2,6),
empty(3,4), empty(3,5), empty(3,6),
empty(4,1), empty(4,2), empty(4,3),empty(4,4), empty(4,5), empty(4,6),empty(4,7), empty(4,8), empty(4,9),
empty(5,1), empty(5,2), empty(5,3),empty(5,4), empty(5,6),empty(5,7), empty(5,8), empty(5,9),
empty(6,1), empty(6,2), empty(6,3),empty(6,4), empty(6,5), empty(6,6),empty(6,7), empty(6,8), empty(6,9),
empty(7,4), empty(7,5), empty(7,6),
empty(8,4), empty(8,5), empty(8,6),
empty(9,4), empty(9,5), empty(9,6)]).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%% PRINT FUNCTION %%%%%%%%%%%%%%%%%%%%%%%%%%%
printExec([]).
printExec([A,E|T]) :- write("Action performed: "),
write(A),nl,
write("Situation: "),
write(E),nl,
printExec(T).
writeExec([I|T]):- write("Initial Situation"),
write(I),nl,
printExec(T),
write("Goal: "),
objectivos(G),
write(G),
write(" satisfied."),nl.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%% AUXILIAR FUNCTIONS %%%%%%%%%%%%%%%%%%%%%%%%%
member(E,[E|_]).
member(E,[_|T]):-member(E,T).
sub([],_).
sub([H|T],L):- member(H,L),
sub(T,L).
remove(_,[],[]):-!.
remove(E1, [E2|T], T):- E1 == E2, !.
remove(E,[H|T1],[H|T2]):- remove(E,T1,T2).
add(E,[],[E]):-!.
add(E1,[E2|T],[E1,E2|T]):- E1 \== E2, !.
add(E,[H|T1],[H|T2]):-add(E,T1,T2).
effects([],S,S).
effects([-H|Fx],S,N) :-!,
remove(H,S,NS),
effects(Fx,NS,N).
effects([H|Fx],S,N) :- !,
add(H,S,NS),
effects(Fx,NS,N).
restriction([]).
restriction([R|T]) :- R,
restriction(T).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%% PLAN EXECUTE %%%%%%%%%%%%%%%%%%%%%%%%%%%
planExecute(P):-testPlan(P,E),writeExec(E),!.
satisfiedGoal(E):- objectivos(Fn),!,
sub(Fn,E).
testPlan(Plan,[I|Exec]) :- inicial(I),
testPlan(Plan,I,Exec,Fn),
satisfiedGoal(Fn).
testPlan([],Fn,[],Fn).
testPlan([H|T],S,[H,N|Exec],Fn) :- accao(nome:H, condicoes:C,efeitos:E, restricoes:R),
sub(C,S),
effects(E,S,N),
restriction(R),
testPlan(T,N,Exec,Fn).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%% FIND PLAN %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
plano(P) :- progressivePlan(P, 0).
progressivePlan(P, N) :- createPlan(P,_,0,N).
progressivePlan(P, N) :- \+ createPlan(P,_,0,N),
NewN is N + 1,
progressivePlan(P, NewN).
createPlan(Plan,[I|Exec],N,Max) :- inicial(I),
createPlan(Plan,I,Exec,Fn,N,Max),
satisfiedGoal(Fn).
createPlan([],Fn,[],Fn,Max,Max):- !.
createPlan([H|T],S,[H,N|Exec],Fn,Acc, Max) :- accao(nome:H, condicoes:C, efeitos:E, restricoes:R),
sub(C,S),
effects(E,S,N),
restriction(R),
NewAcc is Acc+1,
createPlan(T,N,Exec,Fn,NewAcc, Max).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%`
我尝试通过做一两个动作来简化目标，这个动作适用于一个动作，当两个动作不相互矛盾时，就像将一个大理石移动到已经移动过的大理石上一样，当他们这样做时，用两个动作进入循环，就像所说的目标一样：
objectivos（[球（4,5），空（3,5），空（5,5），空（6,5）]）。
我尝试过跟踪和调试，但我似乎无法找到问题，虽然我认为它位于问题的表述而不是Planner本身。任何想法？
答案 0 :(得分：0)
您的代码中至少存在一个逻辑错误，并且可以进行一些简单的性能调整。这为您的问题提供了部分解决方案。
首先，对于逻辑错误：目标objectivos([ball(4,5), empty(3,5), empty(5,5), empty(6,5)])的预期解决方案似乎是计划P = [move(3, 5, 5, 5), move(6, 5, 4, 5)]。但是，根据您对restricoes的定义，这些举措中的第二步是不合法的：对于此移动，您有Xi = 6，Xf = 4以及需要6 =< Xm和{{1}的条件}，但这是不可能的。这些约束的想法是确保Xm <= 4在移动中的其他两个球之间。这是另一种确保这一点的配方：
ball(Xm,Ym)这也排除了以前在跟踪代码时让我困惑的案例：以前使restricoes : [abs(Xf-Xi)+abs(Yf-Yi) =:= 2,
abs(Xf-Xi)*abs(Yf-Yi) =:= 0,
abs(Xf-Xm)+abs(Yf-Ym) =:= 1,
abs(Xi-Xm)+abs(Yi-Ym) =:= 1]
合法是合法的。
其次，要提高效果，请在ball(Xi,Yi) = ball(Xm,Ym)的定义中交换目标effects(E,S,N)和restriction(R)。以前你在检查合法性之前计算了移动的效果！因为策划者提出的大多数动作都是非法的，这浪费了很多时间。
然后，为了使整个事情更好用，您可以将createPlan/6和plano/1的定义更改为：
createPlan/4这比之前的定义更简单，而且行为也更好。我们可以通过一个完整的计划来检查它是否合法，或者只是传入一个固定长度的列表来询问该长度的计划是什么：
plano(P) :-
length(P, PlanLength),
createPlan(P, _, 0, PlanLength).
createPlan(Plan,[I|Exec],N,Max) :- inicial(I),
N =< Max,
createPlan(Plan,I,Exec,Fn,N,Max),
satisfiedGoal(Fn).
根据您的定义，这将继续循环并计算?- P = [_,_], plano(P).
P = [move(3, 5, 5, 5), move(6, 5, 4, 5)] ;
false. % no more solutions
计数器，搜索不存在的其他解决方案。
通过这种表述，我们可以切换到您的大目标，并尝试寻找解决方案（这部分特定于SWI-Prolog）：
Max此时我不得不打断搜索，它变得太慢了。当然可以进行更多的调整，我可能会继续关注它。