Prolog 有一个;“超出全局堆栈”;在序言中
嘿,伙计们,关于序言我有一个相当简单的问题Prolog 有一个;“超出全局堆栈”;在序言中,prolog,planning,failure-slice,Prolog,Planning,Failure Slice,嘿,伙计们,关于序言我有一个相当简单的问题 %on(Block,Object). % clear(Object). block(b1). block(b2). block(b3). place(p1). place(p2). place(p3). place(p4). state1([clear(p2),clear(p4),clear(b2),clear(b3),on(b1,p1),on(b2,p3),on(b3,b1)]). % visual state1 % b3 % b1 b2 %
%on(Block,Object).
% clear(Object).
block(b1).
block(b2).
block(b3).
place(p1).
place(p2).
place(p3).
place(p4).
state1([clear(p2),clear(p4),clear(b2),clear(b3),on(b1,p1),on(b2,p3),on(b3,b1)]).
% visual state1
% b3
% b1 b2
% = = = =
% 1 2 3 4 <----Positions
% can(Action,Condition).
% adds(Action,AddRelationship).
% deletes(Action,DeleteRelationship).
% move(Block,From,To).
can( move( Block, From, To), [ clear( Block), clear( To), on( Block, From)]) :-
block( Block), % Block to be moved
object( To), % "To" is a block or a place
To \== Block, % Block cannot bå moved to itself
object( From), % "From" is a block or a place
From \== To, % Move to new position
Block \== From. % Block not moved from itself
adds(move(X,From,To),[on(X,To),clear(From)]).
deletes(move(X,From,To),[on(X,From),clear(To)]).
object(X):-
place(X)
;
block(X).
% plan(State,Goals,Plan,FinalState).
plan(State,Goals,[],State):-
satisfied(State,Goals).
plan(State,Goals,Plan,FinalState) :-
append(PrePlan,[Action|PostPlan],Plan),
select(State,Goals,Goal),
achieves(Action,Goal),
can(Action,Condition),
plan(State,Condition,PrePlan,MidState1),
apply(MidState1,Action,MidState2),
plan(MidState2,Goals,PostPlan,FinalState).
% satisfied(State,[]).
satisfied(State,[Goal|Goals]):-
member(Goal,State),
satisfied(State,Goals).
select(State,Goals,Goal):-
member(Goal,Goals),
not(member(Goal,State)).
achieves(Action,Goal):-
adds(Action,Goals),
member(Goal,Goals).
apply(State,Action,NewState):-
deletes(Action,DelList),
delete_all(State,DelList,State1),!,
adds(action,AddList),
append(AddList,State1,NewState).
delete_all([],_,[]).
delete_all([X|L1],L2,Diff):-
member(X,L2),!,
delete_all(L1,L2,Diff).
delete_all([X|L1],L2,[X|Diff]):-
delete_all(L1,L2,Diff).
(块、对象)上的%。
%清除(对象)。
区块(b1)。
区块(b2)。
块(b3)。
地点(p1)。
地点(p2)。
地点(p3)。
地点(p4)。
状态1([清除(p2),清除(p4),清除(b2),清除(b3),打开(b1,p1),打开(b2,p3),打开(b3,b1)])。
%视觉状态1
%b3
%b1 b2
% = = = =
%1 2 3 4你只需要看看这个:
plan(State,Goals,[],State):- false,
satisfied(State,Goals).
plan(State,Goals,Plan,FinalState) :-
append(PrePlan,[Action|PostPlan],Plan), false,
select(State,Goals,Goal),
achieves(Action,Goal),
can(Action,Condition),
plan(State,Condition,PrePlan,MidState1),
apply(MidState1,Action,MidState2),
plan(MidState2,Goals,PostPlan,FinalState).
?- plan(state1,on(b1,b2),Plan,FinalState).
计划(状态,目标,[],状态):-false,
满足(状态、目标)。
计划(状态、目标、计划、最终状态):-
追加(预计划、[行动|后计划]、计划),false,
选择(状态、目标、目标),
实现(行动、目标),
can(动作、条件),
计划(状态、条件、预先计划、中期计划1),
应用(中间状态1、行动、中间状态2),
计划(中期2、目标、后期计划、最终状态)。
?计划(状态1,on(b1,b2),计划,最终状态)。
由于此程序已经循环,同一个程序将循环并删除附加的false
目标。你需要先解决这个问题。有关详细信息,请参阅。这是大量代码。你不能提供一个简单的例子来说明这个问题吗?如果你使用SWI,只需说check。
你会得到一个警告,adds(action)
总是会失败。你为什么这么早就调用append
来处理所有变量参数?嘿,谢谢你的回复,看起来你完全正确,但是,由于这是我第一次“接触”prolog,我在这方面非常糟糕。你能建议我如何解决这个问题吗