prolog - 我解决 3 壶水谜题的序言程序有什么问题？

Question

谁能找到为什么我在这段代码中的“开始”无法得到任何真正的答案？例如，我写go(7,3,l)了，我想它应该将 3 升水移到第二个水罐中，但根据序言，这是错误的。怎么了？

:- dynamic go/3.
:- dynamic cur_state/1,init/5.
:- dynamic end_state/1, final/5.

cur_state(State):-State = state(10,0,0,7,l).
end_state(State):-State = state(0,3,3,0,r).

pour(state(D1,D2,D3,n,l),move(D,C,r),state(D,C,D3,n,r)) :-
        D is D1-n,
        C is D2+n.
pour(state(D1,D2,D3,n,r),move(D,C,l),state(D,C,D3,n,l)) :-
        D is D1-n,
        C is D2.
pour(state(D1,D2,D3,n,l),move(D,C,r),state(D,D2,C,n,r)) :-
        D is D1-n,
        C is D3+n.
pour(state(D1,D2,D3,n,l),move(D,C,r),state(D1,D,C,n,r)) :-
        D is D2-n,
        C is D3+n.
pour(state(D1,D2,D3,n,r),move(D,C,l),state(D1,D,C,n,l)) :-
        D is D2-n,
        C is D1+n.
pour(state(D1,D2,D3,n,r),move(D,C,l),state(D1,D,c,n,l)) :-
        D is D2-n,
        C is D3.
pour(state(D1,D2,D3,n,l),move(D,C,r),state(C,D2,D,n,r)) :-
        D is D3-n,
        C is D1.
pour(state(D1,D2,D3,n,r),move(D,C,l),state(D1,C,D,n,l)) :-
        D is D3-n,
        C is D2+n.
pour(state(D1,D2,D3,n,r),move(D,C,l),state(C,D2,D,n,l)) :-
        D is D3-n,
        C is D1+n.

carry(7,0).
carry(3,0).
carry(10,0).
carry(7,0).
carry(7,3).

legal(10,X,Y):-X+Y<10.
legal(X,Y,Z):-X+Y+Z<10.
legal(X,7,Y):-X+Y=<3.
legal(X,Y,3):-X+Y=<7.

newstate(state(D1,D2,D3,n,l),state(D11,D22,D33,n1,r)):-
        carry(M,C),
        M=<7,C=<3,
        D22 is D2+n,
        D11 is D1-n,
    D3 is D33,
    n1 is n,
        D2=<7,D1=<10,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,r),state(D11,D22,D33,n1,l)):-
        carry(M,C),
        M=<10,C=<100,
        D11 is D1-n,
    D22 is D2,
    D33 is D3,
        D1=<10,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,l),state(D11,D22,D33,n1,r)):-
        carry(M,C),
        M=<10,C<3,
        D11 is D1-n,
        D33 is D3+n,
    D22 is D2,
        D1=<10,D3=<3,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,r),state(D11,D22,D33,n1,l)):-
        carry(M,C),
        M=<7,C=<3,
        D22 is D2-n,
        D33 is D1+n,
        D11 is D1,
    D2=<7,D1=<10,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,l),state(D11,D22,D33,n1,r)):-
        carry(M,C),
        M=<7,C=0,
        D22 is D2-n,
        D33 is D3+n,
        D11 is D1,
    D2=<7,D3=<3,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,r),state(D11,D22,D33,n1,l)):-
        carry(M,C),
        M=<7,C=<100,
        D22 is D2-n,
    D33 is D3,
    D11 is D1,    
    D2=<7,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,r),state(D11,D22,D33,n1,l)):-
        carry(M,C),
        M=<3,C=<7,
        D22 is D2+n,
        D33 is D3-n,
        D11 is D1,
    D3=<3,D2=<7,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,r),state(D11,D22,D33,n1,l)):-
        carry(M,C),
        M=<3,C=<100,
        D11 is D1+n,
        D33 is D3-n,
        D22 is D2,
    D3=<3,D1=<10,
    legal(D1,D2,D3).

newstate(state(D1,D2,D3,n,l),state(D11,D22,D33,n1,r)):-
        carry(M,C),
        M=<3,C=<100,
        D33 is D3-n,
        D22 is D2,
    D11 is D1,  
    D3=<3,
    legal(D1,D2,D3).


eisodos(_):- cur_state(State),write(State),nl.

init(S1,S2,S3,S4,S5):-assert(cur_state(State):-State = state(S1,S2,S3,S4,S5)),write('Arxikh:'),
   write(state(S1,S2,S3,S4,S5)),retractall(init(S1,S2,S3,S4,S5)),nl.

final(S1,S2,S3,S4,S5):-assert(end_state(State):-State = state(S1,S2,S3,S4,S5)),write('Telikh:'),
   write(state(S1,S2,S3,S4,S5)),retractall(init1(S1,S2,S3,S4,S5)),nl.

go(Move1,Move2,Move3):-cur_state(State),newstate(State,NextState),
        pour(State,move(Move1,Move2,Move3), NextState),
        retractall(cur_state(State):-State = state(_,_,_,_,_)),asserta(cur_state(NextState)),
        ((end_state(NextState),write('Bravo!!!!')) ;(write(' ---*Eiste sthn katastash --- :'),write(NextState))),nl.

score 3 · Accepted Answer

扩展@Mog 的答案，如果您想找到最短的解决方案，我建议使用迭代深化。以下基于@Mog 发布的代码（从而解决了与OP 发布的问题略有不同的相同问题）。由于我们要描述一个（移动的）列表，DCG 表示法很方便：

solution(Path) :- length(Path, _), phrase(path([0-3, 0-5, 8-8]), Path).

path(State) --> { equivalent(State, [0-3, 4-5, 4-8]) }.
path(State0) --> [From-To],
        { move(State0, State), State = [_-From, _-To, _] },
        path(State).

equivalent(State1, State2) :- forall(member(X, State1), member(X, State2)).

move(State, [NewX-From, NewY-To|NewRest]) :-
    select(X-From, State, Rest),
    X \== 0,
    select(Y-To, Rest, NewRest),
    Fillable is To - Y,
    ToFill is min(X, Fillable),
    NewY is Y + ToFill,
    NewX is X - ToFill.

示例查询：

?- solution(Ps).
Ps = [8-5, 5-3, 3-8, 5-3, 8-5, 5-3, 3-8] .

score 2 · Accepted Answer

好吧，正如我在评论中所说，我不知道如何帮助您完成当前的工作，因为其中有很多错误的东西。我建议您阅读有关 Prolog 的精彩教程（例如Learn Prolog now），以便您掌握语言基础知识。如果您有兴趣，这里有一个简单的方法来解决您的问题。如果您不希望您的问题被破坏，请不要进一步阅读：]（我发布的内容是关于 3/5/8 水罐和 4/4 拆分）。

go(Path) :-
    solve([0-3, 0-5, 8-8], [], [], Temp),
    reverse(Temp, Path).

solve(State, _Visited, Path, Path) :-
    equivalent(State, [0-3, 4-5, 4-8]).
solve(State, Visited, Acc, Path) :-
    move(State, NewState),
    NewState = [_-From, _-To|_],
    forall(member(Past, Visited), \+ equivalent(Past, NewState)),
    solve(NewState, [NewState|Visited], [From-To|Acc], Path).

equivalent(State1, State2) :-
    forall(member(X, State1), member(X, State2)).

move(State, [NewX-From, NewY-To|NewRest]) :-
    select(X-From, State, Rest),
    X \== 0,
    select(Y-To, Rest, NewRest),
    Fillable is To - Y,
    ToFill is min(X, Fillable),
    NewY is Y + ToFill,
    NewX is X - ToFill.

如果您在阅读了有关 prolog 的更多内容后需要有关代码的解释，请不要犹豫！

score 0 · Accepted Answer

您需要更正它的基本语法的第一件事：变量必须以大写开头，因此例如您必须更改此规则

pour(state(D1,D2,D3,n,l),move(D,C,r),state(D,C,D3,n,r)) :-
    D is D1-n,
    C is D2+n.

至

pour(state(D1,D2,D3,N,l),move(D,C,r),state(D,C,D3,N,r)) :-
    D is D1-N,
    C is D2+N.

和

newstate(state(D1,D2,D3,n,l),state(D11,D22,D33,n1,r)):-
        carry(M,C),
        M=<7,C=<3,
        D22 is D2+n,
        D11 is D1-n,
    D3 is D33,
    n1 is n,
        D2=<7,D1=<10,
    legal(D1,D2,D3).

至

newstate(state(D1,D2,D3,N,l),state(D11,D22,D33,N1,r)):-
    carry(M,C),
    M=<7,C=<3,
    D22 is D2+N,
    D11 is D1-N,
    D3 is D33,
    N1 is N,
    D2=<7,D1=<10,
    legal(D1,D2,D3).

您应该意识到 N1 仅在第一个 newstate/2 过程中被评估：然后更正该规则的其他实例。

然后你应该删除无用的动态断言：想想你的实际断言/收回所有用法：你需要的是在任何步骤改变状态在“水罐”之间移动“水”：所以你应该在循环时断言/收回 state/5 go，从初始状态（在启动程序时断言）到最终可接受的状态，在 go 中进行测试以停止循环。

但是这种基于状态变化的风格导致调试非常困难。相反，尝试删除所有动态、断言/撤回，并在循环中传递状态。

prolog - 我解决 3 壶水谜题的序言程序有什么问题？

3 回答 3

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