我正在尝试用Prolog实现一个块世界程序。块世界是人工智能中一个著名的难题，本身相当简单。以下是我目前的代码：

% Define the blocks in your worldblocks([a, b, c, d, e, f]).% A block X is a member of the list of blocksblock(X) :-    blocks(BLOCKS),  % Extract the list BLOCKS    member(X, BLOCKS).% Given predicatesmove(X, Y, Z, S1, S2):-    member([clear, X], S1),    member([on, X, Y], S1),    block(Y),    member([clear, Z], S1),    notequal(X, Z),    substitute([on, X, Y], [on, X, Z], S1, INT),    substitute([clear, Z], [clear, Y], INT, S2).notequal(X, X):-!, fail.notequal(_, _).substitute(X, Y, [X|T], [Y|T]).substitute(X, Y, [H|T], [H|T1]):-    substitute(X, Y, T, T1).% Additional predicates to be implemented% (i) move from a block onto the tablemove_onto_table(X, Y, S1, S2):-    member([clear, X], S1),    member([on, X, Y], S1),    substitute([on, X, Y], [on, X, 'table'], S1, INT),    substitute([clear, Y], [clear, X], INT, S2).% (ii) move from the table onto a blockmove_onto_block(X, Y, S1, S2):-    member([clear, X], S1),    member([on, X, 'table'], S1),    substitute([on, X, 'table'], [on, X, Y], S1, INT),    substitute([clear, X], [clear, Y], INT, S2).% Define start and goal statesstart([[on, d, 'table'], [on, c, d], [on, a, c], [on, b, 'table'], [clear, a], [clear, b]]).goal([[on, d, 'table'], [on, c, 'table'], [on, a, b], [on, b, 'table'], [clear, a], [clear, c], [clear, d]]).% Define notYetVisitednotYetVisited(State, PathSoFar):-    permutation(State, PermuteState),    \+ member(PermuteState, PathSoFar).% Define path and connect predicatespath(S1, S2):-    move(X, Y, Z, S1, S2).path(S1, S2):-    move_onto_table(X, Y, S1, S2).path(S1, S2):-    move_onto_block(X, Y, S1, S2).connect(S1, S2) :- path(S1, S2).connect(S1, S2) :- path(S2, S1).% Define DFS predicate with added debugging statements and iterative deepeningdfs(X, [X], _):-    goal(X),    write('Goal reached: '), write(X), nl.% else expand X by Y and find path from Ydfs(X, [X|Ypath], VISITED):-    connect(X, Y),    notYetVisited(Y, VISITED),    write('Current state: '), write(X), nl,    write('Moving to: '), write(Y), nl,    dfs(Y, Ypath, [Y|VISITED]).

我使用的查询是这样的：start(Start), dfs(Start, Path, []).

现在，输出的结果只是一个循环。尽管我实现了对重复状态的检查，但dfs似乎只是在两个相同的状态之间来回振荡，如下所示：

Current state: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]Moving to: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Current state: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Moving to: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]Current state: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]Moving to: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Current state: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Moving to: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]Current state: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]Moving to: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Current state: [[on, d, table], [on, c, d], [on, a, b], [on, b, table], [clear, a], [clear, c]]Moving to: [[on, d, table], [on, c, d], [on, a, c], [on, b, table], [clear, a], [clear, b]]

我尝试了多种解决方案。显然，期望的是它能够通过所需的多个状态到达目标状态（该目标状态在脚本中是硬编码的）。以下是我迄今为止所做的尝试：

1. 改变notYetVisited的工作方式，尝试使用简单的等式而不是使用排列。然而，这导致堆栈限制被超出。
2. 在各个点添加调试语句。这些语句已包含在提供的代码中。
3. 对dfs过程使用DFID（深度优先迭代加深）方法。然而，这同样几乎立即超出了堆栈限制。
4. 更改dfs以在回溯时更新状态。
5. 更改移动过程（move_onto_table和move_onto_block），以确保满足’clear’条件。

回答：