在Prolog中从列表的每个子列表中删除最大元素

有人能帮我写一个函数，从列表的每个子列表中删除最大元素吗？我不知道。。。 到目前为止，我只知道如何编写delete子句：

delete(_,[],[]).
delete(E,[H|T],L):-
    delete(E,T,L).
delete(E,[H|T],[H|T1):-
    delete(E,T,T1).

---

对于列表[1,2,4,3,4,5,6,7]==>[1,2,3,5,6]]

这应该适用于任意嵌套的子列表。在这方面，这可能有些过分，但要求并不明确

% second list is the first list with sublists removed
remove_sublists([H|T], L) :-
    (   is_list(H)
    ->  L = T1
    ;   L = [H|T1]
    ),
    remove_sublists(T, T1).
remove_sublists([], []).

% max value in list ignoring sublists
% if L members are all lists, then this predicate fails
mymax(L, Max) :-
    remove_sublists(L, F),
    F \= [],
    max_list(F, Max).

% delete maximum value from every sublist,
% and sublists of sublists
delete_max_subs([H|T], [H1|T1]) :-
    (   is_list(H)
    ->  (   mymax(H, M)
        ->  delete(H, M, HD)
        ;   HD = H
        ),
        delete_max_subs(HD, H1)
    ;   H1 = H
    ),
    delete_max_subs(T, T1).
delete_max_subs([], []).

一些结果：

| ?- delete_max_subs([1,[2,4,3,4],5,[6,7]], L).

L = [1,[2,3],5,[6]]

yes
| ?- delete_max_subs([[[1,2],2],[2,4,3,4],5,[6,7]], L).

L = [[[1]],[2,3],5,[6]]

yes
| ?- delete_max_subs([[[1,2,3],2,5],[2,4,3,4],5,[6,8,7]], L).

L = [[[1,2],2],[2,3],5,[6,7]]

yes

附录

一个更加直观/Turbo Prolog友好的版本：

% second list is the first list with sublists removed
remove_sublists([H|T], T1) :-
    list(H),
    remove_sublists(T, T1).
remove_sublists([H|T], [H|T1]) :-
    not(list(H)),
    remove_sublists(T, T1).
remove_sublists([], []).

% max value in list ignoring sublists
mymax(L, Max) :-
    remove_sublists(L, F),
    F \= [],
    max_list(F, Max).    % I don't know if something like this exists in TP

% delete max recursively from L and all sublists
delete_max(L, R) :-
    mymax(L, M),
    delete(L, M, L1),
    delete_max_subs(L1, R).
delete_max(L, R) :-
    delete_max_subs(L, R).
delete_max([], []).

% delete maximum only from sublists
delete_max_subs([H|T], [H1|T1]) :-
    list(H),
    delete_max(H, H1),
    delete_max_subs(T, T1).
delete_max_subs([H|T], [H|T1]) :-
    not(list(H)),
    delete_max_subs(T, T1).
delete_max_subs([], []).

---

第三个filter/3谓词减少为：filterX、[H|T]、[H|C]：-X\=H、filterX、T、C。第二个和第三个deletemax/2子句也是如此。当第一个参数是单个元素列表时，不需要使用append/3。这是一个不错的解决方案+1，但它不能处理嵌套子列表的所有情况，例如，[[1,2]，2]，[2,4,3,4]，5，[6,7]]。但是OP是否需要任意的子列表嵌套尚不清楚。是否需要处理多个级别的子列表，如[1、[2,4,3、[7,1]]、5、[[6]、[7,8]]？您好。你知道turbo prolog是否有->/2吗？@capelical我找不到任何证据表明turbo prolog支持->/2，所以我添加了一个不使用它的版本。真不错！我们可以理解->/2是多么有用@我同意/在我看来，2在很多情况下都能澄清逻辑，当然，如果使用正确的话：。

% second list is the first list with sublists removed
remove_sublists([H|T], T1) :-
    list(H),
    remove_sublists(T, T1).
remove_sublists([H|T], [H|T1]) :-
    not(list(H)),
    remove_sublists(T, T1).
remove_sublists([], []).

% max value in list ignoring sublists
mymax(L, Max) :-
    remove_sublists(L, F),
    F \= [],
    max_list(F, Max).    % I don't know if something like this exists in TP

% delete max recursively from L and all sublists
delete_max(L, R) :-
    mymax(L, M),
    delete(L, M, L1),
    delete_max_subs(L1, R).
delete_max(L, R) :-
    delete_max_subs(L, R).
delete_max([], []).

% delete maximum only from sublists
delete_max_subs([H|T], [H1|T1]) :-
    list(H),
    delete_max(H, H1),
    delete_max_subs(T, T1).
delete_max_subs([H|T], [H|T1]) :-
    not(list(H)),
    delete_max_subs(T, T1).
delete_max_subs([], []).