list - 从列表中删除每个第 n 个元素的 Prolog 程序

Question

你能帮我解决以下问题吗？

Write a ternary predicate delete_nth that deletes every n-th element from a list.

示例运行:

?‐ delete_nth([a,b,c,d,e,f],2,L).
L = [a, c, e] ;
false
?‐ delete_nth([a,b,c,d,e,f],1,L).
L = [] ;
false
?‐ delete_nth([a,b,c,d,e,f],0,L).
false

我试过这个:

listnum([],0).
listnum([_|L],N) :-
   listnum(L,N1),
   N is N1+1.

delete_nth([],_,_).
delete_nth([X|L],C,L1) :- 
   listnum(L,S),
   Num is S+1,
   (  C>0
   -> Y is round(Num/C),Y=0
   -> delete_nth(L,C,L1)
   ;  delete_nth(L,C,[X|L1])
   ).

Answer 1

我有点奢侈的变体:

delete_nth(L, N, R) :-
    N > 0, % Added to conform "?‐ delete_nth([a,b,c,d,e,f],0,L). false"
    ( N1 is N - 1, length(Begin, N1), append(Begin, [_|Rest], L) ->
        delete_nth(Rest, N, RestNew), append(Begin, RestNew, R)
    ;
        R = L
    ).