检查列表是否是回文。如果没有，插入元素使其成为回文。 （序言）

Question

我写了下面的代码来检查它是否是回文。我还创建了在列表不是回文时插入元素的逻辑

reverse_list(Inputlist, Outputlist) :-
   reverse(Inputlist, [], Outputlist).    

reverse([], Outputlist, Outputlist).    
reverse([Head|Tail], List1, List2) :-
   reverse(Tail, [Head|List1], List2).

printList([]).
printList([X|List]) :-
   write(X),
   write(' '),
   printList(List).

palindrome(List1) :-
   reverse_list(List1, List2),
   compareLists(List1, List1, List2, List2).

compareLists(L1, [], [], L2) :-
   write("\nList is Palindrome").    
compareLists(L1, [X|List1], [X|List2], L2) :-
   compareLists(L1, List1, List2, L2),
   !.        
compareLists(L1, [X|List1], [Y|List2], [Z|L2]) :-
   write("\nList is not Palindrome. "),
   append(L1, L2, L),
   printList(L).

代码为

提供了正确的输出

palindrome([a,b,c,a]).
List is not Palindrome. a b c a c b a 

palindrome([a,b,c]).
List is not Palindrome. a b c b a 

然而，对于诸如

之类的输入

palindrome([a,b,c,b]).
List is not Palindrome. a b c b c b a 

然而，最佳解决方案应该是

 a b c b a

我应该加入哪些更改才能实现此目标？

Answer 1

DCG的前3个方程捕获回文模式。 添加第四个，涵盖不匹配，以完成规范：

p([]) --> [].
p([T]) --> [T].
p([T|R]) --> [T], p(P), [T], {append(P,[T],R)}.
p([T|R]) --> [T], p(P), {append(P,[T],R)}.

?- phrase(p(L), [a,b,c,b]).
L = [a, b, c, b, a] ;
L = [a, b, c, c, b, a] ;
L = [a, b, c, b, c, b, a] ;
L = [a, b, c, b, b, c, b, a] ;
false.

Answer 2

我认为你需要一个带有两个Args的谓词，In和Out：

pal([], []).
pal([X], [X]).
pal(In, Out) :-
    % first we check if the first and last letter are the same
    (   append([H|T], [H], In)
        % we must check that the middle is a palindrome
    ->  pal(T, T1),
        append([H|T1], [H], Out)
    ;   % if not, we remove the first letter
        % and we work with the rest
        In = [H|T],
        % we compute the palindrome from T
        pal(T,T1),
        % and we complete the palindrome to
        % fit the first letter of the input
        append([H|T1], [H], Out)).

<强> EDIT1 这段代码看起来不错，但

有一个错误

? pal([a,b,c,a], P).
P = [a, b, c, b, a] .

应该是[a，b，c，a，c，b，a] 我会尝试解决它。​​

<强> EDIT2 看起来正确：

build_pal([H|T], Out):-
    pal(T,T1),
    append([H|T1], [H], Out).


pal([], []).
pal([X], [X]).
pal(In, Out) :-
    (   append([H|T], [H], In)
    ->  pal(T, T1),
        (   T = T1
        ->  append([H|T1], [H], Out)
        ;   build_pal(In, Out))
    ;   build_pal(In, Out)).

带输出：

 ?- pal([a,b,c], P).
P = [a, b, c, b, a] .

 ?- pal([a,b,a], P).
P = [a, b, a] .

 ?- pal([a,b,c,b], P).
P = [a, b, c, b, a] .

 ?- pal([a,b,c,a], P).
P = [a, b, c, a, c, b, a] .

 ?- pal([a,b,a,c,a], P).
P = [a, b, a, c, a, b, a] .