反向/回文的递归Prolog谓词 [英] Recursive Prolog predicate for reverse / palindrome
问题描述
-
我能否获得一个具有两个参数的递归Prolog谓词，称为反向，该参数返回列表的反向:
Can I get a recursive Prolog predicate having two arguments, called reverse, which returns the inverse of a list:
示例查询和预期结果:
?- reverse([a,b,c], L).
L = [c,b,a].
具有两个称为palindrome的参数的递归Prolog谓词，如果给定列表为回文，则返回true.
A recursive Prolog predicate of two arguments called palindrome which returns true if the given list is palindrome.
示例查询具有预期结果:
Sample query with expected result:
?- palindrome([a,b,c]).
false.
?- palindrome([b,a,c,a,b]).
true.
推荐答案
没有一种有效的方法来使用单个递归定义来定义reverse/2而不使用一些辅助谓词.但是，如果仍然允许这样做，则不依赖append/3之类的内置方法(并且应适用于大多数Prolog实现)的简单解决方案将是使用累加器列表，如下:
There isn't an efficient way to define reverse/2 with a single recursive definition without using some auxiliary predicate. However, if this is nevertheless permitted, a simple solution which doesn't rely on any built-ins like append/3 (and should be applicable for most Prolog implementations) would be to use an accumulator list, as follows:
rev([],[]).
rev([X|Xs], R) :-
rev_acc(Xs, [X], R).
rev_acc([], R, R).
rev_acc([X|Xs], Acc, R) :-
rev_acc(Xs, [X|Acc], R).
rev/2是逆向谓词，它简单地委托"到(或包装)基于累加器的版本rev-acc/2，该版本将输入列表的元素以相反的顺序递归添加到累加器中.
rev/2 is the reversal predicate which simply 'delegates' to (or, wraps) the accumulator-based version called rev-acc/2, which recursively adds elements of the input list into an accumulator in reverse order.
运行此:
?- rev([1,3,2,x,4],L).
L = [4, x, 2, 3, 1].
确实，@ false已经指出(+1)，
And indeed as @false has already pointed out (+1),
palindrome(X) :- rev(X,X).
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