在Prolog中查找列表的长度 [英] Finding Length of List in Prolog

问题描述

我正在运行Prolog并尝试编写一个返回列表长度的小函数:

I'm running Prolog and trying to write a small function returning the length of a list:

len([],0).
len([XS], Y) :-
    len([X|XS], M),
    Y is M+1.

我的逻辑是，递归调用应包括列表的尾部(XS)，并将1增加到先前的长度(Y为M + 1.)

My logic is that the recursive call should include the tail of the list (XS) and increase 1 to the previous length (Y is M+1.)

这总是返回false.

This always returns false.

有什么想法吗?

推荐答案

以下是调试和测试Prolog谓词的通用方法:

Here is a general methodology for debugging and testing Prolog predicates:

考虑一下:在Prolog中，您不需要组成一些测试数据.您甚至根本不需要了解谓词:只需输入自由变量即可！那总是专业的举动！

Think of it: In Prolog you do not need to make up some test data. You don't even need to understand a predicate at all: Just hand in free variables! That is always a professional move!

因此，就您而言，

?- len(L,N).
L = [],
N = 0 ;
*LOOPS**

您的定义并不像您声称的那么糟糕:至少对于空白列表而言，它是正确的.

Your definition is not that bad as you claim: At least, it is true for the empty list.

现在，也许看看您可能收到的编译器警告:

Now, maybe look at the compiler warnings you probably received:

Warning: user://1:11:
        Singleton variables: [X]

接下来，按照箭头:- 的方向(即从右到左)读取递归规则:

Next read the recursive rule in the direction of the arrow :- that is, right-to-left:

提供的len([X|Xs], M)是真实的，而Y is M+1是真实的，只要所有这些都是真实的，我们就可以得出结论

Provided len([X|Xs], M) is true and Y is M+1 is true, provided all that is true, we can conclude that

len([XS], Y)也是如此.因此，您总是得出关于长度为1([Xs])的列表的结论.

len([XS], Y) is true as well. So you are always concluding something about a list of length 1 ([Xs]).

您需要将其重新配置为len([X|Xs], M) :- len(Xs, N), Y is M+1.

You need to reformulate this to len([X|Xs], M) :- len(Xs, N), Y is M+1.

这是另一种策略:

通过消除目标，我们可以推广程序 ¹ .这是我最喜欢的方式.通过添加谓词(*)/1像这样:

By removing goals, we can generalize a program¹. Here is my favorite way to do it. By adding a predicate (*)/1 like so:

:- op(950,fy, *).

*_.

现在，让我们从程序中删除所有目标:

Now, let's remove all goals from your program:

len([],0).
len([XS], Y) :-
    * len([X|XS], M),
    * Y is M+1.

我们现在所拥有的是概括.再次，我们将看看最普遍的查询的答案:

What we have now is a generalization. Once again, we will look at the answers of the most general query:

?- len(L, N).
L = [],
N = 0 ;
L = [_].

什么? len/2仅适用于长度为0和1的列表.这意味着，即使len([1,2], N)也会失败！因此，现在我们可以确定:必须修复程序可见剩余部分中的某些内容.实际上，[XS]仅描述了长度为1的列表.因此，必须将其删除...

What? len/2 is only true for lists of length 0 and 1. That means, even len([1,2], N) fails! So now we know for sure: something in the visible remaining part of the program has to be fixed. In fact, [XS] just describes lists of length 1. So this has to be removed...

精美打印:

_{1某些限制适用.本质上，您的程序必须是纯粹的单调程序.}

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