Prolog如何将列表中的前3个元素存储到另一个列表中
我尝试过这样的事情： 说path = [[1,2],[2,3],[3,4],[5,6]]
first_three(N,[H|T],[H|_]):- not(N=0),N is N-1, first_three(N,T,L).
然后我调用first_third(3, path, Y) 但它返回no。
I tried something like this:
said path = [[1,2],[2,3],[3,4],[5,6]]
first_three(N,[H|T],[H|_]):- not(N=0),N is N-1, first_three(N,T,L).
then I call first_three(3, path, Y) but it return no.
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您需要添加一个基本情况（当 N = 0 时）。并且还需要实例化一个新的fresh变量来给N的前驱赋值，最后还需要返回递归的结果（子句头部的[H|L]）：
当然，你也可以只需编写如下内容：
并调用它：
You need to add a base case (when N = 0). And you also need to instantiate a new fresh variable to assign the predecesor of N, and finally you also need to return the result of the recursion (the [H|L] in the head of the clasuse):
Of course, you can also just write something like this:
and call it:
首先，
N is N - 1不可能成功 -N不是命令式编程语言意义上的变量，因此您不会在其中存储值，它只能被赋值一次，可以这么说 - 所以你可能想引入一个新变量并调用类似N1 is N - 1的东西，然后在中使用N1对first_third的递归调用。然后，该递归调用的最后一个参数不知从何而来，因此您可能也需要注意它。作为最后一个建议，尝试将谓词分成两个子句：第一个子句管理基本情况，其中N为0，第二个子句管理重要情况，构建在另一个之上。哦，由于
N是一个可变数量，我建议为您的谓词使用不同的名称，而不是误导性的first_third。First of all,
N is N - 1can't possibly succeed -Nis not a variable in the imperative programming language sense, so you don't store values in it, it can be assigned only once, so to speak - so you may want to introduce a new variable and call something likeN1 is N - 1, and subsequently useN1in the recursive call tofirst_three. Then, the last argument to that recursive call comes out of nowhere, so you may need to pay attention to it too. As a last suggestion, try to split the predicate in two clauses: the first to manage the elementary basic case whereNis0, the second to manage the meaty cases, building on top of the other.Oh, and since
Nis a variable quantity, I would suggest to use a different name rather than the misleadingfirst_threefor your predicate.