问题描述
我需要编写一个Prolog谓词take(L, N, L1)，如果列表L1以相同的顺序包含列表L的前N个元素，则该谓词成功.例如:
I need to write a Prolog predicate take(L, N, L1) which succeeds if list L1 contains the first N elements of list L, in the same order. For example:
?- take([5,1,2,7], 3, L1).
L1 = [5,1,2]
?- take([5,1,2,7], 10, L1).
L1 = [5,1,2,7]
到目前为止，序言对我来说意义不大，我很难将其分解.这是我到目前为止的内容:
Prolog thus far is making little sense to me, and I'm having a hard time breaking it down. Here is what I have so far:
take([H|T], 0, []).
take([H|T], N, L1) :-
take(T, X, L2),
X is N-1.
您能在这里解释我做错了什么吗?
Can you please explain what I did wrong here?
推荐答案
此处是使用Haskell 之类的功能语言实现与take的关系副本的定义.首先，参数顺序应该不同，这有利于部分应用.有一个切入点，但是只有在生成内置instantiation_error的内置错误(=<)/2进行错误检查之后，该参数才能包含变量.
Here is a definition that implements the relational counterpart to take in functional languages like Haskell. First, the argument order should be different which facilitates partial application. There is a cut, but only after the error checking built-in (=<)/2 which produces an instantiation_error should the argument contain a variable.
take(N, _, Xs) :- N =< 0, !, N =:= 0, Xs = [].
take(_, [], []).
take(N, [X|Xs], [X|Ys]) :- M is N-1, take(M, Xs, Ys).
| ?- take(2, Xs, Ys).
Xs = [],
Ys = [] ? ;
Xs = [_A],
Ys = [_A] ? ;
Xs = [_A,_B|_C],
Ys = [_A,_B] ? ;
no
请注意上述查询的读取方式:
Note how above query reads:
有3个不同的答案.如果Xs为空，则Ys也为空.如果Xs是具有一个元素的列表，则Ys也是.如果Xs至少包含2个元素，那么这两个元素就是Ys.
And there are 3 different answers. If Xs is empty, then so is Ys. If Xs is a list with one element, then so is Ys. If Xs has at least 2 elements, then those two are Ys.
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