[英]Lists size multiplication
I'm new to Prolog and I'm trying to get my head around lists.我是 Prolog 的新手,我正在尝试了解列表。 The problem I'm struggling with is: Given numbers in the form of lists (1 : [x], 3: [x, x, x]), implement the 'times' predicate /3.
我正在努力解决的问题是:给定列表形式的数字 (1 : [x], 3: [x, x, x]),实现'times'谓词 /3。 Eg: times([x, x], [x, x, x], R).
例如:times([x, x], [x, x, x], R)。 R = [x, x, x, x, x, x].
R = [x, x, x, x, x, x]。
The plus, and successor predicates where 2 previous points of the exercise.加号和后继谓词表示练习的前 2 点。 I know I'm not using the successor predicate, but it didn't seem that useful later on.
我知道我没有使用后继谓词,但后来似乎没有那么有用。
This is what i've tried so far这是我迄今为止尝试过的
successor([], [x]).
successor([X|T], R) :-
append([X|T], [X], R).
plus(L1, L2, R) :- append(L1, L2, R).
times([], _, []).
times(_, [], []).
times([_], L, L).
times(L, [_], L).
times([_|T], L2, R) :- plus(L2, R, RN),
times(T, L2, RN).
The output is: R is [].输出是:R是[]。
I think you make things too complicated here.我认为你在这里把事情搞得太复杂了。 You can define
successor
as:您可以将
successor
定义为:
successor(T, [x|T]).
We can define plus/3
as:我们可以将
plus/3
定义为:
plus([], T, T).
plus([x|R], S, [x|T]) :-
plus(R, S, T).
This is more or less the implementation of append/3
, except that here we check if the first list only contains x
.这或多或少是
append/3
的实现,除了这里我们检查第一个列表是否只包含x
。
For times/3
we know that if the first item is empty, the result is empty:对于
times/3
我们知道如果第一项为空,则结果为空:
times([], _, []).
and for a times/3
where the first item has shape [x|R]
, we need to add the second item to the result of a call to times/3
with R
:对于第一项具有形状
[x|R]
的times/3
,我们需要将第二项添加到使用R
调用times/3
的结果中:
times([x|R], S, T) :-
times(R, S, T1),
plus(S, T1, T).
So putting it all together, we obtain:所以把它们放在一起,我们得到:
successor(T, [x|T]).
plus([], T, T).
plus([x|R], S, [x|T]) :-
plus(R, S, T).
times([], _, []).
times([x|R], S, T) :-
times(R, S, T1),
plus(S, T1, T).
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.