Prolog 比较并有条件地返回 true 或 false [英] Prolog comparing and returning true or false conditionally

问题描述

我是 prolog 的新手，正在尝试解决一些问题.我有两个列表，例如 List1 和 List2 下面分别是:

I am new to the prolog and trying to solve some problems. I have two lists for example List1 and List2 respectively below:

List1: [[0.1, 0.6], [0.1, 0.5], [0.3, 0.5]]
List2: [0.2, 0.7, 0.4]

我需要检查 List2 的第 n 个元素是否在 List1 中相应(第 n 个)数字对的范围之间.如果List2中的数字在范围内，则检查List2的第(n+1)个元素是否满足相同条件.如果一个数字不满足条件，那么我需要返回 false.如果 List2 中的所有数字都满足条件，那么我需要返回 true.

I need to check if nth element of the List2 is between the range of the corresponding (nth) number pair in the List1. And if the number in List2 is in the range, then check if the (n+1)th element of the List2 satisfies the same condition. If a number does not satisfy the condition, then I need to return false. If all of the numbers in List2 satisfies the condition, then I need to return true.

我想出了这个主意，但它行不通，我也找不到原因.

I came up with this idea, but it does not work and I could not find why.

bet(L, R, B) :-
    Factor1 is B-L,
    Factor2 is R-B,
    Factor1 > 0,
    Factor2 > 0.

comp_lims([], []).
comp_lims([H1|T1], [H2|T2]) :-
    member([Left|T], H1),
    member(Right, T),
    bet(Left, Right, H2) -> comp_lims(T1, T2); fail.

bet 实际上是我为浮点数编写的比较谓词，在如下测试时效果很好:

bet is actually a comparison predicate i wrote for the floating point numbers and it works fine when testing it like below:

?- bet(0.3, 0.7, 0.4).
true.

?- bet(0.3, 0.7, 0.8).
false.

但是我找不到 comp_lims 谓词有什么问题，我被卡住了.非常感谢您的帮助！

But I cannot find what is wrong with the comp_lims predicate and I am stuck. A help will be much appreciated!

推荐答案

comp_lims([], []).
comp_lims([H1|T1], [H2|T2]) :-
    H1 = [Left, Right],
    ( bet(Left, Right, H2) -> comp_lims(T1, T2); fail ).

更好的版本:

comp_lims([], []).
comp_lims([ [Left, Right] | T1], [H2|T2]) :-
    bet(Left, Right, H2), comp_lims(T1, T2).

1. 模式匹配不只是[Head|Tail]形式，你可以用[A, B]来匹配两个元素列表.
2. 你可以写A ->乙;失败 和 A, B 一样(也有一个隐含的削减，但你现在可以忽略它).

1. Pattern matching need not just be [Head|Tail] form, you can just use [A, B] to match two element lists.
2. You can write A -> B; fail as just A, B(there is an implicit cut too, but you can ignore it for now).

同样在 bet/3 中，你可以这样做:

Also in bet/3, you can just do :

bet(L, R, B) :-
    B-L > 0,
    R-B > 0.

> 隐式计算双方，因此您不需要为它使用单独的 is.

> implicitly evaluates both sides, so you do not need to use separate is for it.

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