问题描述
我目前正在阅读一本Prolog书籍，并且被困在其中一项挑战练习中.我打算用一个参数创建一个谓词.当此参数为变量时，它将返回以下内容并进行回溯，并且X将继续无限制地递增.
I am currently reading a Prolog book and I am stuck on one of the challenge exercises. I am meant to create a predicate with one argument. When this argument is a variable, it will return the following with backtracking, and X will continue to increment with no limit.
X = 0，X = 1，X = 2，X = 3，X = ...
X = 0, X = 1, X = 2, X = 3, X = ...
我做了一个简单的谓词，在该谓词下回溯了0-2，但我想不出一种使它无限继续的方法.
I made the simple predicate below which backtracks through 0-2 but I can't figure out a way to make it continue infinitely.
backtracking_exercise(X) :-
X = 0;
X = 1;
X = 2.
我正在考虑使用between/3谓词，但这只会给出有限数量的数字.我还尝试了plus/3谓词和递归，但是没有运气.这是我想出的，但是正如您所知，它目前没有用.
I was considering using the between/3 predicate but this would be only give a finite amount of numbers. I also experimented with the plus/3 predicate and recursion but no luck. This is what I have come up with, but as you can tell it is currently useless.
backtracking_excercise2(X) :-
X = 0,
plus(X,1,Y),
backtracking_excercise2(Y).
任何有关继续操作的技巧将不胜感激.
Any tips on how to proceed would be greatly appreciated.
先谢谢您.
推荐答案
Jim解决方案的尾部递归变体:
A tail-recursive variant of Jim's solution:
plus(N) :-
next_integer(1, N).
next_integer(I, I).
next_integer(I, K) :-
J is I + 1,
next_integer(J, K).
为防止在使用实例化参数调用plus/1谓词时进入无限循环，即使谓词仅是生成器，可以将第一个子句修改为:
To prevent going into an infinite loop when the plus/1 predicate is called with an instantiated argument, i.e. to make the predicate be only a generator, the first clause can be modified to:
plus(N) :-
var(N),
next_integer(1, N).
这篇关于如何在Prolog中使用回溯来无限增加变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！