Prolog中的逆阶乘

本文介绍了Prolog中的逆阶乘的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

有人可以帮助我找到在Prolog中获得逆阶乘的方法吗...

Can someone helping me to find a way to get the inverse factorial in Prolog...

例如inverse_factorial(6,X) ===> X = 3.

我已经花了很多时间了.

I have been working on it a lot of time.

我目前有阶乘，但我必须使其可逆.请帮助我.

I currently have the factorial, but i have to make it reversible. Please help me.

推荐答案

Prolog的谓词是关系，因此一旦定义了阶乘，就也隐式定义了逆.但是，常规算术是在Prolog中设置的，也就是说，必须在运行时知道(is)/2或(>)/2中的整个表达式，否则，将发生错误.约束克服了这个缺点:

Prolog's predicates are relations, so once you have defined factorial, you have implicitly defined the inverse too. However, regular arithmetics is moded in Prolog, that is, the entire expression in (is)/2 or (>)/2 has to be known at runtime, and if it is not, an error occurs. Constraints overcome this shortcoming:

:- use_module(library(clpfd)). n_factorial(0, 1). n_factorial(N, F) :- N #> 0, N1 #= N - 1, F #= N * F1, n_factorial(N1, F1).

此定义现在可以双向使用.

This definition now works in both directions.

?- n_factorial(N,6). N = 3 ; false. ?- n_factorial(3,F). F = 6 ; false.

由于SICStus 4.3.4和SWI 7.1.25也终止了以下内容:

Since SICStus 4.3.4 and SWI 7.1.25 also the following terminates:

?- n_factorial(N,N). N = 1 ; N = 2 ; false.

有关更多信息，请参见手册.

See the manual for more.