Prolog中的逆阶乘 [英] Inverse factorial in Prolog

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.

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