在 Prolog 中将 peano 数 s(N) 转换为整数 [英] Convert peano number s(N) to integer in Prolog
问题描述
我在教程中遇到了这种逻辑数的自然数评估,这让我有些头疼:
I came across this natural number evaluation of logical numbers in a tutorial and it's been giving me some headache:
natural_number(0).
natural_number(s(N)) :- natural_number(N).
规则大致规定:如果 N
是 0
这是很自然的,如果不是我们尝试将 s/1
的内容发回递归到规则直到内容为0
,则为自然数,否则为非自然数.
The rule roughly states that: if N
is 0
it's natural, if not we try to send the contents of s/1
back recursively to the rule until the content is 0
, then it's a natural number if not then it's not.
所以我测试了上面的逻辑实现,心想,如果我想将 s(0)
表示为 1
和 s(s(0))
为 2
,但我希望能够将 s(0)
转换为 1
.
So I tested the above logic implementation, thought to myself, well this works if I want to represent s(0)
as 1
and s(s(0))
as 2
, but I´d like to be able to convert s(0)
to 1
instead.
我已经想到了基本规则:
I´ve thought of the base rule:
sToInt(0,0). %sToInt(X,Y) Where X=s(N) and Y=integer of X
这是我的问题:如何将 s(0) 转换为 1 并将 s(s(0)) 转换为 2?
So here is my question: How can I convert s(0) to 1 and s(s(0)) to 2?
已答复
我修改了实现中的基本规则,我接受的答案指向我:
I modified the base rule in the implementation which the answer I accepted pointed me towards:
decode(0,0). %was orignally decode(z,0).
decode(s(N),D):- decode(N,E), D is E +1.
encode(0,0). %was orignally encode(0,z).
encode(D,s(N)):- D > 0, E is D-1, encode(E,N).
所以我现在可以随心所欲地使用它了,谢谢大家!
So I can now use it like I wanted to, thanks everyone!
推荐答案
这是另一个使用 library(clpfd)
的 SWI、YAP 或 SICStus
Here is another solution that works "both ways" using library(clpfd)
of SWI, YAP, or SICStus
:- use_module(library(clpfd)).
natsx_int(0, 0).
natsx_int(s(N), I1) :-
I1 #> 0,
I2 #= I1 - 1,
natsx_int(N, I2).
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