在列表 Prolog 中查找 2 的幂 [英] Find powers of 2 in a list Prolog

问题描述

我正在尝试在 Prolog (SWI Prolog) 中创建一个列表并检查哪些数字是 2 的幂，然后找出特定数字在列表中的次数(在本例中，我试图找出有多少次数 3 在列表中).例如，如果你问

?- check([0,2,3,-5,-2,1,8,7,4], MULT2, THREE).

你应该看到

MULT2=[2,8,4]三=1

我第一次尝试找到解决方案是使用 head 搜索列表并执行 head mod 2 = 0 以找到所有 2 的幂的数字，但是出了点问题，我只得到false"作为答案.

解决方案

这里是如何以纯逻辑的方式找到2 的幂"！

使用 sicstus-prolog4.3.5、library(reif)和library(clpz):

<前>:- use_module([图书馆(reif), 图书馆(clpz)]).power_of_two_t(I, T) :-L #= min(I,1),M #= I/ (I-1),调用((L = 1，M = 0)，T).% 使用 (=)/3 和 (',')/3 of library(reif)

示例查询¹使用元谓词tfilter/3 与 power_of_two_t/2 结合:

?- tfilter(power_of_two_t, [0,2,3,-5,-2,1,8,7,4], Ps).Ps = [2,1,8,4].% 确定性地成功

这是评论建议的更一般的查询:

?- tfilter(power_of_two_t, [X], Ps).Ps = [X], 0#=X/\_A, _A+1#=X, X in 1..sup, _A in 0..sup;Ps = [], dif(_A,0), _A#=X/\_B, _B+1#=X, X in 1..sup, _B in 0..sup;Ps = [], dif(_A,1), _A#=min(X,1), _B#=X/\_C, _C+1#=X, X#>=_A, _A inf..1.

<小时>

^{脚注 1: 上面显示的答案序列被刷过以表明调用的确定性.}

^{脚注 2: 要重现结果，请使用 call_det/2，其定义如下:}

<前>call_det(G_0, Det) :-call_cleanup(G_0, Flag = set),(非变量(标志)-> 检测 = 真;检测 = 假).

I'm trying to create a list in Prolog (SWI Prolog) and check which numbers are powers of 2 and second find how many times a specific number is in the list (in this example I'm trying to find how many times the number 3 is in the list). For a example, if you ask

?- check([0,2,3,-5,-2,1,8,7,4], MULT2, THREE).

you should see

MULT2=[2,8,4] 
THREE=1 

My first try to find a solution is to search the list with head and doing head mod 2 = 0 to find all numbers which are powers of 2, but something went wrong and I only get "false" as an answer.

解决方案

Here's how you can find the "powers of two" in logically-pure way!

Using sicstus-prolog 4.3.5, library(reif) and library(clpz):

:- use_module([library(reif), library(clpz)]).

power_of_two_t(I, T) :-
   L #= min(I,1),
   M #= I / (I-1),
   call((L = 1, M = 0), T).      % using (=)/3 and (',')/3 of library(reif)

Sample query¹ using meta-predicate tfilter/3 in combination with power_of_two_t/2:

?- tfilter(power_of_two_t, [0,2,3,-5,-2,1,8,7,4], Ps).
Ps = [2,1,8,4].                  % succeeds deterministically

Here's a more general query suggested by a comment:

?- tfilter(power_of_two_t, [X], Ps).
   Ps = [X], 0#=X/\_A, _A+1#=X, X in 1..sup, _A in 0..sup
;  Ps = [], dif(_A,0), _A#=X/\_B, _B+1#=X, X in 1..sup, _B in 0..sup
;  Ps = [], dif(_A,1), _A#=min(X,1), _B#=X/\_C, _C+1#=X, X#>=_A, _A in inf..1.

---

^{Footnote 1: The answer sequences shown above were brushed up to indicate the determinism of calls.}

^{Footnote 2: To reproduce the results use call_det/2 which is defined like this:}

call_det(G_0, Det) :-
   call_cleanup(G_0, Flag = set),
   (  nonvar(Flag)
   -> Det = true
   ;  Det = false
   ).

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