Java: true's & 的生成器给出数字 N 的 false 组合； [英] Java: Generator of true's & false's combinations by giving the number N;

问题描述

我尽可能简化任务，所以我可以将它应用到我的算法中.

I tied to simplify the task as much as possible, so I could apply it to my algorithm.

这是数学家和程序员面临的挑战:

And here is the challenge for mathematicians and programmers:

我需要创建一个传递参数 int n 的方法:

I need to create a method where I pass parameter int n:

public void optionality_generator(int n){
  //some kind of loops, or recursions...to make it workable
  System.out.println("current combination: ...");
}

输出应显示真假的所有可能组合.

The output should show all possible combinations of true's and false's.

这里是 N=1 的例子；N=2；N=3；N=4；N=5，其中 x=false，0=true；请注意，空折线只是为了让您更容易识别模式.希望我包括了所有可能的组合):

Here is examples where N=1; N=2; N=3; N=4; N=5 where x=false and 0=true; Please note, empty break lines is just for you to recognise easier the patterns. Hopefully, I included all possible combinations):

Combination of 1:
0
x

Combination of 2:
00
x0
0x
xx

Combination of 3:
000
X00
0X0
00X
XX0
0XX
XXX

Combination of 4:
0000

X000
0X00
00X0
000X

XX00
X0X0
X00X

0XX0
0X0X

00XX

XXX0
XX0X
X0XX
0XXX

XXXX

Combination of 5:
00000
X0000
0X000
00X00
000X0
0000X

XX000
X0X00
X00X0
X000X

X0X00
X00X0
X000X

0XX00
0X0X0
0X00X

00XX0
00X0X

000XX

XXX00
XX0X0
XX00X

X0XX0
X0X0X
X00XX

0XXX0
0XX0X

00XXX

XXXX0
XXX0X
XX0XX
X0XXX
0XXXX

XXXXX

另外，如果您看到输出，这是我识别的模式，所有组合都反转了一半(例如，第一个组合是 00000，最后一个是 XXXXX，第二个是 X0000，最后一个之前的一个是 0XXXX 等..).也许，这种模式将有助于使整个算法更高效，对此不确定.提前谢谢！

Also, If you see the output, here is the pattern I recognized, that all combinations are inverted on half (e.g first combination is 00000 last one will be XXXXX, second one X0000, one before the last one will be 0XXXX etc..). Maybe, this pattern will help to make the whole algorithm more efficient, not sure about this. Thank you in advance!

推荐答案

这是一个仅使用 Java API 的非常基本的方法:

Here is a really basic way using only Java APIs:

final int n = 3;
for (int i = 0; i < Math.pow(2, n); i++) {
    String bin = Integer.toBinaryString(i);
    while (bin.length() < n)
        bin = "0" + bin;
    System.out.println(bin);
}

结果:

000
001
010
011
100
101
110
111

当然，您可以将 n 设置为任何您喜欢的值.并且，有了这个结果，您可以从字符串中选择第 n 个字符作为真/假.

Of course, you can set n to whatever you like. And, with this result, you can pick the nth character from the string as true/false.

如果只需要检查某个位是否为真，则不需要将其转换为字符串.这只是为了说明输出值.

If you only need to check if a bit is true, you don't need to convert it to a string. This is just to illustrate the output values.

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