用于生成下一位翻转格雷码的C代码 [英] C code for generating next bit to flip in a gray code

问题描述

我需要一个函数，该函数返回一个数字，从本质上讲告诉我，当移至格雷代码的第n个元素时，哪一位将成为翻转位.不管是标准(反射)格雷码还是其他一些最小的位切换方法，都没有关系.我可以做到，但似乎不必要.目前我有这个:

I need a function that returns a number essentially telling me which bit would be the one to flip when moving to the nth element of a Gray code. It doesn't matter if it's the standard (reflecting) Gray code or some other minimal bit-toggling approach. I can do it, but it seems unnecessarily unwieldy. Currently I have this:

#include <stdio.h>

int main()
{
        int i;
        for (i=1; i<32; i++)
                printf("%d\n",grayBitToFlip(i));
}


int grayBitToFlip(int n)
{
        int j, d, n1, n2;

        n1 = (n-1)^((n-1)>>1);
        n2 = n^(n>>1);
        d = n1^n2;
        j = 0;
        while (d >>= 1)
                j++;
        return j;
}

main()中的循环仅用于演示函数的输出.

The loop in main() is only there to demonstrate the output of the function.

有更好的方法吗?

仅查看输出，很明显可以更简单地完成此操作.我添加了第二个函数gray2，它可以更简单地完成相同的操作.这会是这样做的方法吗?顺便说一下，这不是生产代码，而是爱好者.

just looking at the output, it's obvious one can do this more simply. I've added a 2nd function, gray2, that does the same thing much more simply. Would this be the way to do it? This is not production code by the way but hobbyist.

#include <stdio.h>

int main()
{
        int i;
        for (i=1; i<32; i++)
                printf("%d  %d\n",grayBitToFlip(i), gray2(i));
}


int grayBitToFlip(int n)
{
        int j, d, n1, n2;

        n1 = (n-1)^((n-1)>>1);
        n2 = n^(n>>1);
        d = n1^n2;
        j = 0;
        while (d >>= 1)
                j++;
        return j;
}

int gray2(int n)
{
        int j;
        j=0;
        while (n)
                {
                if (n & 1)
                        return j;
                n >>= 1;
                j++;
                }
        return j;
}

推荐答案

最容易使用的格雷码是Johnson格雷码(JGC).

The easiest Gray code to use is a Johnson Gray Code (JGC).

BitNumberToFlip = ++BitNumberToFlip  % NumberOfBitsInCode;

JGC = JGC ^ (1 << BitNumberToFlip);         // start JGC = 0;

Johnson代码在表示所需的位数上是线性的.
二进制反射格雷码(BRGC)具有更好的位密度，因为仅 需要一个对数位数来表示BRGC码的范围.

A Johnson code is linear in the number of bits required for representation.
A Binary Reflected Gray Code (BRGC) has a much better bit density since only a logarithmic number of bits are required to represent the range of BRGC codes.

int powerOf2(int n){ return          //   does 16 bit codes
           ( n & 0xFF00 ? 8:0 )  +   //    88888888........
           ( n & 0xF0F0 ? 4:0 )  +   //    4444....4444....
           ( n & 0xCCCC ? 2:0 )  +   //    22..22..22..22..
           ( n & 0xAAAA ? 1:0 )  ; } //    1.1.1.1.1.1.1.1.
                           // much faster algorithms exist see ref.

int BRGC(int gc){ return (gc ^ gc>>1);} 

int bitToFlip(int n){ return powerOf2( BRGC( n ) ^ BRGC( n+1 ) ); }

有关详细信息，请参见ref:

for details see ref:
How do I find next bit to change in a Gray code in constant time?

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