从第(n

本文介绍了从第(n-1)个格雷码派生第n个格雷码的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

有没有办法通过对第(n-1)个格雷码进行位操作来使用第(n-1)个格雷码导出第4位第n个格雷码?

Is there a way to derive the 4-bit nth Gray code using the (n-1)th Gray code by using bit operations on the (n-1)th Gray Code?

例如，第4个格雷码是0010.现在，我想通过对0010进行位运算来获得第5个格雷码0110.

For example the 4th Gray code is 0010. Now I want to get the 5th Gray Code, 0110, by doing bit operations on 0010.

推荐答案

也许是作弊"，但您可以将查找表打包成64位常量值，如下所示:

Perhaps it's "cheating" but you can just pack a lookup table into a 64-bit constant value, like this:

0000 0 -> 1 0001 1 -> 3 0011 3 -> 2 0010 2 -> 6 0110 6 -> 7 0111 7 -> 5 0101 5 -> 4 0100 4 -> C 1100 C -> D 1101 D -> F 1111 F -> E 1110 E -> A 1010 A -> B 1011 B -> 9 1001 9 -> 8 1000 8 -> 0 FEDCBA9876543210 nybble order (current Gray code) | | V V EAFD9B80574C2631 next Gray code

然后，您可以使用移位和遮罩执行查找(取决于您的语言):

Then you can use shifts and masks to perform a lookup (depending on your language):

int next_gray_code(int code) { return (0xEAFD9B80574C2631ULL >> (code << 2)) & 15; }

或者，您可以使用公式将灰色转换为二进制，递增该值，然后从二进制转换为灰度，即n xor(n/2):

Alternatively, you can use the formula for converting from Gray to binary, increment the value, and then convert from binary to Gray, which is just n xor (n / 2):

int next_gray_code(int code) { code = code ^ (code >> 2); code = code ^ (code >> 1); code = (code + 1) & 15; return code ^ (code >> 1); }