汉明权重

汉明权重是一串符号中非零符号的个数。因此它等同于同样长度的全零符号串的汉明距离。在最为常见的数据位符号串中，它是1的个数。
历史及应用.
汉明权重是以理查德·卫斯里·汉明的名字命名的，它在包括信息论、编码理论、密码学等多个领域都有应用。
高效实现.
在密码学以及其它应用中经常需要计算数据位中1的个数，针对如何高效地实现人们已经广泛地进行了研究。一些处理器使用单个的命令进行计算，另外一些根据数据位向量使用并行运算进行处理。对于没有这些特性的处理器来说，已知的最好解决办法是按照树状进行相加。例如，要计算二进制数A=0110110010111010中1的个数，这些运算可以表示为：
这里的运算是用C语言表示的，所以X » Y表示X右移Y位，X &amp; Y表示X与Y的位与，+表示普通的加法。基于上面所讨论的思想的这个问题的最好算法列在这里：
//types and constants used in the functions below
typedef unsigned __int64 uint64; //assume this gives 64-bits
const uint64 m1 = 0x5555555555555555; //binary: 0101...
const uint64 m2 = 0x3333333333333333; //binary: 00110011..
const uint64 m4 = 0x0f0f0f0f0f0f0f0f; //binary: 4 zeros, 4 ones ...
const uint64 m8 = 0x00ff00ff00ff00ff; //binary: 8 zeros, 8 ones ...
const uint64 m16 = 0x0000ffff0000ffff; //binary: 16 zeros, 16 ones ...
const uint64 m32 = 0x00000000ffffffff; //binary: 32 zeros, 32 ones ...
const uint64 hff = 0xffffffffffffffff; //binary: all ones
const uint64 h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...
//This is a naive implementation, shown for comparison,
//and to help in understanding the better functions.
//It uses 24 arithmetic operations (shift, add, and).
int popcount_1(uint64 x) {
 x = (x &amp; m1 ) + ((x » 1) &amp; m1 ); //put count of each 2 bits into those 2 bits 
 x = (x &amp; m2 ) + ((x » 2) &amp; m2 ); //put count of each 4 bits into those 4 bits 
 x = (x &amp; m4 ) + ((x » 4) &amp; m4 ); //put count of each 8 bits into those 8 bits 
 x = (x &amp; m8 ) + ((x » 8) &amp; m8 ); //put count of each 16 bits into those 16 bits 
 x = (x &amp; m16) + ((x » 16) &amp; m16); //put count of each 32 bits into those 32 bits 
 x = (x &amp; m32) + ((x » 32) &amp; m32); //put count of each 64 bits into those 64 bits 
 return x;
//This uses fewer arithmetic operations than any other known 
//implementation on machines with slow multiplication.
//It uses 17 arithmetic operations.
int popcount_2(uint64 x) {
 x -= (x » 1) &amp; m1; //put count of each 2 bits into those 2 bits
 x = (x &amp; m2) + ((x » 2) &amp; m2); //put count of each 4 bits into those 4 bits 
 x = (x + (x » 4)) &amp; m4; //put count of each 8 bits into those 8 bits 
 x += x » 8; //put count of each 16 bits into their lowest 8 bits
 x += x » 16; //put count of each 32 bits into their lowest 8 bits
 x += x » 32; //put count of each 64 bits into their lowest 8 bits
 return x &amp;0xff;
//This uses fewer arithmetic operations than any other known 
//implementation on machines with fast multiplication.
//It uses 12 arithmetic operations, one of which is a multiply.
int popcount_3(uint64 x) {
 x -= (x » 1) &amp; m1; //put count of each 2 bits into those 2 bits
 x = (x &amp; m2) + ((x » 2) &amp; m2); //put count of each 4 bits into those 4 bits 
 x = (x + (x » 4)) &amp; m4; //put count of each 8 bits into those 8 bits 
 return (x * h01)»56; //returns left 8 bits of x + (x«8) + (x«16) + (x«24) + ... 
在最坏的情况下，上面的实现是所有已知算法中表现最好的。但是，如果已知大多数数据位是0的话，那么还有更快的算法。这些更快的算法是基于这样一种事实即X与X-1相"与"得到的最低位永远是0。例如：
减1操作将最右边的符号从0变到1，从1变到0，"与"操作将会移除最右端的1。如果最初X有N个1，那么经过N次这样的迭代运算，X将减到0。下面的算法就是根据这个原理实现的。
//This is better when most bits in x are 0
//It uses 3 arithmetic operations and one comparison/branch per "1" bit in x.
int popcount_4(uint64 x) {
 uint64 count;
 for (count=0; x; count++)
 x &amp;= x-1;
 return count;
//This is better if most bits in x are 0.
//It uses 2 arithmetic operations and one comparison/branch per "1" bit in x.
//It is the same as the previous function, but with the loop unrolled.
int popcount_5(uint64 x) {
 if (x == 0) return 0;
 f( 1) f( 2) f( 3) f( 4) f( 5) f( 6) f( 7) f( 8)
 f( 9) f(10) f(11) f(12) f(13) f(14) f(15) f(16)
 f(17) f(18) f(19) f(20) f(21) f(22) f(23) f(24)
 f(25) f(26) f(27) f(28) f(29) f(30) f(31) f(32)
 f(33) f(34) f(35) f(36) f(37) f(38) f(39) f(40)
 f(41) f(42) f(43) f(44) f(45) f(46) f(47) f(48)
 f(49) f(50) f(51) f(52) f(53) f(54) f(55) f(56)
 f(57) f(58) f(59) f(60) f(61) f(62) f(63)
 return 64;
//Use this instead if most bits in x are 1 instead of 0