请解释背后Kernighan的比特计数算法逻辑 [英] Please explain the logic behind Kernighan's bit counting algorithm

问题描述

到<一读之后，此问题直接如下href="http://stackoverflow.com/questions/12380478/bits-counting-algorithm-brian-kernighan-in-an-integer-time-complexity">位在整数时间复杂度计数算法（布莱恩Kernighan的）。 Java的code的问题是

  INT COUNT_SET_BITS（INT N）{
  诠释计数= 0;
    而（N！= 0）{
      N'安培; =（N-1）;
      算上++;
    }
 }
 

我想明白 N'放大器; =（N-1）正在实现吗？我已经看到了类似的一种结构的另一个漂亮的算法，用于检测一个数是否是2就像一个功率：

 如果（N及（N-1）== 0）{
    的System.out.println（的数目是2的幂）;
}
 

解决方案

通过code在调试器步进帮助了我。

如果你开始

  N = 1010101＆放大器; N-1 = 1010100 =＆GT; 1010100
N = 1010100＆放大器; N-1 = 1010011 =＆GT; 1010000
N = 1010000＆放大器; N-1 = 1001111 =＆GT;百万
ñ= 1000000＆放大器; N-1 = 0111111 =＆GT; 0000000
 

所以这个迭代4次。每次迭代递减其被设置为1的至少显著位消失以这样的方式的值。

递减被一个翻转的最低位和每一位到第一个。例如如果你有1000 .... 0000 -1 = 0111 .... 1111不管有多少位有翻转，并停在那里留下设置不变的其他位。当你和这个与 N 的最低位设置，只有最低位变为 0

This question directly follows after reading through Bits counting algorithm (Brian Kernighan) in an integer time complexity . The Java code in question is

int count_set_bits(int n) {
  int count = 0;
    while(n != 0) {
      n &= (n-1);
      count++;
    }
 }

I want to understand what n &= (n-1) is achieving here ? I have seen a similar kind of construct in another nifty algorithm for detecting whether a number is a power of 2 like:

if(n & (n-1) == 0) {
    System.out.println("The number is a power of 2");
}

解决方案

Stepping through the code in a debugger helped me.

If you start with

n = 1010101 & n-1=1010100 => 1010100
n = 1010100 & n-1=1010011 => 1010000
n = 1010000 & n-1=1001111 => 1000000
n = 1000000 & n-1=0111111 => 0000000

So this iterates 4 times. Each iteration decrements the value in such a way that the least significant bit that is set to 1 disappears.

Decrementing by one flips the lowest bit and every bit up to the first one. e.g. if you have 1000....0000 -1 = 0111....1111 not matter how many bits it has to flip and it stops there leaving any other bits set untouched. When you and this with n the lowest bit set and only the lowest bit becomes 0

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