Question

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");
}

Answer 1

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