Counting the number of set bits in an integer

Most natural ways to do this is to loop through all bits in the integer increment the count for each set bit.

int NumberOfSetBits(int n){
  int count = 0;
  while(n){
    count += n & 1;
    n >>= 1;
  }
  return count;
}

But we can do better. The following code works perfectly fine for counting the number of bits set.

//for 64 bit numbers
int NumberOfSetBits64(long long i)
{
    i = i - ((i >> 1) & 0x5555555555555555);
    i = (i & 0x3333333333333333) +
        ((i >> 2) & 0x3333333333333333);
    i = ((i + (i >> 4)) & 0x0F0F0F0F0F0F0F0F);
    return (i*(0x0101010101010101))>>56;
}
//for 32 bit integers
int NumberOfSetBits32(int i)
{
    i = i - ((i >> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
    i = ((i + (i >> 4)) & 0x0F0F0F0F);
    return (i*(0x01010101))>>24;
}

The magic numbers might seem very confusing. But they are not random (obviously!).
Let us start with 0X55555555. Binary representation of 5 is 0101. So 0X55555555 has 16 ones, 16 zeros and the ones,zeros take alternate positions. Similarly 0X33333333 has 16 ones, 16 zeros and 2 consecutive ones, 2 consecutive zeros alternate.

F = 0x55555555 = 01010101010101010101010101010101
T = 0x33333333 = 00110011001100110011001100110011
O = 0x0f0f0f0f = 00001111000011110000111100001111

Now split the number into 16 sets of 2 bits
Ex: 9 = 1001 = 00,00,…..00,10,01
First step that algorithm does is to count the number of set bits in each of these 2-bit numbers. As each 2-bit number can have a maximum of 2 set bits. Each of these counts fit into a 2-bit number. That is what the below snippet does.

i = (i & 0x55555555) + ((i >> 1) & 0x55555555);

‘i & F’ preserves only the odd positioned ones. (i>>1) & F preserves only the even positioned ones and gets the even ones to odd positions. Adding them results in each pair of 2-bits having the number of set bits in their corresponding positions.
Similarly the after next 2 steps we end up having a number which can be seen as 4 8-bit numbers whose sum is the required count.

i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
i = (i & 0x0F0F0F0F) + (i >> 4)) & 0x0F0F0F0F);

Multiplying this number with 0X01010101 will result in a number whose most significant 8 bits contain the required count. Finally on summing up all these steps we get the following code.

int NumberOfSetBits32(int i)
{
    i = (i & 0x55555555) + ((i >> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
    i = (i & 0x0F0F0F0F) + (i >> 4)) & 0x0F0F0F0F);
    return (i*(0x01010101))>>24;
}

Here some more micro optimizations are possible. After giving some thought it can be understood that the 2 lines of code in the below snippet produce the same result.

i = (i & 0x55555555) + ((i >> 1) & 0x55555555);
i = i - ((i >> 1) & 0x55555555);

In general ‘&’ operator is not distributive over ‘+’. But in this case it does. In i and i>>4, each of the eight 4-bit numbers has a maximum value of 4 (fits in 3 bits). And adding these numbers will not cause an overflow in any of the 4-bit slots. So it works in this case. (Think of why it will not work int the previous steps).

i = (i & 0x0F0F0F0F) + (i >> 4)) & 0x0F0F0F0F);
i = (i & + (i >> 4)) & 0x0F0F0F0F;

So finally we arrive at the code for NumberOfSetBits32 as in the 2nd snippet ( NumberOfSetBits64 can be implemented in similar lines).

About these ads

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s