Devise a fast algorithm for computing the number of 1-bits in an unsigned integer. If the machine supports n-bit integers, can we compute the number of 1-bits in O(log n) machine instructions? Can we compute the number of 1-bits in O(b) machine instructions where b is the number of 1-bits in the integer?

Commonly asked in job interviews for Software Engineers. I first heard it in 1996 when I was a graduate student.

This article presents six solutions to this problem. Source code in C is available.

int bitcount (unsigned int n) {
   int count = 0;
   while (n) {
      count += n & 0x1u;
      n >>= 1;
   }
   return count;
}

int bitcount (unsigned int n)  {
   int count = 0 ;
   while (n)  {
      count++ ;
      n &= (n - 1) ;
   }
   return count ;
}

int bitcount (unsigned int n)   {
   int count = 8 * sizeof(int) ;
   n ^= (unsigned int) - 1 ;
   while (n)  {
      count-- ;
      n &= (n - 1) ;
   }
   return count ;
}

Sparse Ones and Dense Ones were first described by Peter Wegner in "A Technique for Counting Ones in a Binary Computer", Communications of the ACM, Volume 3 (1960) Number 5, page 322.

static int bits_in_char [256] ;

int bitcount (unsigned int n)  {
  // works only for 32-bit ints

  return bits_in_char [n  & 0xffu]
      +  bits_in_char [(n >>  8 ) & 0xffu]
      +  bits_in_char [(n >> 16) & 0xffu]
      +  bits_in_char [(n >> 24) & 0xffu] ;
}

static char bits_in_16bits [0x1u << 16] ;

int bitcount (unsigned int n)  {
   // works only for 32-bit ints

   return bits_in_16bits [n         & 0xffffu]
       +  bits_in_16bits [(n >> 16) & 0xffffu] ;
}

#define TWO(c)     (0x1u << (c))
#define MASK(c) \
  (((unsigned int)(-1)) / (TWO(TWO(c)) + 1u))
#define COUNT(x,c) \
  ((x) & MASK(c)) + (((x) >> (TWO(c))) & MASK(c))

int bitcount (unsigned int n)  {
   n = COUNT(n, 0) ;
   n = COUNT(n, 1) ;
   n = COUNT(n, 2) ;
   n = COUNT(n, 3) ;
   n = COUNT(n, 4) ;
   /* n = COUNT(n, 5) ;    for 64-bit integers */
   return n ;
}

#define MASK_01010101 (((unsigned int)(-1))/3)
#define MASK_00110011 (((unsigned int)(-1))/5)
#define MASK_00001111 (((unsigned int)(-1))/17)
 int bitcount (unsigned int n) {
   n = (n & MASK_01010101) + ((n >> 1) & MASK_01010101) ;
   n = (n & MASK_00110011) + ((n >> 2) & MASK_00110011) ;
   n = (n & MASK_00001111) + ((n >> 4) & MASK_00001111) ;
   return n % 255 ;
}

According to Don Knuth (The Art of Computer Programming Vol IV, p 11), in the first textbook on programming, The Preparation of Programs for an Electronic Digital Computer by Wilkes, Wheeler and Gill (1957, reprinted 1984), pages 191--193 presented Nifty Parallel Count by D B Gillies and J C P Miller.

int bitcount(unsigned int n) {
   /* works for 32-bit numbers only    */
   /* fix last line for 64-bit numbers */

   register unsigned int tmp;

   tmp = n - ((n >> 1) & 033333333333)
           - ((n >> 2) & 011111111111);
   return ((tmp + (tmp >> 3)) & 030707070707) % 63;
}

MIT HAKMEM Count is funky. Consider a 3 bit number as being 4a+2b+c. If we shift it right 1 bit, we have 2a+b. Subtracting this from the original gives 2a+b+c. If we right-shift the original 3-bit number by two bits, we get a, and so with another subtraction we have a+b+c, which is the number of bits in the original number. How is this insight employed? The first assignment statement in the routine computes tmp. Consider the octal representation of tmp. Each digit in the octal representation is simply the number of 1's in the corresponding three bit positions in n. The last return statement sums these octal digits to produce the final answer. The key idea is to add adjacent pairs of octal digits together and then compute the remainder modulus 63. This is accomplished by right-shifting tmp by three bits, adding it to tmp itself and ANDing with a suitable mask. This yields a number in which groups of six adjacent bits (starting from the LSB) contain the number of 1's among those six positions in n. This number modulo 63 yields the final answer. For 64-bit numbers, we would have to add triples of octal digits and use modulus 1023. This is HACKMEM 169, as used in X11 sources. Source: MIT AI Lab memo, late 1970's.

GNU compiler allows for

int __builtin_popcount (unsigned int x);

                  No          Some        Heavy
            Optimization Optimization Optimization

  Precomp_16 52.94 Mcps   76.22 Mcps   80.58 Mcps
   Precomp_8 29.74 Mcps   49.83 Mcps   51.65 Mcps
    Parallel 19.30 Mcps   36.00 Mcps   38.55 Mcps
         MIT 16.93 Mcps   17.10 Mcps   31.82 Mcps
       Nifty 12.78 Mcps   16.07 Mcps   29.71 Mcps
      Sparse  5.70 Mcps   15.01 Mcps   14.62 Mcps
       Dense  5.30 Mcps   14.11 Mcps   14.56 Mcps
    Iterated  3.60 Mcps    3.84 Mcps    9.24 Mcps

  Mcps = Million counts per second

Which of the several bit counting routines is the fastest? Results of speed trials on an i686 are summarized in the table on left. "No Optimization" was compiled with plain gcc. "Some Optimizations" was gcc -O3. "Heavy Optimizations" corresponds to gcc -O3 -mcpu=i686 -march=i686 -fforce-addr -funroll-loops -frerun-cse-after-loop -frerun-loop-opt -malign-functions=4.

Thanks to Seth Robertson who suggested performing speed trials by extending bitcount.c. Seth also pointed me to MIT_Hackmem routine. Thanks to Denny Gursky who suggested the idea of Precompute_11bit. That would require three sums (11-bit, 11-bit and 10-bit precomputed counts). I then tried Precompute_16bit which turned out to be even faster.

If you have niftier solutions up your sleeves, please send me an e-mail or write comments below!

1. HAKMEM (bit counting is memo number 169), MIT AI Lab, Artificial Intelligence Memo No. 239, February 29, 1972.
2. Bit Twiddling Hacks by Sean Anderson at Stanford University.
3. Bitwise Tricks and Techniques by Don Knuth (The Art of Computer Programming, Part IV).

On 15 March 2009, this article was discussed on Reddit — you might learn quite a bit by reading the comments therein.