Comparing arbitrary bit sequences in a byte array

Question

I have a couple uint8_t arrays in my c code, and I'd like to compare an arbitrary sequence bits from one with another. So for example, I have bitarray_1 and bitarray_2, and I'd like to compare bits 13 - 47 from bitarray_1 with bits 5-39 of bitarray_2. What is the most efficient way to do this?

Currently it's a huge bottleneck in my program, since I just have a naive implementation that copies the bits into the beginning of a new temporary array, and then uses memcmp on them.

Answer 1

bits 13 - 47 of bitarray_1 are the same as bits 5 - 39 of bitarray_1 + 1.
Compare the first 3 bits (5 - 7) with a mask and the other bits (8 - 39) with memcmp().

Rather than shift and copy the bits, maybe representing them differently is faster. You have to measure.

/* code skeleton */
static char bitarray_1_bis[BIT_ARRAY_SIZE*8+1];
static char bitarray_2_bis[BIT_ARRAY_SIZE*8+1];
static const char *lookup_table[] = {
    "00000000", "00000001", "00000010" /* ... */
    /* 256 strings */
    /* ... */ "11111111"
};

/* copy every bit of bitarray_1 to an element of bitarray_1_bis */
for (k = 0; k < BIT_ARRAY_SIZE; k++) {
    strcpy(bitarray_1_bis + 8*k, lookup_table[bitarray_1[k]]);
    strcpy(bitarray_2_bis + 8*k, lookup_table[bitarray_2[k]]);
}
memcmp(bitarray_1_bis + 13, bitarray_2_bis + 5, 47 - 13 + 1);

You can (and should) limit the copy to the minimum possible.

I have no idea if it's faster, but it wouldn't surprise me if it was. Again, you have to measure.

Answer 2

Here is my unoptimized bit sequence comparison function:

#include <stdio.h> 
#include <stdint.h> 
                  //  01234567    01234567
uint8_t bitsA[] = { 0b01000000, 0b00010000 };
uint8_t bitsB[] = { 0b10000000, 0b00100000 };

int bit( uint8_t *bits, size_t bitpoz, size_t len ){
  return (bitpoz<len)? !!(bits[bitpoz/8]&(1<<(7-bitpoz%8))): 0;
}

int bitcmp( uint8_t *bitsA, size_t firstA, size_t lenA,
            uint8_t *bitsB, size_t firstB, size_t lenB ){
  int cmp;
  for( size_t i=0; i<lenA || i<lenB; i++ ){
    if( (cmp = bit(bitsA,firstA+i,firstA+lenA) -
               bit(bitsB,firstB+i,firstB+lenB))  ) return cmp;
  }
  return 0;
}

int main(){
  printf( "cmp: %i\n", bitcmp( bitsA,1,11, bitsB,0,11 ) );
}

EDIT: Here is my (untested) bitstring equality test function:

#include <stdlib.h> 
#include <stdint.h> 

#define load_64bit(bits,first) (*(uint64_t*)bits<<first | *(bits+8)>>(8-first)) 
#define load_32bit(bits,first) (*(uint32_t*)bits<<first | *(bits+4)>>(8-first)) 
#define load_16bit(bits,first) (*(uint16_t*)bits<<first | *(bits+2)>>(8-first)) 
#define load_8bit( bits,first) (           *bits<<first | *(bits+1)>>(8-first)) 

static inline uint8_t last_bits( uint8_t *bits, size_t first, size_t size ){
  return (first+size>8?load_8bit(bits,first):*bits<<first)>>(8-size);
}

int biteq( uint8_t *bitsA, size_t firstA,
           uint8_t *bitsB, size_t firstB, size_t size ){
  if( !size ) return 1;
  bitsA+=firstA/8; firstA%=8;
  bitsB+=firstB/8; firstB%=8;

  for(; size>64;size-=64,bitsA+=8,bitsB+=8)
    if(load_64bit(bitsA,firstA)!=load_64bit(bitsB,firstB)) return 0;

  for(; size>32;size-=32,bitsA+=4,bitsB+=4)
    if(load_32bit(bitsA,firstA)!=load_32bit(bitsB,firstB)) return 0;

  for(; size>16;size-=16,bitsA+=2,bitsB+=2)
    if(load_16bit(bitsA,firstA)!=load_16bit(bitsB,firstB)) return 0;

  for(; size> 8;size-= 8,bitsA++, bitsB++ )
    if(load_8bit( bitsA,firstA)!=load_8bit( bitsB,firstB)) return 0;

  return !size || 
         last_bits(bitsA,firstA,size)==last_bits(bitsB,firstB,size);
}

I made a simple measurement tool to see how fast is it:

#include <unistd.h> 
#include <stdio.h> 
#include <signal.h> 

#define SIZE 1000000 
uint8_t bitsC[SIZE];

volatile int end_loop;
void sigalrm_hnd( int sig ){ (void)sig; end_loop=1; }

int main(){
  uint64_t loop_count; int cmp;
  signal(SIGALRM,sigalrm_hnd);
  loop_count=0; end_loop=0; alarm(10);
  while( !end_loop ){
    for( int i=1; i<7; i++ ){
      loop_count++;
      cmp = biteq( bitsC,i, bitsC,7-i,(SIZE-1)*8 );
      if( !cmp ){ printf( "cmp: %i (==0)\n", cmp ); return -1; }
    }
  }
  printf( "biteq: %.2f round/sec\n", loop_count/10.0 );
}

Result:

bitcmp:   8.40 round/sec
biteq:  363.60 round/sec

EDIT2: last_bits() changed.

Answer 3

What about writing the function that will calculate the offsets from both arrays, apply the mask, shift the bits and store the result to the int so you may compare them. If the bits count (34 in your example) exceeds the length of the int - recurse or loop.

Sorry, the example will be pain in the ass.

Answer 4

three words: shift, mask and xor.

shift to get the same memory alignment for both bitarray. If not you will have to shift one of the arrays before comparing them. Your exemple is probably misleading because bits 13-47 and 5-39 have the same memory alignment on 8 bits addresses. This wouldn't be true if you were comparing say bits 14-48 with bits 5-39.

Once everything is aligned and exceeding bits cleared for table boundaries a xor is enough to perform the comparison of all the bits at once. Basically you can manage to do it with just one memory read for each array, which should be pretty efficient.

If memory alignment is the same for both arrays as in your example memcmp and special case for upper and lower bound is probably yet faster.

Also accessing array by uint32_t (or uint64_t on 64 bits architectures) should also be more efficient than accessing by uint8_t.

The principle is simple but as Andrejs said the implementation is not painless...

Here is how it goes (similarities with @caf proposal is no coincidence):

/* compare_bit_sequence() */
int compare_bit_sequence(uint8_t s1[], unsigned s1_off, uint8_t s2[], unsigned s2_off,
    unsigned length)
{
const uint8_t mask_lo_bits[] =
    { 0x00, 0x01, 0x03, 0x07, 0x0f, 0x1f, 0x3f, 0x7f, 0xff };
const uint8_t clear_lo_bits[] =
    { 0xff, 0xfe, 0xfc, 0xf8, 0xf0, 0xe0, 0xc0, 0x80, 0x00 };
uint8_t v1;
uint8_t * max_s1;
unsigned end;
uint8_t lsl;
uint8_t v1_mask;
int delta;

/* Makes sure the offsets are less than 8 bits */
s1 += s1_off >> 3;
s1_off &= 7;

s2 += s2_off >> 3;
s2_off &= 7;

/* Make sure s2 is the sequence with the shorter offset */
if (s2_off > s1_off){
    uint8_t * tmp_s;
    unsigned tmp_off;
    tmp_s = s2; s2 = s1; s1 = tmp_s;
    tmp_off = s2_off; s2_off = s1_off; s1_off = tmp_off;
}
delta = s1_off;

/* handle the beginning, s2 incomplete */ 
if (s2_off > 0){
    delta = s1_off - s2_off;
    v1 = delta
       ? (s1[0] >> delta | s1[1] << (8 - delta)) & clear_lo_bits[delta]
       : s1[0];
       if (length <= 8 - s2_off){
           if ((v1 ^ *s2)
                & clear_lo_bits[s2_off]
                & mask_lo_bits[s2_off + length]){
                return NOT_EQUAL;
           }
           else {
               return EQUAL;
           }
        }
        else{
            if ((v1 ^ *s2) & clear_lo_bits[s2_off]){
                return NOT_EQUAL;
        }
        length -= 8 - s2_off;
    }
    s1++;
    s2++;
}

/* main loop, we test one group of 8 bits of v2 at each loop */
max_s1 = s1 + (length >> 3);
lsl = 8 - delta;
v1_mask = clear_lo_bits[delta];
while (s1 < max_s1)
{
    if ((*s1 >> delta | (*++s1 << lsl & v1_mask)) ^ *s2++)
    {
        return NOT_EQUAL;
    }
}

/* last group of bits v2 incomplete */
end = length & 7;
if (end && ((*s2 ^ *s1 >> delta) & mask_lo_bits[end]))
{
    return NOT_EQUAL;
}

return EQUAL;

}

All possible optimisations are not yet used. One promising one would be to use larger chunks of data (64 bits or 32 bits at once instead of 8), you could also detect cases where offset are synchronised for both arrays and in such cases use a memcmp instead of the main loop, replace modulos % 8 by logical operators & 7, replace '/ 8' by '>> 3', etc., have to branches of code instead of swapping s1 and s2, etc, but the main purpose is achieved : only one memory read and not memory write for each array item hence most of the work can take place inside processor registers.

Answer 5

The easiest way to do this is to convert the more complex case into a simpler case, then solve the simpler case.

In the following code, do_compare() solves the simpler case (where the sequences are never offset by more than 7 bits, s1 is always offset as much or more than s2, and the length of the sequence is non-zero). The compare_bit_sequence() function then takes care of converting the harder case to the easier case, and calls do_compare() to do the work.

This just does a single-pass through the bit sequences, so hopefully that's an improvement on your copy-and-memcmp implementation.

#define NOT_EQUAL 0
#define EQUAL 1

/* do_compare()
 *
 * Does the actual comparison, but has some preconditions on parameters to
 * simplify things:
 *
 *     length > 0
 *     8 > s1_off >= s2_off
 */
int do_compare(const uint8_t s1[], const unsigned s1_off, const uint8_t s2[],
    const unsigned s2_off, const unsigned length)
{
    const uint8_t mask_lo_bits[] =
        { 0xff, 0x01, 0x03, 0x07, 0x0f, 0x1f, 0x3f, 0x7f, 0xff };
    const uint8_t mask_hi_bits[] =
        { 0x00, 0x80, 0xc0, 0xe0, 0xf0, 0xf8, 0xfc, 0xfe, 0xff };
    const unsigned msb = (length + s1_off - 1) / 8;
    const unsigned s2_shl = s1_off - s2_off;
    const unsigned s2_shr = 8 - s2_shl;
    unsigned n;
    uint8_t s1s2_diff, lo_bits = 0;

    for (n = 0; n <= msb; n++)
    {
        /* Shift s2 so it is aligned with s1, pulling in low bits from
         * the high bits of the previous byte, and store in s1s2_diff */
        s1s2_diff = lo_bits | (s2[n] << s2_shl);

        /* Save the bits needed to fill in the low-order bits of the next
         * byte.  HERE BE DRAGONS - since s2_shr can be 8, this below line
         * only works because uint8_t is promoted to int, and we know that
         * the width of int is guaranteed to be >= 16.  If you change this
         * routine to work with a wider type than uint8_t, you will need
         * to special-case this line so that if s2_shr is the width of the
         * type, you get lo_bits = 0.  Don't say you weren't warned. */  
        lo_bits = s2[n] >> s2_shr;

        /* XOR with s1[n] to determine bits that differ between s1 and s2 */
        s1s2_diff ^= s1[n];

        /* Look only at differences in the high bits in the first byte */
        if (n == 0)
            s1s2_diff &= mask_hi_bits[8 - s1_off];

        /* Look only at differences in the low bits of the last byte */
        if (n == msb)
            s1s2_diff &= mask_lo_bits[(length + s1_off) % 8];

        if (s1s2_diff)
            return NOT_EQUAL;
    }

    return EQUAL;
}

/* compare_bit_sequence()
 *
 * Adjusts the parameters to match the preconditions for do_compare(), then
 *  calls it to do the work.
 */
int compare_bit_sequence(const uint8_t s1[], unsigned s1_off,
    const uint8_t s2[], unsigned s2_off, unsigned length)
{
    /* Handle length zero */
    if (length == 0)
        return EQUAL;

    /* Makes sure the offsets are less than 8 bits */
    s1 += s1_off / 8;
    s1_off %= 8;

    s2 += s2_off / 8;
    s2_off %= 8;

    /* Make sure s2 is the sequence with the shorter offset */
    if (s1_off >= s2_off)
        return do_compare(s1, s1_off, s2, s2_off, length);
    else
        return do_compare(s2, s2_off, s1, s1_off, length);
}

To do the comparison in your example, you'd call:

compare_bit_sequence(bitarray_1, 13, bitarray_2, 5, 35)

(Note that I am numbering the bits from zero, and assuming that the bitarrays are laid out little-endian, so this will start the comparison from the sixth-least-significant bit in bitarray2[0], and the sixth-least-signifcant bit in bitarray1[1]).