On finding 1-bit sequences

March 6, 2014 by David 2 Comments

Principles

Given is an arbitrary integer variable. How to find the index of the least significant bit (LSB) of the first 1-bit sequence of length >= n? Assuming n=4, let’s consider the following example of a random 32-bit integer value. The index we’re looking for is 10 in this case (indicated by the V character).

    31      24 | 23      16 | 15    V 8 | 7      0
MSB  01000111  |  11111101  |  10111100 | 01101001  LSB

Using a series of bit-wise and and shift-right operations the index of the LSB of the first 1111 sequence in the integer x can be found with the following trick.

x &= x>>1;
x &= x>>2;
index = __builtin_ffs(x)-1; // use _BitScanForward in Visual C++

After the first statement every 1 in x indicates the start of a 11 sequence. After the second statement every 1 in x indicates the start of a 1111 sequence. In the last statement the GCC intrinsic __builtin_ffs() (use _BitScanForward() if you’re on Visual C++) returns the bit position of the first set bit, starting from the LSB.
Note, that it doesn’t work to shift by four bits at once because it’s necessary to combine neighboring 1-bits to make sure that there are no 0-bits in-between. The following example illustrates how shifting by 3 bits wrongly yields two isolated 1-bits. In contrast, shifting by 2 bits correctly yields a sequence of 2 bits which can be further reduced into a single 1-bit indicating the start of the 1111 sequence.

 shift by 2    shift by 3
  01111010      01111010
& 00011110    & 00001111
= 00011010    = 00001010
    ok           wrong

Arbitrary sequence lengths

By cleverly choosing the number of bits to shift, it’s even possible to extend this construction to find bit sequences which length is not a power of two. As the order of the and-shift-right operations has no relevance, the following algorithm can be used to compute the number of bits to shift in order to find the n-bit sequence index. The sum of shifted bits must be equal to n-1 and the number of bits to shift is halved in each iteration. Therefore, the total number of executed iterations is ceil(log2(n)).

int FindBitSeqIndexLsb(int x, int n)
{
    assert(n >= 0 && n <= 32);

    while (n > 1)
    {
        const int shiftBits = n>>1;
        x &= (unsigned)x>>shiftBits; // shift in zeros from left
        n -= shiftBits;
    }

    return __builtin_ffs(x)-1; // use _BitScanForward in Visual C++
}

Exact sequence length

The described method finds bit sequences of length >= n. In case you’re looking for a bit sequence of exactly n bits, the following statement can be inserted right before the LSB scan is performed. This statement masks out any 1-bit which has a 1 on its left or right side. All the remaining 1-bits are isolated and indicate the start of a sequence of exactly n bits.

mask = (~(x<<1))&(~((unsigned)x>>1)); // shift in zeros from left and right
x &= mask;

Sequence alignment

To account for aligned bit sequences, unaligned 1-bits can be simply masked out from x before the LSB scan is performed. For example, to regard only bit sequences starting at nibble boundaries x can be modified with the operation x &= 0x11111111 (it is 0x...11 = 0b...00010001) to clear all bits not starting at an index which is a multiple of four.